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Michael M. Woolfson
University of York, UK
Imperial College Press
. NJ 07601 UK office: 57 Shelton Street. Danvers. including photocopying. Elections. or parts thereof. Ltd. London WC2H 9HE British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. may not be reproduced in any form or by any means. This book. ISBN-13 ISBN-10 ISBN-13 ISBN-10 978-1-84816-031-6 1-84816-031-3 978-1-84816-032-3 (pbk) 1-84816-032-1 (pbk) Printed in Singapore. recording or any information storage and retrieval system now known or to be invented. In this case permission to photocopy is not required from the publisher. please pay a copying fee through the Copyright Clearance Center. USA. Singapore 596224 USA office: 27 Warren Street. Suite 401-402. without written permission from the Publisher. Gambling and War Copyright © 2008 by Imperial College Press All rights reserved. Hackensack. 222 Rosewood Drive. EVERYDAY PROBABILITY AND STATISTICS Health. electronic or mechanical. For photocopying of material in this volume. 5 Toh Tuck Link. Pte. MA 01923. Inc. Covent Garden..Published by Imperial College Press 57 Shelton Street Covent Garden London WC2H 9HE Distributed by World Scientific Publishing Co.
. .Contents Introduction Chapter 1 The Nature of Probability 1. . . 1. . . . . . . Spinning a Coin . . . . . . . 2. . . . Throwing or Spinning Other Objects Problems 1 . . . . . . Genetically Inherited Disease — Just Gene Dependent 2. . . . . . . . . . . Children Leaving a Room . .4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A Dice Game — American Craps . Genetically Dependent Disease — Gender Dependent 2.2. .1. . . . . Kinds of Probability . . Either–or Probability . . . . . Both–and Probability . . . . . . . . .2. . .1. . . . . . . . . . .3. . . . . . . . . . .1. . . . . . . . . .2. . 1 5 5 6 8 10 13 13 15 16 19 21 23 25 25 27 31 33 35 35 Chapter 4 Making Choices and Selections 4. . . v . . . . . . . . Chapter 3 A Day at the Races 3. . . . . . . . . . 3. . . . . . . . . 3. . . . The Best Conditions for a Punter Problem 3 . . . . . . . . . . . . . . . . . Problems 2 . . 2. . .5. . . . . . . . .3. . . . . . . . . . . . . . . . Chapter 2 Combining Probabilities 2. . . 1. . .1. . . . . . . . . . . . . . Betting on a Horse . . . . . . . . . . Probability and Everyday Speech . . .3. . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . 45 49 52 54 55 55 58 64 67 67 70 72 75 77 77 79 82 84 89 91 93 . American Craps with Loaded Dice 8.4. . . . . . . . . . . . . . . . Loaded Dice. . . . . . . . . . . . . .1. . . . . . . . . . . . 5. . . . . . . . . . . . . . . . . . . . . . . . . 8. . . . . . . . . . . . . . . Chapter 6 Probability and Health 6. 8. The Throwers Chance of Winning at American Craps Problems 7 . . . . . . . . . . . . . . . . . . . . 7. . . . . . . . . . . .6. . . . . . . . . . . . Testing for a Loaded Die . . . The Need to Test for Fairness . . . . . . . . . . . . . The UK National Lottery . . . .3. .vi Everyday Probability and Statistics 4. . . . . . .1. . . . Picking a Team . . . .3. . . . . . . . . . . . . . . . . . . . The UK National Lottery . . . A Simple Probability Machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems 4 . . . . . . . . . . . . . . . . . . .3. . . . Chapter 8 The UK National Lottery. . Problems 6 . .4. . . . . . . . . . . . . . . . . . . . . . 37 40 41 44 45 Chapter 5 Non-Intuitive Examples of Probability 5. . . . . . . .2. . . . . . . . . . . . . Choosing an Email Username 4. 5. . . Chapter 7 Combining Probabilities. . . . . . . . .2. Pontoon — A Card Game . . . . . 6. . . . . . .2. . . . . 8. . . . . . .2. . . . . . . . . . . . . . . .2. . . . . . . . . .1. 4. . . . . . To Switch or Not to Switch — That is the Question Problems 5 . . . . . . Crown and Anchor . . . The Roulette Wheel . . . . . Problems 8 . . Testing Drugs . . . . . .5. . . . . . . . . . . . . . and Crooked Wheels 8. The Birthday Problem . . Finding the Best Treatment . . . 7. . . . . . Testing Random Numbers . . . . . The Craps Game Revealed 7. .1. 8. .3.
. 95 95 96 99 101 104 105 105 108 110 111 114 114 115 117 117 119 119 120 123 Chapter 10 The Normal (or Gaussian) Distribution 10. . . . . . . .Contents vii Chapter 9 Block Diagrams 9. . . . . Electric Light Bulbs . . . .7. Histogram Shapes . . . . . . . . .3. 10. . 9. . . . . . . . . . . . . .4. .2.1. . . . . . . . . . . . . . . .1. . .2. 10.2. . . . 12. . . . . A Little Necessary Mathematics . . . .1. . . The Form of the Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Random and Systematic Errors . . . . . .6. . . Probability Distributions . 10. . . . . . . . .2. . . . . . . . . . . . . . . . . . . . . . . . . . . 131 131 133 . . . . . . . . . . . . . . . . . People on Trolleys and Under-Used Resources Problems 10 . . 10. . 9. .3. . . . . . . . . . . A Shoe Manufacturer . . . . . .2. . Typing a Manuscript . . . . . . . . . . . . . . Problem 9 . . . . . . . . . . . . 10. Some Examples of the Normal Distribution . 10. . Too Much Information . . . . 10. . .5. . Lofty and Shorty . . . . . . . . . . . . 10. . . . . . . .1. . . . . The Variance and Standard Deviation . . . . . . . Powers of Numbers . . . . . . . . . . . . . . . . . . . . . Some Special Numbers . .8. . . . . .5. . . . . . . . . .1. .2. . . Properties of Normal Distributions . . . . . Chapter 12 The Poisson Distribution and Death by Horse Kicks 12. . . . . . . . . . Variation in Almost Everything 9. 11. . . . . . . . . . . . . . . . . . . . . . . . . .1. . . . .8. . . The Normal Distribution . . . . .5. .8. . . . .3. 10. . . 125 125 126 127 130 11. . . . . . . . . .4. . . . . . . . Chapter 11 Statistics — The Collection and Analysis of Numerical Data . . . . Another Way of Finding the Variance 11. . . . . From Regional to National Statistics Problems 11 . . . . . . 10. . . . . Rare Events . . . 10. . . . .
. . . . . . . . . and Violent Crime . . 16. . . . . . . . . . . . . . . . . . . . . General Comments on Sampling 14. . . . . . . .3. . . . . . . . . . . . . . . . . . . . . The Poisson Distribution as a Formula .5. . . . . . . . 15.6. . . .1. 155 156 158 160 160 162 164 165 165 167 170 171 171 172 174 Chapter 15 Differences — Rats and IQs 15. . . Vehicle Crime. . An Illustrative Example . . Finding out from Samples . . Crime and the Reporting of Crime . . . . . . . . . . . . . . .4. . . . . . . . . . Problem 15 . 14. . . . . . . 13. . . . . . . . . . . . . . . . . . 14. . . . Problems 12 . . . .2. . . Problems 14 . . . .5. . . . Death by Horse Kicks . Some Other Examples of the Poisson Distribution 12. . . 14. . . . . . . . . . . . . . . . . . Problems 13 . . . 12. . . . . . .3. Why do we Sample? . . . . . . . . . . . . . . . .1. . . . .2. Burglary. . . . . 145 146 149 151 153 154 Chapter 14 Taking Samples — How Many Fish in the Pond? 155 14. . . . . . . . . .3. . .5. . . How Many Fish in the Pond? . . . . . . . .3.1.5. . . . . Clustering of a Disease .5. . . . Polling Statistics . . . . 135 137 139 139 141 142 143 145 . .viii Everyday Probability and Statistics 12. . . Significantly Different — So What! . . . . . . . . . Combining Polling Samples . . . . 16. . . . . 13. . .2. Quality Control . . . . . Flying Bomb Attacks on London . . . . . . . . . . . . . . .1. . . . 12. . Chapter 13 Predicting Voting Patterns 13. . . . . . . . . . . . . . . . . . . . . . . 14. . . . . . . . . . . . .2.3. . . The Significance of Differences . . . . . . . . . . . . . . 13. . . .2. . . . . . . . . . . . 13. . . . . . . . . . . .4. . . . . . The Trend for Overall Crime in England and Wales . . . Some Further Examples .4. . . . . . . . . . . 12. . Chapter 16 Crime is Increasing and Decreasing 16. . . . . . . . . . . .1. . . . . Election Polls . . . . . . .5. . . . 12. . . . . . Factors Affecting Polls and Voting . Polling with More than Two Parties . . . . . .
. . . . and Making Decisions 18. . . . . . . . Genetics and Disease . . . . The Winds of Chance . 177 179 181 183 183 185 187 189 191 191 194 195 197 200 204 205 221 . . . . 18. . . . . . . . . . . Solutions to Problems Index . Luck.3.2. . . . Chapter 18 Chance. .2. . . . . . . . . . . . Homicide . . . 16. . . . Crime and Politicians . The Mathematics of War . . 18. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4. . . . . . . . Problem 17 . . . . . . . . . . . . . . . . . . . . . . Choices . . . . . . . . . . . . . . I Want a Lucky General . . . . The Incidence of Smoking in the United Kingdom 17. . To Fight or Not to Fight — That is the Question 18.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem 16 . . . . . . . . . . . . . . . . . . . . . . . Chapter 17 My Uncle Joe Smoked 60 a Day 17.1. . . . . . . . . . . . .5. . . . 18. . The Smoking Lottery .Contents ix 16. . .5. . . . . . 17. . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem 18 . . .1. . . .4. . .
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radio. Greek. almost with pride. health. social services and all other aspects of national life. and through the internet and there is much greater awareness of the factors that affect our daily lives. statistics plays a dominant role. television. Those that held this view. and their strengths and weaknesses. and English classical literature. The deeds and misdeeds of those that govern. In this game of claim and counterclaim. would often announce.Introduction There was a time. Indeed. where the great and the good in society would hold that to be considered a "cultured man" one would have to be knowledgeable in Latin. are exposed as never before. 1 . or even overemphasize. Times have changed — society is much more complex than it was a century ago. politicians can rely on the lack of statistical understanding of those they address. unfortunately. affecting education. There is universal suffrage and a greater spread of education. or even overemphasize. the United Kingdom has had a Chancellor of the Exchequer who claimed that he could only do calculations relating to economics with the aid of matchsticks. the virtues of one's own party and to emphasize. It is inevitable that in a multi-party democracy there is a tendency to emphasize. that they had no ability to handle numbers or any mathematical concepts. security. The judicious use of statistics can be very helpful in making a case and. during the 19th century and for much of the 20th century. and held themselves to be members of the cultured elite. even within living memory. the inadequacies of one's opponents. Information is more widely disseminated by newspapers.
To make sense of the barrage of numbers with which society is bombarded. but the choices cannot be labeled as "bad" or "good" — they are just alternatives. damned lies and statistics" and in this he succinctly summarized the role that statistics plays in the hands of those who wish to use it for political advantage. However. To give a hypothetical example — the party in power wishes to expand a public service. The party in power say. politician. Many medical decisions are made on a statistical basis since individuals differ in their reactions to medications or surgery in an unpredictable way. These are hard decisions that have to be made by . The party in opposition say that the nation is getting poor value for money since for 44% more expenditure it is getting only 20% more service. and post-Victorian. one needs to understand what they mean and how they are derived and then one can make informed decisions. Leonard Henry Courtney (1832–1913). When resources are limited then allocating those resources to give the greatest benefit to the greatest number of people may lead to some being denied help that. Neither party is lying — they are telling neither lies nor damned lies — but they are using selective statistics to argue a case. Courtney knew what he was talking about — he was the President of the Statistical Society from 1897–1899. with pride. in a speech about proportional representation used the phrase "lies. Do you want 20% more service for 44% more expenditure or do you want to retain what service you have with no more expenditure? That is a clear choice and a preference can be made.2 Everyday Probability and Statistics An eminent Victorian. in principle. that it has invested a great deal of money in the service and that 20% more personnel are delivering it to the nation. statistics plays a role in life outside the area of politics. To attract more people in to deliver the service it has to increase salaries by 20% and in this way it increases the number of personnel by 20%. it would be possible for them to receive. In that case the treatment applied is based on getting the best outcome for as many patients as possible — although some individual patients may not get the best treatment for them.
supported the Conservative party and in the few days before the actual vote presented headlines with no factual or relevant content that were thought to have swayed a significant number of voters. the Sun newspaper. in school . of 10 dentists hand-picked by the company. In 1997. All choices are not in their nature between good and bad — they are often between bad and worse — and a mature society should understand this. premium bonds (in the United Kingdom). statistics is the governing factor that controls gambling of any kind — horse-racing. At another level entirely. at best. which duly won a landslide election victory. Many adults who were mediocre. particularly the so-called tabloid press — the type of newspaper that headlines the antics of an adulterous pop-idol while relegating a major famine in Africa to a small item in an inside page. In the 1992 UK general election. with the largest circulation in the United Kingdom. card playing. football pools. for example. It is in this area that the public at large seems much more appreciative of the rules of statistics. Alternatively it could mean that. dice throwing. Perhaps the antiseptic fluid does kill 99% of all germs but what about the other 1% — are they going to kill you? Another rich source of manipulated statistics is the press. These newspapers are particularly effective in influencing public opinion and the skilful presentation of selected statistics is often part of this process. 9 recommend the toothpaste — and perhaps the dissenting dentist was only chosen to make the claim seem more authentic. the Sun switched its allegiance to the Labour party. and the national lottery.Introduction 3 politicians and those who run public services. Advertisers are adept in making attractive claims for various products but those claims should be treated with scepticism. roulette wheels. How often have you seen the advertisement that claims that 9 dentists out of 10 recommend toothpaste X? What does it mean? If it means that nine-tenths of all the dentists in the country endorse the product then that is a formidable claim and one that should give reason to consider changing one's brand of toothpaste.
Worked out solutions are given so that. even if the reader has not been successful in solving a problem. not only has a wide range of applicability but also has a large number of component topics within it. within the capabilities of many people. may find it helpful to test their new knowledge by tackling some problems set at the end of each chapter. To master the subject completely requires all the abilities of a professional mathematician. involving an intuitive appreciation of the applications of statistics.4 Everyday Probability and Statistics mathematics acquire amazing skills. . Statistics. as a branch of mathematics. In order to understand statistics one needs also to know something about probability theory and this too forms a component of this book. when it comes to gambling. to comprehend some of its main ideas and what they mean is. reading the solution may help to strengthen his. understanding. More mature readers who long ago lost contact with formal mathematics. However. something that is available to comparatively few people. Here the aim is to explain how statistics impinges on everyday life and to give enough understanding to at least give the reader a fighting chance of detecting when organizations and individuals are trying to pull the wool over the public's eyes. with a little effort. or her. or younger ones who struggle somewhat with the subject. In the 21st century a cultured man should understand something about statistics otherwise he will be led by the nose by those who know how to manipulate statistics for their own ends.
To the extent that we can. be predictable. some of which are virtually synonymous with others. 5 . Some of these events. going to work is normally a predictable and frequent event but the mishaps. we have available a battery of words with different shades of meaning. Most of these words are so basic that they can best be defined in terms of each other. on any day. sometimes on a regular. Probability and Everyday Speech The life experienced by any individual consists of a series of events within which he or she plays a central role.1. we try to compensate for the undesirable uncertainties of life — by making sure that our homes are reasonably secure against burglary — a comparatively rare event despite public perception — or by taking out insurance against contingencies such as loss of income due to ill health or car accidents. such as illnesses. For example. (Aristotle.Chapter 1 The Nature of Probability Probable impossibilities are to be preferred to improbable possibilities. or might not. like the rising and setting of the sun. Others occur often. that occasionally prevent someone from going to work are events that are to be expected from time to time but can be predicted neither in frequency nor timing. To express the likelihoods of the various events that define and govern our lives. we are certain that the sun will set. 384–322 BC) 1. outside the polar regions. if not daily. If we say that something is certain then we mean that the event will happen without a shadow of doubt. basis and might. occur without fail each day.
6 Everyday Probability and Statistics We can qualify certain with an adverb by saying that something is almost certain meaning that there is only a very small likelihood that it would not happen. with the exception of the extreme words. 1. By attaching various qualifiers to these words — almost impossible as an example — we can obtain a panoply of overlapping meanings but at the end of the day. certain and impossible. the event is incapable of happening under any circumstances. . Something much more objective. Nevertheless. However. Spinning a Coin We are all familiar with the action of spinning a coin — it happens at cricket matches to decide which team chooses who will bat first and at football matches to decide which team can choose the end of the pitch to play the first half. we imply that the chance of it happening is greater than it not happening. tail. and numerically defined. or standing on an edge. There are rare years when it does not rain in January but these represent freak conditions. While these fuzzy descriptions of the likelihood that events might occur may serve in everyday life. they are clearly unsuitable for scientific use. but in some contexts it could be taken to mean that the likelihood is not very great — or that the event is unlikely. Finally. head. The word possible or feasible could just mean that an event is capable of happening without any connotation of likelihood. it is probable that there will be some rain in August because that happens in most years. when we say that an event is likely. There are three possible outcomes to the event of spinning a coin. impossible is a word without any ambiguity of meaning.1). which is a thin disk (Fig. That comes from the shape of a coin. or probable. 1.2. August is usually sunny and warm and it is not unusual for there to be no rain in that month. there is a subjective element in both their usage and interpretation. It is almost certain that rain will fall sometime during next January because that month and February are the wettest months of the year in the United Kingdom. is needed.
or something fairly close to that result. We now say that the probability of getting a head is 1 because that is the fraction of the occasions we expect that outcome.1. the probability of getting a head is 1/2. of the occurrence of particular outcomes. Since we expect a tail 50% of the times we spin the coin. Discounting the possibility that the coin will end up standing on one edge (unlikely but feasible in the general language of probabilities). the one with a head on both sides. we would suspect that something was wrong — either that it was a trick coin with a head on each side or one that was so heavily biased it could only come down one way. We have taken the first step in assigning a numerical value to the likelihood. that of symmetry. .The Nature of Probability 7 Fig. the shape of the coin contains another element. we deduce from symmetry that the probability of a head facing upward is the same as that for a tail facing up. it happens 100% of the time. we say that the probability of getting a tail is 1/2. Supposing that we repeated the above experiment of spinning a coin but this time it was with the trick coin. From an instinctive feeling of the symmetry of the event we would expect that the two outcomes had equal probability so that the most likely result of spinning the coin 100 times would be 50 heads and 50 tails. The three possible outcomes for spinning a coin. Similarly. If we were to spin a coin 100 times and we obtained a head each time. However. or probability. Every time we spin the coin we get a head. we get a tail on 0% of the times we flip the coin. because that is the fraction of the occasions that we expect that outcome. Getting a head is certain and that is what is meant by a probability of 1. Conversely. 1.
if we throw a die. 1.1).. 3. 1.2 shows this assignment of probabilities in a graphical way. Similarly.2. If the probabilities of getting a head or a tail are ph and pt . Analogous to the coin equation (1. The range shown for probability in Fig.1) 1. or 6. i. we have for the probability of each of the . without numbers marked on them. 5. no probability can be less that 0. However. From the symmetry of the die. A probability cannot be greater than 1 because no event can be more certain than certain.3. since no event can be less possible than impossible. 2 (1. it would be expected that the fraction of throws yielding a particular number. Figure 1. The numerical probability range with some notional verbal descriptions of regions. negative.3). 1.5 probable 1 certain description Fig. all the faces are similar and similarly disposed with respect to other faces (Fig. 4. would be 1/6 so that the probability of getting a 4 is p4 = 1/6 and that would be the same probability of getting any other specified number.8 Everyday Probability and Statistics probability 0 impossible possible 0. Throwing or Spinning Other Objects Discounting the slight possibility of it standing on an edge there are just two possible outcomes of spinning a coin. 2. something that comes from the symmetry of a disk. then there are six possible outcomes — 1. say a 4. A die is a cubic object with six faces and. then we can write 1 ph = pt = . respectively.2 is complete. We are now in a position to express the probabilities for spinning an unbiased coin in a mathematical form. Getting a tail is impossible and that is what is meant by a probability of 0. the probability of getting a tail is 0. head or tail.e.
each with the same probability. Fig. The two faces that we cannot see have two and four spots on them. in principle. it would give.The Nature of Probability 9 Fig.5 for a device giving numbers 1–5 with equal .3. respectively. This object would not tumble very well if thrown onto a flat surface unless thrown quite violently but. we see a regular tetrahedron.2) It is possible to produce other symmetrical objects that would give other numbers of possible outcomes. six possible outcomes 1 p1 = p2 = p3 = p4 = p5 = p6 = . 4 (1. A die showing three of the six faces. with equal probability. the numbers 1–4. 6 (1.4. 1. This is shown in Fig. A regular tetrahedron — a "die" with four equal-probability outcomes. 1. 1.3) A better device in terms of its ease of use is a regular shaped polygon mounted on a spindle about which it can be spun.4. each of which is an equilateral triangle. so that 1 p1 = p2 = p3 = p4 = . In Fig. a solid object with four faces. 1.
. 5 probability. 1. Low pressure is expected to affect northern and western parts of the UK throughout the period. The pentagon is spun about the spindle axis like a top and eventually comes to rest with one of the straight boundary edges resting on the supporting surface. rather warm conditions are expected. Next we will consider slightly more complicated aspects of probability when combinations of different outcomes can occur. Problems 1 1. The following is a Meteorological Office forecast for the United Kingdom covering the period 23 September to 2 October 2006.10 Everyday Probability and Statistics Fig.5. The spindle is through the center of the pentagon and perpendicular to it. along with some periods of strong winds too. A device for giving p1 = p2 = p3 = p4 = p5 = 1 .1. More central and western parts of the UK are likely to be rather unsettled with showers and some spells of rain at times. We have now been introduced to the idea of probability expressed as a fractional number between 0 and 1. There is a risk of some showery rain over south-eastern parts over the first weekend but otherwise much of eastern England and possibly eastern Scotland should be fine. with warm weather in any sunshine in the east. However. which indicates the number for that spin. with a southerly airflow dominating. Meteorology is not an exact science and hence weather forecasts have to be couched in terms that express that lack of precision. the only useful way for a scientist or mathematician.
If the faces are marked with the numbers 1 to 12. which will be the most and least probable ways for the object to come to rest? 1.2. and it is known that 123 out of 4205 patients in a recent epidemic died. As deduced from this information. and each face is similarly disposed with respect to the other 11 faces. that a given patient contracting the disease will die? . expressed to three decimal places.The Nature of Probability 11 Identify all those parts of this report that indicate lack of certainty. i. each a regular pentagon. Make drawings showing the possible ways that the object can come to rest if it is thrown onto the floor. a cone with the top sliced off with a cut parallel to the base. A certain disease can be fatal. The figure below is a dodecahedron. Based on your intuition. 1.3. what is the probability. then what is the probability of getting a 6 if the dodecahedron is thrown? 1.4. The object shown below is a truncated cone.e. It has 12 faces.
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a 1 and a 6 — one-third of the possible outcomes — satisfy our requirement so the probability of obtaining a 1 or a 6 is p1 or 6 = 1 2 = 6 3 (2. 1692–1752) 2.1. Two of these.2) says that "the probability of getting either a 1 or a 6 is the sum of the probabilities of getting a 1 and getting a 6". we consider the six possible outcomes. all of equal probability. Either–or Probability Let us consider the situation where a die is thrown. 6 6 6 2 (2. (2. Eq.Chapter 2 Combining Probabilities But to us probability is the very guide of life. and we wish to know the probability that the outcome will be either a 1 or a 6. or 3 when throwing the die is p1 or 2 or 3 = p1 + p2 + p3 = 13 1 1 1 1 + + = . How do we find this? First. This idea can be extended so that the probability of getting one of 1. 6 6 3 (2. (Bishop Butler. 2.2) In words.1) Another way of expressing this result is to say that p1 or 6 = p1 + p6 = 1 1 1 + = .3) .
To explore this idea further consider a standard pack of 52 cards.5) Now. In either–or probability if a 1 is obtained then a 2 or a 3 is excluded. the probability of getting either a head or a tail when flipping a coin is 1 1 ph or t = ph + pt = + = 1. jack. It is a general rule that the probability of having an outcome which is one or other of a set of mutually exclusive outcomes is the sum of the probabilities for each of them taken separately. 52 52 52 52 52 13 (2.e.. The separate probabilities of outcomes for jacks.1. a 2 or a 3 (Fig. If we are interested in the outcome being a 1. if we obtained a 2 the outcomes 1 and 3 would have been excluded. queen or king) from the pack. and kings are all 1/13 but the outcomes If then Fig. or 3 then. 2. we exclude the possibility of having obtained either of the other outcomes of interest. 2. we want to know the probability of picking a court card (i. (2.1). Four of the cards are jacks so the probability of picking a jack is pJ = 1 1 1 4 1 1 + + + = = . throwing a die.14 Everyday Probability and Statistics Similarly. In considering these combinations of probability we are taking alternative outcomes of a single event — e. 2.g. if we obtain a 1. queens.4) 2 2 which corresponds to certainty since the only possible outcomes are either a head or a tail. The outcomes for which the probabilities are being combined are mutually exclusive. The probability of choosing a particular card by a random selection is 1/52. . Similarly.
we imagine that two events occur — a coin is spun and a die is thrown.6) Of course one could have lumped court cards together as a single category and since there are 12 of them in a pack of 52 the probability 3 of selecting one of them could have been found directly as 12 = 13 52 but having a mathematically formal way of considering problems is sometimes helpful in less obvious cases. we consider the probability of obtaining a head — which is 1/2. a queen. that is not a mathematical restriction and this type of probability combination can be applied to any number of mutuallyexclusive outcomes. This kind of combination of probabilities has been called either– or and a pedantic interpretation of the English language would give the inference that only two possible outcomes could be involved. Both–and Probability Now. 13 13 13 13 (2.Combining Probabilities 15 of obtaining a jack. First.2. we list all the outcomes that are possible: h+1 h+2 t+1 t+2 h+3 t+3 h+4 t+4 h+5 t+5 h+6 t+6 There are 12 possible outcomes. We now ask the question "What is the probability that we get both a head and a 6?". and we are concerned with the one marked with an arrow. However. Hence the probability of picking a court card is pJ or Q or K = pJ + pQ + pK = 1 1 3 1 + + = . This probability can be considered in two stages. Clearly the probability of having both a head and a 6 is 1/12. The result of spinning the coin can have no conceivable influence on the result of throwing the die. 2. First. The two outcomes are certainly not mutually exclusive — indeed they are independent outcomes. each of equal probability. This corresponds to the outcomes . or a king are mutually exclusive.
For example.16 Everyday Probability and Statistics on the top row of our list. and picking a card from a pack gives a jack is ph and 6 and J = ph × p6 × pJ = 1 1 1 1 = .000 such diseases known and particular diseases tend to be prevalent in particular ethnic groups. which are responsible for carrying oxygen from the lungs to muscles in the body and carbon dioxide back from the muscles to the lungs. which will include many of the black populations of the Caribbean and North America and also of the United Kingdom. The hemoglobin forms long rods within the cells. the combined probability that spinning a coin gives a head.3. we see that pboth h and 6 = ph × p6 = 1 1 1 × = . Looked at in two stages. our description of this probability combination has done violence to the English language.8) Once again. sickle-cell anemia is mainly present in people of West African origin. × × 156 2 6 13 (2. Thus. Genetically Inherited Disease — Just Gene Dependent Within any population there will exist a number of genetically inherited diseases. throwing a die gives a 6. The combination both–and should formally be applied only to two outcomes but we stretch it to describe the combination of any number of independent outcomes. These rules of "either–or" and "both–and" combinations of probability can themselves be combined together to solve quite complicated probability problems. 2. This disease affects the hemoglobin molecules contained within red blood cells. we consider the probability that we also have a 6 that restricts us to one in six of the combinations in the top row since the probability of getting a 6 is 1/6. Now. There are about 4. 12 2 6 (2. .7) This rule can be extended to find the combined probability of any number of independent events.
000 to hundreds of thousands of the base units that make up DNA. including humans. gives a tendency to produce shorter individuals. so that. In addition. one gene pair might control stature so that gene A predisposes toward height while the other member of the pair. Another genetic disease is Tay–Sachs disease which affects people of Jewish origin.Combining Probabilities 17 distorting them into a sickle shape and making them less flexible so they flow less easily.000.000–40. For example. and is always fatal. the cells live less time than the normal 120 days for a healthy red cell and so the patient suffers from a constant state of anemia. Each person has two of these stature genes in his cells. One can only talk about tendency in this instance since other factors influence stature. To understand how genetically transmitted diseases are transmitted. large numbers of genes. The number of genes controlling human characteristics is somewhere in the range 30. in particular diet. If they are both A then the person will have a tendency to be tall and if they are both a then there will be a tendency to be short. Genes occur in pairs which usually correspond to contrasting hereditary characteristics. we need to know something about the gene structure of living matter. destroying brain and nerve cells. for example. Contained within each cell of a human being there are a large number of chromosomes. thread like bodies which contain. usually at the infant stage. This attacks the nervous system. A child inherits one "stature" gene from each parent and which of the two genes he gets from each parent is purely random. Aa will tend . Each gene is a chain of DNA of length anything from 1. gene a. we show some of the possibilities for various parental contributions: Father Aa AA AA Mother Aa aa AA Child (all pairs of equal probability) AA Aa aA aa Aa Aa Aa Aa AA AA AA AA where an individual has a contrasting gene pair then sometimes the characteristics will combine. Here. strung out along them.
01 = 0. possessing the gene pair Dd. we note that: the probability that both gene 1 is D and gene 2 is d is 0. no monitoring of parents. i. again.e. the probability that this one is d is 0. we take it that d is a very rare gene in the community and D is dominant. Thus.01 because that is the proportion of the d gene. then.0099. All Chinese are BB so that all Chinese children must inherit the gene pair BB from their parents. What is the probability that. or her.. if B is a "brown-eye" gene and b is a "blue-eye" gene then BB gives an individual with brown eyes.01 = 0. or one chance in 10. such a person is known as a carrier. with random mating. Hence the probability that both the first gene is d and the second gene is d is 0. The probability that the first gene will be d is 0. Thus.99 × 0.000.01 × 0. and Bb (equivalent to bB) will give brown eyes because B is the dominant gene. using both–and probability.01 × 0. but in other cases one of the genes may be dominant. Anyone who happens to be Dd will be free of the disease but may pass on the harmful gene d to his.01. children. bb gives an individual with blue eyes. The allocation of the second gene of the pair is independent of what the first one happens to be so. . i. Now. However. The gene d predisposes toward the disease and someone who inherits a pair dd will certainly get the disease and die before maturity.99 = 0.0099 the probability that both gene 1 is d and gene 2 is D is 0. Sometimes genes become "fixed" in a population.18 Everyday Probability and Statistics to give a medium-height individual. all Chinese have brown eyes. we consider a genetically related disease linked to the gene pair D.0001. d.e. a baby born in that population will have the disease? We can consider this problem by considering the allocation of the gene pair to the baby one at a time. If we were interested in how many babies would be carriers of the faulty gene. Let us suppose that in this particular population the ratio of d:D is 1:100..
which is thought to have a genetically transmitted element.0099 = 0. Some genetically related diseases are very rare indeed because the incidence of the flawed gene is low. Henry VIII divorced two wives because they did not give him a son but now we know that he was to blame for this! There are some defective genes. for example which leads to hemophilia. On the other hand diabetes. The female always contributes the same chromosome to her offspring. Son with X Y will have the disease because there is no accompanying X to compensate for the X .0099 + 0.0198 so that about one in 50 babies born is a carrier.4. X.000 only one child in a million would contract the disease although about one in 500 of them would be carriers of the disease. which only occurs in the X chromosome. Genetically Dependent Disease — Gender Dependent There are genetically transmitted diseases where the gender of the individual is an important factor. If we call a chromosome with the defective X gene X then the following situations can occur: Daughter with chromosome combination XX' will be a carrier of hemophilia but will not have the disease because the presence of X compensates for the X . a female having the chromosome pair XX and a male XY. is much more common and the number of carriers is probably quite high. for d:D = 1:1. The sex of an individual is determined by two chromosomes. but the man contributes X or Y with equal probability thus giving a balance between the numbers of males and females in the population.Combining Probabilities 19 Since Dd and dD are mutually exclusive the probability of the baby carrying the gene pairs either Dd or dD is 0. . X and Y. 2.
Half the sons are X Yand so are hemophiliacs. Half of the daughters are XX' and so are carriers but the other half are XX and are not carriers. Half the daughters are X X and so will have the disease while the other half are X X and so are carriers. Half the sons are X Y and so are hemophiliacs. (iv) Here. However. we can see various outcomes from different parental chromosome compositions (i) All children are free of hemophilia and none are carriers (ii) The father suffers from the disease but all children are free of hemophilia.000 males are born with .20 Everyday Probability and Statistics Now. The incidence of hemophilia does not seem to have any strong correlation with ethnicity and about 1 in 5. Sons are completely unaffected because they just receive a Y chromosome from their fathers. The other half of the sons are XY and so are free of the disease. The other half of the sons are XY and so are free of the disease. all daughters are XX' and so all are carriers. (iii) Here. we have a hemophiliac father with a carrier mother. the father is free of the disease but we have a carrier mother.
If the thrower gets a sum of either 7 or 11 in his first throw that is called "a natural" and he immediately wins the game." The player then continues throwing until either he gets his "point" again. became influential in the Russian court because of his apparent ability to ameliorate the symptoms of the disease in the young Alexis. A Dice Game — American Craps American craps is a dice game that is said to originate from Roman times and was probably introduced to America by French colonists. (a) Snake-eyes (b) A natural (first throw) but losing throw later. if he gets 2 (known as "snakeeyes").5. born in 1904.Combining Probabilities 21 the disease. in which case he wins.2. The most famous case of a family history of hemophilia is concerned with Queen Victoria who is believed to have been a carrier. Any other sum is the players "point. However. 3. or 12 he immediately loses (Fig. The probabilities of obtaining various sums are clearly the essence of this game.2). A coarse. . It involves throwing two dice and the progress of the game depends on the sum of the numbers on the two faces. Rasputin. 2. She had nine children and 40 grandchildren and several male descendants in Royal houses throughout Europe suffered from the disease. 2. 2. There are many who believe that his malign influence on the court was an important contributory factor that led to the Russian revolution of 1917. priest. or until he gets a 7. The most notable example was that of Tsarevitch Alexis. the long-awaited male heir to the throne of Russia. and rather dissolute. (a) (b) Fig. in which case he loses.
9) 1 4 = . 3. e. each combination has a probability of 1/36. If the player has to play to his "point" then his chances of winning depends on the value of that "point" and how the probability of obtaining it compares with the probability of obtaining 7.g. or 12. 3 or 12 = p1+1 + p1+2 + p2+1 + p6+6 = (2. These combinations are mutually exclusive. Point 4 5 6 8 9 10 3+1 4+1 5+1 6+2 6+3 6+4 Combinations 2+2 3+2 4+2 5+3 5+4 5+5 1+3 2+3 3+3 4+4 4+5 4+6 1+4 2+4 3+5 3+6 1+5 2+6 Probability 3 36 4 36 5 36 5 36 4 36 3 36 = = 1 12 1 9 = = 1 9 1 12 . or 12 is psnake-eyes = p1 × p1 = p2. (2. Below. 6 6 36 Hence the probability of getting 2. as a sum requires one or other of the combinations 6+1 5+2 4+3 3+4 2+5 1+6 6+5 5 + 6. we consider the probability of obtaining 2. we show for each possible "point. 3.22 Everyday Probability and Statistics First. each with a probability of 1/36 so that the probability of a "natural" is 8/36 = 2/9.10) 36 9 Getting 7 or 11. 1 1 1 × = . The two dice must show one of 1+1 1+2 2+1 6+6 and these are mutually exclusive combinations." the combinations that can give it and the probability of achieving it per throw. Since the numbers on each of the dice are independent of each other.
Later.3? 2.6.) 2.5. the latter gene leads to a particular disease although F is the dominant gene. and (ii) carriers of the disease? .1. is higher than that of any of the points. If a card is drawn from a pack of cards then what is the probability that it is an ace? 2.2 and a normal six-sided die are thrown together. If the ratio of the incidence of the genes is f :F = 1:40 then what proportion of the population is expected to be (i) affected by the disease.2 and a normal six-sided die are thrown together. For a particular gene pair F and f . This is an extremely popular gambling game in the United States and those that play it almost certainly acquire a good instinctive feel for probability theory as it affects this game. The dodecahedron from problem 1.2 what is the probability that a throw will give a number less than 6? 2. Is this probability greater than or less than that found in problem 2. when we know a little more about probabilities.Combining Probabilities 23 The total of 7 can be achieved in six possible ways (see above ways of obtaining a "natural") and its probability of attainment. we shall calculate the overall chance of the thrower winning — it turns out that the odds are very slightly against him. The dodecahedron from problem 1.4. Problems 2 2. Find the combined probability of getting a jack from a pack of cards and getting four heads from spinning four coins.3. What is the probability that the sum of the numbers on the sides is 6? (Hint: Consider the number of ways that a total of six can be obtained and the probability of each of those ways. For the dodecahedron described in problem 1.2. What is the probability that both the dodecahedron and the die will give a 5? 2. although few would be able to express their instinctive knowledge in a formal mathematical way. 1/6.
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A die is a symmetrical object. Thus.Chapter 3 A Day at the Races Do not trust the horse. from experiment. the only guide is previous experience — knowledge of the death-rate from the disease. into six equal parts and associate a probability of 1/6 with each of the faces. or reason. each face of which is related geometrically in the same way to the other five faces so again we divide the probability 1. is known as empirical probability and it is the more important kind of probability from the point of view of everyday life — unless one is an inveterate gambler in casinos where many games are based on logical probabilities. That was the basis of problem 1. if one is planning a 2-week holiday in Florida one could predict the probability of 25 . Usually this is based on the concept of symmetry.4.1. a coin has two sides neither of which is special in a probabilistic sense so each has an associated probability of 1/2. Here. Such assessments of probability are both logical and intuitive. Kinds of Probability The kind of probability we have considered numerically thus far can be regarded as logical probability. corresponding to certainty. which is to say that we can estimate probability by an exercise of logic. This kind of probability. 70–19 BC) 3. assessed from those that have contracted it previously in similar circumstances. (Virgil. If we wish to know the likelihood that an infected person will die from a potentially fatal disease we cannot use logical probability. where appropriate. gained from observation or.
the unpredictable interaction of horses as they jump together over a fence can cause upsets in the race result. fractions of a tornado do not exist but what the answer means is that the probability of having a tornado within a distance of 10 miles is 0.011 or about 1 in 90.000 square miles so. will win a particular race. of course. In races like the Grand National. There is. The extra dimension that must be involved here in assessing probabilities is judgment and we can .09 × 314 2 × = 0.09 per year per 10.000 52 Of course. 10.000—so very small. The total area within 10 miles of any spot is the area of a circle of that radius and is A = π × (10)2 square miles = 314 square miles. assuming that all times of the year have the same probability. probably more like 1 in 30. the number expected within an area of 314 square miles in a particular 2-week period is N = 9.011. The kind of probability involved here is certainly neither logical nor even empirical since each case being considered has its own individual features and cannot be accurately assessed from past performance. would have taken place under different conditions both in the state of the track and the characteristics of the opposition. The average frequency of tornadoes in Florida is 9. or perhaps a greyhound. is in assessing the likelihood that a horse. horses have their good days and bad days and these cannot be predicted. A kind of probability.26 Everyday Probability and Statistics being within 10 miles of a tornado during the holiday by consulting the weather statistics for that state. which involves jumping over high fences. the previous record of the contesting animals but there are many other factors to take into account. like humans. Again. which is empirical in essence but has an extra dimension. For a tornado to be a risk to an individual he would have to be in its direct path and the risk of that is much smaller than 1 in 90. Each of the previous races. on which the form would be assessed.
depending on the number of horses in the race.2. If the horse backed has odds 3/1 (three-to-one) then a successful £10 bet will bring a return of the original £10 plus 3 times £10.000 on the horse then our punter would receive a fraction 10/20. Thus. However. or four places. A betting shop or a bookmaker. a fraction of the odds for winning. Betting on a Horse There are three main ways that a punter can bet on the results of a race — in a betting shop either locally or at the track. £475.000.. if the horse finishes in one of the first two. 000 The Tote is actually a little more complicated than we have described because "place bets" can be made where the punter will get a reduced return. i. say £500. Thus his return will be £475. by contrast. the essentials of the Tote system are described by just considering bets to win. and lose heavily. . say £25. or by the Tote. A bookmaker can lose. Getting this right is of critical importance to bookmakers who take bets on the outcome of races and who could be ruined by consistently making faulty judgements.000. The Tote organizers will take the total money staked on the race.50. 000 × 10 = £237. At the time a punter places a £10 bet on "My Fair Lady" with the Tote he has only a rough idea of what he would win if that horse won. 20. if the people backing "My Fair Lady" staked a total of £20. three.A Day at the Races 27 define this kind of probability assessment as "judgmental probability". through a bookmaker at the track.e. cannot lose money.000 of the returned stake money. The Tote is a system which. between those that backed 'My Fair Lady' to win in proportion to how much they staked. by its very nature. to cover expenses and to create a profit and will divide the remaining money. take a proportion of this. enters into a contract with the punter to pay particular odds at the time the bet is placed. £40 in all.000. 3.
his return from the race would be £3.001) (£1.001) (£1. He bets on every horse and places bets as follows: Diana Dawn Lady Fairy Princess Olive Green Mayfly £250 £143 £143 £91 £91 (£1. whichever horse won. This case is easy to see without analyzing it in detail but more subtle examples of bad odds-setting can occur.001) . offers the following odds on a race for fillies: Diana Dawn Lady Fairy Princess Olive Green Mayfly Dawn Chorus Missy Lovelorn Helen of Troy Piece of Cake 3/1 6/1 6/1 10/1 10/1 15/1 15/1 20/1 25/1 30/1 Although it is not obvious from a casual inspection of the list of odds this bookmaker is heading for certain ruin and any punter worth his salt could make a profit on this race.000.000) (£1. Consider a two-horse race where the bookmaker sets odds of 2/1 on each horse. Let us take a hypothetical race with 10 runners and the bookmaker. A punter would not take long to work out that if he staked £1. Let us take an example of odds set by a bookmaker that would obviously be foolish.28 Everyday Probability and Statistics on a particular race but if he knows his craft and assesses odds skillfully he will make a profit over the long term.001) (£1.000 but.000 on each horse his total stake on the race would be £2. Let us see how he does this. assessing the relative merits of the various runners.
This condition gives the . the stake is £1.000 — although bookmakers do not accept stakes involving pennies and fractions of pennies.000 or a small amount more.008) (£1.000 and £1. with n = 3. planning to get £1. in what follows we shall assume that he adjusts his stake to receive exactly £1.023) In parentheses. after the bets. n2 /1· · ·n10 /1. Clearly no bookmaker would offer such odds on a race and we shall now see what the principles are for setting the odds that ensure a high probability of profit for the bookmaker in the longer run.000 the amount staked is £1. To check this.000. For simplicity in analyzing the situation. is S= 1000 1000 1000 + ··· + .000/(n + 1). are shown what the punter will receive if that particular horse wins — anything between £1.000 returned no matter which horse wins.023. in pounds. S must be greater than 1. Now let us take a general case where the odds for a ten-horse race are indicated as n1 /1. We notice that what the punter has done is to set his stake on each horse at a whole number of pounds that will give a return (stake money plus winnings) of £1. then his total stake.A Day at the Races 29 Dawn Chorus Missy Lovelorn Helen of Troy Piece of Cake £63 £63 £48 £39 £33 (£1.000 then the punter is bound to win.008) (£1. we see that for Diana. + n10 + 1 n1 + 1 n2 + 1 (3.000/(3 + 1) = £250. However. If the punter backs every horse in the race.014) (£1. Thus for the bookmaker not to be certain to lose money to the knowledgeable punter. if the bets are added they come to £964.008) (£1. the punter is a certain winner and he will win between £36 and £59 depending on which horse wins the race.1) If S is less than £1. If the horse has odds of n/1 then the punter will receive n + 1 times his stake money so to get a return of £1.
n7 + 1 n8 + 1 n9 + 1 n10 + 1 If relationship (3.3) is the condition that a punter cannot guarantee to win — although of course he may win just by putting a single bet on the winning horse. Put into a mathematical notation. The summation symbol m means that you are going to add together m quantities i=1 with the value of i going from 1 to m. If he fixed the odds on a horse at 10/1 when the horse actually has a one-in-five chance of winning then astute professional gamblers would soon pick this up and take advantage of it.3) were not true and the summation was less than 1.2) and for the bookmaker not to lose to the clever punter the left-hand side must be greater than 1. The skill of a bookmaker is not just in fixing his odds to satisfy the golden rule — anyone with modest mathematical skills could do that. ni + 1 (3.000 then we have S 1 1 1 + + ··· + . (3.30 Everyday Probability and Statistics bookmaker's golden rule which we will now express in a mathematical form. He must also properly assess the likelihood of each horse winning the race. If we take Eq. and i runs from 1 to 10 so that 10 i=1 1 1 1 1 1 1 1 = + + + + + ni + 1 n1 + 1 n2 + 1 n3 + 1 n4 + 1 n5 + 1 n6 + 1 + 1 1 1 1 + + + .3) The symbol > means "greater than" and Eq. as in the hypothetical case we first considered. the golden-rule condition is m i=1 1 > 1. = n10 + 1 1000 n1 + 1 n2 + 1 (3. the number of horses. which we first give and then explain. . then the punter could guarantee to win. (3. In our specific case m = 10.1) and divide both sides by 1.
then the larger its value the larger is the likely profit that the bookmaker will make on the race. (3.111) (0.048) (0. The Best Conditions for a Punter We have seen that the bookmaker's golden rule. which is usually the case. assuming that the relative chances of the horses winning is properly reflected in the odds offered.143) (0.062) (0. Let us consider the following two races with horses and odds as shown.200) (0.3. the higher is the summation the better it is for the bookmaker. In brackets is shown the value of 1/(n + 1) Race 1 Diana Dawn Lady Fairy Princess Olive Green Mayfly Dawn Chorus Missy Lovelorn Helen of Troy Piece of Cake Pompeii Noble Lad Gorky Park Comet King Valliant Park Lane 2/1 4/1 6/1 10/1 10/1 15/1 15/1 20/1 25/1 30/1 3/2 2/1 5/1 8/1 12/1 16/1 (0.032) (0. Eq.400) (0. then the smaller is the summation the less are the odds stacked against the punter. will ensure that no punter can place bets so as to guarantee to win on a particular race.A Day at the Races 31 3. If the quantity on the left-hand side was less than 1 then the punter could guarantee to win and if it is equal to 1 then the punter could so place bets as to guarantee to get exactly his stake money returned — but.167) (0. that would be a pointless exercise.059) Race 2 (=1 1 /1) 2 .333) (0.333) (0.038) (0. Logically.091) (0.077) (0.3).091) (0.062) (0. Conversely. Given that the left-hand side is greater than 1. of course. what is good for the bookmaker is bad for the punter so we may make the proposition that.
and the Grand National. Other things being equal. the punter would have a better chance of winning by betting on his fancied horse in the first race. by carefully assessing the odds. One hundred pounds won on a minor race at a minor racetrack is the same to them as one hundred pounds won on a classic race — and if the classic race does not offer them a good opportunity then they will simply ignore it. who normally never gamble on horses. to place bets either individually or through a sweepstake organized by a social club or at work. manage to make a living. or girlfriend called Sheila. by whims or hunches. and assuming that the relative chances of the horses winning are accurately reflected in the odds. There are a number of classic races during the year.32 Everyday Probability and Statistics The value of the summation for race 1 is 1. . the Derby. Most people placing bets on such occasions know nothing of the horses but may be attracted by the fact that they are trained locally or that their names are the same or similar to that of a friend or family member. Sheila's Cottage won the Grand National at odds of 50/1 and was backed by large numbers of punters lucky enough to have a wife.147. It must always be remembered that bookmakers also make a living. By contrast to the occasional or recreational punter there are some professional gamblers who. for the average punter. studying form. for example. that attract large numbers of people. and usually a good one so. In 1948. For them the big classic races have no particular attraction. and betting only when conditions are most favorable.100 and for the second race it is 1. gambling is best regarded as a source of entertainment and betting limited to what can be afforded. daughter. the St Leger. One factor that works to the advantage of bookmakers is that much betting is done on an illogical basis.
1. The runners and odds offered by a bookmaker for three races at a particular track are as follows: 2:30 pm Bungle Crazy Horse Memphis Lad Eagle Galileo Mountain Air Diablo Copper Boy Dangerous Boxer Kicker Zenith Turbocharge Tantrum Argonaut Firedance Dural 1/1 4/1 6/1 8/1 12/1 18/1 2/1 3/1 3/1 4/1 6/1 8/1 2/1 3/1 4/1 8/1 12/1 (Evens) 3:15 pm 3:45 pm Do the odds for all these races conform to the bookmakers golden rule? Which is the race that best favors the bookmaker's chance of winning? .A Day at the Races 33 Problem 3 3.
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Amelia. ABCD BACD CABD DABC ABDC BADC CADB DACB ACBD BCAD CBAD DBAC ACDB BCDA CBDA DBCA ADBC BDAC CDAB DCAB ADCB BDCA CDBA DCBA While it is possible to generate these 24 orders of leaving in a systematic way it gets more and more complicated as the number of children increases. Again.1. and Christine. You can easily check that there are no more than these six possibilities. and choose the good (Isaiah 7:14) 4. one-by-one. The time comes for them to leave and they emerge from the room. there is not much point in generating the detailed orders of leaving if all that is required is to know how many different orders there are.1. Barbara. In how many different orders can they leave the room? Using just their initial letters.Chapter 4 Making Choices and Selections Refuse the evil. 4. Let us see how this could be derived without 35 . For three children there were six orders of leaving and for four children there were 24. playing in a room. then the number of different orders of leaving would have been considerably greater — viz. Daniella. we show the possible orders in Fig. If there had been another child in the room. Children Leaving a Room We imagine that there are three children.
Once the first child has left then there are two children remaining in the room and hence two possibilities for the exit of the second child — corresponding to the second letter for each of the orders. 4. two for the third. who finally leaves.1.1.1) If we repeat this pattern for four children — there are four possibilities for the first to leave. or C — corresponding to the different rows of Fig. The six orders in which three girls can leave a room. This leaves one child in the room. three for the second. Seen in this way the number of possible patterns of departure is given by N3 = 3 × 2 × 1 = 6. and then the last one leaves — the number of possible patterns of departure for four children is given by N4 = 4 × 3 × 2 × 1 = 24. B.2) . For three children there are three different possibilities for the first child to leave — A. (4. (4.36 Everyday Probability and Statistics Fig. writing down each of the orders as has been done above. 4.
but very complicated to write down each individual order as was done for three children and four children. A computer generating one billion (one thousand million) of these orders per second would take more than eight million billion years to complete the job! 4. . that has entered a national bridge tournament in which they will be represented by a team of four players. consisting of 12 members. which is a number with 33 digits. Notice that.4) where 7! is called "factorial 7". For factorials of large numbers. so as a shorthand we write 7 × 6 × 5 × 4 × 3 × 2 × 1 = 7! (4. we can find the number of orders even when determining and listing the individual orders would not be practicable.Making Choices and Selections 37 This kind of approach gives the results that we found by listing the orders in detail and extending it to. 040. Picking a Team We consider a bridge club. The first member of the team can be picked in 12 ways. the number of orders in which a class of 30 pupils could leave a classroom is 30!. The members of the club are all of equal merit so it is decided to pick the team by drawing lots — which means that all possible selections of four players have the same probability of occurring. say.2. the conciseness of the notation is of great benefit.3) It would be possible. Fully writing out these products of numbers for larger numbers of children is tedious and takes up a great deal of space. say 50!. much as we considered children leaving the room. The question we ask is "How many different compositions of team is it possible to select from the members of the club?" This problem can be considered in stages by picking members of the team one-by-one. For example. seven children the number of different orders of leaving is N7 = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5. (4.
(4. From this. then the value of S given by (4. This is precisely the problem of the number of different orders in which four girls can leave a room. without regard to the order of their selection. then the second in three ways. and finally the fourth in one way so the number of ways of ordering four letters is 4 × 3 × 2 × 1 = 4!.38 Everyday Probability and Statistics For each of those 12 ways there are 11 ways of picking the next team member — so there are 12 × 11 = 132 ways of picking the first two members of the team. 3. without regard to order of selection. = = 8! 8×7×6×5×4×3×2×1 . We note that 12 × 11 × 10 × 9 12! 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 . one must divide S by the number of different ways of ordering four letters.6) It is possible to express this in a more succinct way. and 4 to decide on how they play in pairs against opponents. Representing individuals by letters the value of S comes about if selection ABCD is regarded as different from ACBD and all the other different ways of ordering the four letters. the first letter can be chosen in four ways. The next member can be picked in 10 ways and the final member in 9 ways giving a total number of different ordered selections: S = 12 × 11 × 10 × 9 = 11. as T= S 12 × 11 × 10 × 9 = = 495.880 ordered selections there are groups of the same four individuals and to find the number of teams. we find the number of teams.5) This will be the answer if the players have to be designated in some order because the tournament requires that players of a team must be labeled 1.5) is too great. However. 2. Within the 11. without regard to order. if the team is just a group of four individuals. 4! 4! (4. then the third in two ways.880.
If we take r of them out of the bag.8) By making n = 12 and r = 4. numbered balls contained in a bag. However. (n − r)!r! (4.7) We can generalize from this example. if the order is not taken into account.9) which. with n = 12 and r = 4 is the number of different teams of four people one can select from 12 without order being taken into account. (n − r)! (4.9) is important enough to have its own symbol so we can write the number of combinations of r objects taken from n as n Cr = n! .Making Choices and Selections 39 giving S= and then T= S 12! = . and take the order of selection into account then the number of different selections is S= n! .10) It is obvious that the number of combinations of n objects that can be taken from n is 1 — you take all the objects and that is the . (n − r)!r! (4. distinguishable from each other — say. the number of ways of selecting an ordered team of four people from 12. then the number of different selections is T= n! . 8! (4. we get (4. corresponding to putting your hand in the bag and taking a handful of r balls.7). one-at-atime. 4! 8!4! 12! . Let us suppose that there are n objects. The expression on the right-hand side of (4.
Inserting r = n in (4. 0!n! from which we conclude the interesting result that 0! = 1. and Z rarely occur. So he adds the digit 1 to his username and finds that it is still unavailable.11) 4. First. 17. Choosing an Email Username The most common number of initials designating an individual in western societies is three. corresponding to a family name plus two given names. which is not at all obvious. He then tries the digit 2 with the same result — as happens for 3.12) . Hence the expected number of times the combination GLJ would appear in one million customers is N= 1.576 (4. when Gareth signs up to an internet provider and asks for the username GJL he is told that it is already in use. (4. However. with which he would sanction alterations on his checks or which he would put at the bottom of notes to colleagues at work. Why is this? Let us assume that the internet provider has one million customers wishing to use a set of three initials as a username.10) gives n Cn = n! = 1.576.3. X.40 Everyday Probability and Statistics only combination. we ask ourselves how many different sets of three initials we could have on the assumption that all letters of the alphabet are equally likely — which is not actually true since letters such as Q. 4. Each initial has 26 possibilities so the number of possible combinations is 26 × 26 × 26 = 17. Thus Gareth Llewelyn Jones would be recognized by his initials GLJ.000. and 5.000 = 57.
If Gareth were to try something like GLJ123 then he would be likely to find a unique username. . and J are fairly common initials the number of people in the provider's customer list sharing Gareth's combination may well be closer to 100.2. What are the chances of winning a first prize with a single entry? All combinations of six different numbers are equally likely and 6 winning numbers Bonus ball 2 41 18 23 15 8 33 (a) (b) (c) (d) (e) 2 2 8 2 5 8 15 15 6 8 15 18 23 8 14 23 23 33 29 15 33 33 35 33 33 41 41 41 41 48 Fig. the bonus ball. The UK National Lottery A single entry in the UK national lottery consists of picking six different numbers in the range 1–49. an appended digit might still be needed since 264 (= 456. If these match the numbers on the balls picked at random out of a drum then the first prize is won. (a) First prize — six correct. A seventh ball is also selected from the drum. L. Even if four-letter usernames were normal. There are also smaller prizes for getting five correct without the bonus-ball number. (d) Fourth prize — four correct. Figure 4. 4.4. (b) Second prize — five correct plus bonus ball. if you have the bonus-ball number plus any five of the six main winning numbers then you win the second prize. 4.2 shows some examples for winning each kind of prize. Actually. given the fact that G. (c) Third prize — five correct. The selected numbered balls and some winning lines. and also for four or three correct numbers.Making Choices and Selections 41 to the nearest whole number. (e) Fifth prize — three correct.976) is less than one million.
2 these are 2 2 2 2 2 8 8 8 8 8 15 15 15 15 15 23 23 23 23 23 33 33 33 33 33 41 41 41 41 41 Since there are six different ways of having five of the six winning numbers so there are six different combinations — five winning numbers + bonus-ball number — that can win the second prize.15) We have seen that there are six different ways of having five winning numbers and the number of ways of having a sixth number which is neither a winning number nor the bonus-ball number is .983. (4. that is. divided by the total number of different combinations.983. 13.13) 43!6! 6×5×4×3×2×1 Since there is only one correct combination the probability of winning the first prize is Pfirst = 1 . 6. The number of different ways of having five of the six winning numbers is six.42 Everyday Probability and Statistics the number of combinations of six that can be selected from 49 is obtained from (4. Thus the probability of winning this way is the ratio of the number of ways of having a winning combination.983.14) It is probably sensible to await the draw before ordering the Rolls Royce! To win a second prize you must have five of the six winning numbers plus the bonus-ball number.816 2. giving 49 C6 = 49! 49 × 48 × 47 × 46 × 45 × 44 = = 13.816.636 (4.10) with n = 49 and r = 6.816 (4.330. 4. 13. Psecond = 1 6 = . For the example of the six winning numbers given in Fig.
983.820.Making Choices and Selections 43 49–number of winning numbers–bonus-ball number = 49 − 6 − 1 = 42.820 = .545 = . The number of ways of doing this is 6 C3 × 43 C3 = 6! 43! × = 246.545.032.983.816 56.816 1. 3!3! 40!3! so the chance of winning this way is Pfifth = 1 246. giving the probability of winning as Pfourth = 1 13.18) .816 55. C2 = 2!41! Hence the number of combinations with just four correct numbers is 6 C4 × 43 C2 = 15 × 903 = 13. Thus.16) The number of ways of winning a prize with four correct numbers is the number of combinations of having four out of six winning numbers with two out of 43 nonwinning numbers.17) Finally.983.33 (4. the number of combinations for winning with five correct is 6 × 42 = 252 and hence the chance of winning this way is Pthird = 1 252 = . 13. to have just three winning numbers you need three out of six winning numbers together with three out of 43 nonwinning numbers.66 (4. 13. 13.4 (4. 4!2! and the number of ways of having two out of 43 nonwinning numbers is 43! 43 = 903. The number of ways of having four out of the six winning numbers is 6 C4 = 6! = 15.491.
and have the dream that one day the big prize will be theirs." There are ways to contribute to good causes while having a "flutter" that are more cash efficient. In how many different ways can the team be chosen? 4. A darts club has 10 of its members who are better than all the other members and are all equally proficient. one at a time. A bag contains six balls all of different colors. green.2. green. Some of it is returned to the prize-winners. Three balls are taken out of the bag. Problems 4 4. and blue in that order? The three selected balls are returned to the bag and then a handful of three balls is taken out. some provides a profit for the company that organizes and runs the lottery and the remainder goes to "good causes.44 Everyday Probability and Statistics The money staked on the national lottery is used for three purposes. Four winning numbers are selected from a drum-full of numbered balls plus a bonus ball. In how many different orders can the children leave? 4. performed on television with flashing lights and theatrical music. the second prize is for three correct numbers plus the bonus ball and the third prize is for three correct numbers. and blue? 4.4. but there is no doubt that many enjoy the anticipation of the periodic drawing of the winning balls one-by-one from a drum. There are only 20 numbers from which to pick. A town decides to run a local lottery based on the same model as the national lottery. and only four numbers are picked. Five children are in a room and leave in single file. 1 to 20. What is the probability that the three balls are red.3. What are the probabilities of winning each of these prizes with a single entry in the lottery? . What is the probability that the balls selected will be red. The top prize is for four correct numbers.1. It is required to select a team of four from these ten to represent the club in a tournament and the selection is done by drawing lots.
1. One way of thinking about it is to consider that one of them tells you his birthday.5 and that of getting a 6 when throwing a die is 1/6 but sometimes intuition can lead you astray. The Birthday Problem Ignoring the complication of leap-days there are 365 days on which a birthday can occur. (Henry Wadsworth Longfellow.Chapter 5 Non-Intuitive Examples of Probability And things are not what they seem. 5. our intuition tells us that the probability of this must be very close to 1 — indeed we should be astonished if no two people shared a birthday with such numbers. to most people. then while it is not certain that at least one pair share a birthday. clearly. Then you ask the other one for his 45 . If the number of people were reduced further then. 1807–1882) We have already mentioned that it is intuitively obvious that the probability of obtaining a head when flipping an unbiased coin is 0. the probability would reduce but would still remain high down to 200 people or so. Here. Let us now start at the other end with two people in a room. we shall deal with three situations where. The probability that they have the same birthday is low and is easy to calculate. the probabilities of particular outcomes are not-at-all obvious. If the number of people were reduced to 365 or 364. A moment of thought tells us that if there are 366 people in a room then it is certain (probability = 1) that at least one pair of them must share a birthday.
the second has only 364 possibilities so the answer is 365 × 364. 365 365 (5.2) the answer obtained previously. Hence the probability of the birthdays being the same is 1/365. Now consider three people in a room. The . 365 × 365 365 (5. The number of ways they can have birthdays. the obvious one to ask — "In how many ways can the two people have different birthdays?". the first one has 365 possibilities but once his birthday is fixed then. we can find the probability that they have different birthdays as pdiff = = Number of ways of having different birthdays Total number of ways of having birthdays 364 365 × 364 = . The other possible way to look at this problem is to first ask the question "In how many possible ways can two people have birthdays without any restriction?". Now. without any restriction. Now. or psame = 1 − pdiff = 1 − 1 364 = .46 Everyday Probability and Statistics birthday. There are 365 possible answers he can give but only one of them will match the birthday given by the first person. is 365 × 365 × 365. perhaps. we are going to ask a question that is not. Since there are 365 possibilities for each of them then the answer to this is 365 × 365 — every date in the calendar for one combined with every date in the calendar for the other. for the birthdays to be different. The reason we have introduced this apparently convoluted way of getting the answer is because it better lends itself to dealing with the problem of when there are three or more people in the room.1) Now. either the two individuals have the same birthday or they do not and these are the only possibilities and they are mutually exclusive so that pdiff + psame = 1. Well.
027136. 365 × 365 × 365 × 365 × 365 (5. All birthdays being different is a single outcome. three with the same birthday with the other two different. and many other combinations. is now evident. for the second to be different from the first there are 364 ways and for the third to be different from the other two there are 363 ways. We would have to consider one pair the same and the others all different. two pairs with the same birthday.9917958. (5. Hence the probability that they have different birthdays is pdiff = = number of ways birthdays can be different number of ways they can have birthdays 365 × 364 × 363 = 0. Suppose that. .3) If the birthdays are not all different then at least two of them must be the same and the probability of this is 3 pnot diff = 1 − pdiff = 0.5) The advantage of calculating the probability in the way that we do. we tried to directly find the probability that two or more people in a set of five shared a birthday.Non-Intuitive Examples of Probability 47 number of ways they can have different birthdays is found by noting that the first person can have a birthday in 365 ways.4) The prefix 3 before the p is to indicate that the answer is for three people. the probability of which is comparatively easy to find. by finding the probability that all birthdays are different and then subtracting from 1. We should now be able to see by extending the principle that the probability that at least two people out of five people in a room would share a birthday is 5 pnot diff =1− 365 × 364 × 363 × 362 × 361 = 0. 365 × 365 × 365 (5.0082042. Hence the total number of ways they can have different birthdays is 365 × 364 × 363.
443688 0. Table 5. we ask the question that is really the basis of this section — "How many people must there be in a room for there to be a probability greater than 0.5 and hence when there is a probability greater than 0. 5.5. we give the probability of having at least two people with the same birthday with increasing n.194410 0.002740 0.379119 0.5 that two or more people have the same birthday.1.027136 0.411438 0.141141 0. 5.223103 0. 365 364 363 362 361 360 359 × × ··· × × × × × 365 365 365 365 365 365 365 The value of n for which the answer first falls below 0. We see that the probability just exceeds 0.5 is when the probability of all birthdays being different is less than 0.056236 0.016356 0. What is your intuitive assessment of n at this point? In Table 5. We work out the following product term-by-term until the answer we get is less than 0.283604 0.1.538344 n 4 7 10 13 16 19 22 25 Probability 0.094624 0.252901 0.475695 0. There is greater than 50% chance that two of these will share a birthday! .167025 0.040462 0.5 that at least two of them would share a birthday?". This can be done with a calculator. A visual impression of what this result gives is illustrated in Fig.116948 0.1. This is the number of people that fulfils our requirement.1. n 2 5 8 11 14 17 20 23 Probability 0.074335 0.5 when 23 individuals are present — a result that most people find surprisingly low.346911 0.568700 Fig.507297 n 3 6 9 12 15 18 21 24 Probability 0.48 Everyday Probability and Statistics Now.008204 0. The probability that two or more people have the same birthday with n people present.315008 0.
Non-Intuitive Examples of Probability 49 For someone teaching a class of 30 students the probability that two or more will share a birthday is 0. A set of three Crown and Anchor dice. for 40 students 0. However.2). He explains that since the probability of getting a 6 with one die is 1/6 then the probability of getting a 6 when throwing three dice is 3/6 = 1/2. the essence of the game is preserved if the dice are normal ones with the numbers 1 to 6 on their faces and we shall consider them so. It was played with three dice each of which had on its six faces a crown.2.706316.2. the banker who runs the game and the thrower who rolls the dice. 5. The banker first justifies the fairness of the game to his intended non-mathematical victim. who is the thrower-to-be. 5. We start by considering a game which is not Crown and Anchor. . that is complete nonsense since adding probabilities is only justified when the outcomes are mutually exclusive — but the outcomes from the three dice are actually independent since getting a 6 on one die has no influence on what happens with the other two Fig. an anchor and the symbols for the card suits — club.891232.970374. 5. diamond. Crown and Anchor Crown and Anchor is a dice game that was once popular with sailors in the British navy. heart. Of course. There are two players. and for 50 students 0. and spade (Fig.
as seen in the final column of Table 5.50 Everyday Probability and Statistics dice. he must just be having an unlucky day. is played 216 times with a unit stake then on 125 occasions the banker wins and on the remaining 91 occasions the thrower wins giving a net loss of 34 stake units. The first thing we must determine is what outcomes are possible from throwing three dice. our thrower-to-be has not read this book.2. Is he really unlucky or has he been duped? Let us analyze the game. However. in which the thrower is not rewarded for multiple 6s. the probabilities of the outcomes listed above are given in the second column of Table 5. albeit a little slower than previously. Representing not-a-6 by the symbol 0 these possibilities for the three dice are 000 600 060 006 660 606 066 666 The probability of getting a 6 with any die is 1/6 and the probability of getting not-a-6 is 5/6. With a show of reluctance the banker agrees and they continue to play what is now truly the game of Crown and Anchor. The thrower still loses.2 by multiplying probabilities. accepts the argument and throws on the basis that if he gets a 6 the banker pays him £1 and if he fails to get a 6 then he will pay the banker £1. but he concludes that since the game is now fair. However. He remonstrates with the banker and asks to be paid £2 if he gets two 6s and £3 if he gets three 6s. Sometimes he throws two 6s and once in a while three 6s but he still gets paid just £1. he still has a loss of . and being played under the rules he demanded. It will be found that the sum of the probabilities in the second column add to 1 as indeed they must since the mutually exclusive outcomes listed are the totality of possible outcomes. when the game is played according to the rules of Crown and Anchor the thrower receives two stake units when he throws two 6s and three stake units when he throws three 6s but. This is shown in the third column of the table. The thrower loses heavily and it gradually dawns on him that something is wrong. Since the outcomes from the three dice are independent. When the first game.
Non-Intuitive Examples of Probability Table 5. 108.e. The thrower still wins 108 units. Now let us suppose that when 6s do appear they only appear singly so that two 6s and three 6s never happen. one sixth of which.. one unit for each appearance of a 6. Now imagine that two of the expected 108 6s do appear together and the other 106 6s appear singly. In this case the thrower will win 108 times — each time a 6 appears — and the banker will win 108 times — each time a six does not appear — so there will be no profit or loss on either side. Outcome 5 6 1 6 5 6 5 6 1 6 1 6 5 6 1 6 Probability 5 6 5 6 1 6 5 6 1 6 5 6 1 6 1 6 5 6 5 6 5 6 1 6 5 6 1 6 1 6 1 6 125 216 25 216 25 216 25 216 5 216 5 216 5 216 1 216 Profit (loss) on first game (−125) 25 25 25 5 5 5 1 34 Profit (loss) on C&A (−125) 25 25 25 10 10 10 3 17 51 000 600 060 006 660 606 066 666 × × × × × × × × × × × × × × × × = = = = = = = = Net loss 17 stake units. are expected to be 6s.2. but . The game seemed fair — why is the thrower still losing? If three dice are thrown 216 times then altogether 3 × 216 = 648 faces are seen. The probabilities for various outcomes of throwing three dice and the expected profits and losses for the thrower for each outcome playing the first game 216 times and then playing Crown and Anchor 216 times. i.
and C. The banker is now one unit in profit. The variation from what has been described is that the thrower could choose the symbol on which he was staking — for example. We now describe one such game. the anchor — but that made no difference to the probabilities involved. All is not what it seems in Crown and Anchor. 5. In the early years of the 20th century. the conditions are created for extra gains by the banker. bankers would set up street games of Crown and Anchor in London and other big cities. This is the secret of the game — every time that 6s appear twice or three times. no doubt welcomed jubilantly by the thrower. To Switch or Not to Switch — That is the Question Television game shows in which contestants select boxes containing various rewards are very popular. B. Pick the right box and a fortune is yours. Now. Crown and Anchor is a game so heavily biased against the thrower that it is tantamount to robbing the innocent and so should be banned even under present more liberal legislation. The contestant is offered three boxes which we shall indicate as A.3. and perhaps earlier.000 while the other two boxes each contains a sausage. The games never involve the single simple act of selecting a box but are usually accompanied by a set of intermediate actions or decisions before the box is finally opened. he archly asks the contestant whether he would like to stick with his original box or to switch to . In one of the boxes is £1. The contestant picks box A but does not open it.52 Everyday Probability and Statistics he has won this on 107 throws so now there are 109 occasions when there is no 6. This was an illegal activity since the only form of gambling that was legal at that time was betting on horses or greyhounds at registered tracks and in registered casinos and clubs. pick the wrong box and the prize is derisory or even something unpleasant. The game-show host then opens one of the other two boxes which is seen to contain a sausage.
First. then he will win the money. our contestant will have picked the other sausage and so by switching he has lost out.Non-Intuitive Examples of Probability 53 the other unopened box. he has made the wrong choice. The three possible arrangements of money and sausages. and one of them contains the money.. whichever sausage the host decides to display. let us assume that the contestant initially chooses box A. A surprising result — but let us see how we get to it. the probability of the money being in each unopened box is 0. if the arrangement is one of the other two he will find a sausage in his box. 5.3. If the arrangement is the top one in Fig. Now. 1/3. 5. Next. His chance of winning is actually doubled by deciding to change to the other unopened box. The contestant is suspicious of this offer. He concludes that since there are two boxes unopened. each with probability 1/3.3. in fact. 5. . i. The argument seems impeccable but.3. But for the Box Probability A B C 1/3 1/3 1/3 Fig.e. If the arrangement is the top one then. we suppose that the contestant decides not to switch boxes.5 so he might just as well stick with his original choice. However. The critical factor in this game is that the host knows where the money is and will always open a box containing a sausage. we assume that he always decides to switch boxes. His chance of winning is therefore one in three. The possible content of the boxes in terms of money and sausages is as shown in Fig.
Construct a table similar to Table 5. In the terminology of Table 5.1. 0 6.2 showing what the expected profit is for the banker from 36 games for both the first game and the two-dice Crown-and-Anchor game. will have different birthdays? (ii) What is the probability that four Arretians. the switch will give the contestant box B and he will win the money. just double what it was if he did not switch.2. will have different birthdays? (iii) "How many Arretians must there be in a room for there to be a probability greater than 0. He always has a better chance of winning the money if he decides to switch boxes. chosen at random. the result we have found is quite independent of which box the contestant originally chose. B. Since all the boxes are effectively equivalent — it is arbitrary which are chosen as A.54 Everyday Probability and Statistics middle arrangement the host will have opened box C. Problems 5 5. and C.2 the possible outcomes are 0 0. so his probability of winning is 2/3. chosen at random. A surprising result you will agree. .5 that at least two of them would share a birthday?" 5. for the bottom arrangement the host will have opened box B. For two out of the three possible arrangements he wins. our contestant will have switched to box C and again he wins the money. On the planet Arret there are 100 days in a year. 6 0. Likewise. A game of Crown and Anchor is played with two dice. and 6 6. (i) What is the probability that two Arretians.
such as the training.Chapter 6 Probability and Health If you believe the doctors.1. other factors come into play. for example.1. (Lord Salisbury. Of course.4 and in Sec. nothing is wholesome. skill. and experience of the doctor or surgeon and also the characteristics of the patient — age and robustness. When the computer diagnoses for four of these systems were tested against the diagnoses produced by a group of highly experienced and competent doctors for particular sets of symptoms. In view of all these factors there is some element in diagnosis and treatment that is akin to what was called judgmental probability in Sec. Trying to introduce some certainty in diagnosis. 1830–1903) 6. Bodies such as national Medical Councils and the World Health Organization collect and collate information from different sources and so derive empirical fatality rates for various diseases and success rates for different clinical treatments and surgical procedures. 3.1. Finding the Best Treatment Medicine is a field of activity in which the type of probability of interest is empirical probability. in each individual case where diagnosis and treatment is required. the programs gave a correct diagnosis in between one-half and three-quarters of cases. Where there were several possible conditions suggested by the symptoms then the programs gave less than one-half of the possibilities suggested 55 . as described in problem 1. various computer-based diagnostic systems have been devised. 3.
1). as we have already indicated. I'm not really sure what's wrong with you but I know the best way to treat it. that is not always so. Sometimes symptoms may not point uniquely to a particular medical condition but may indicate one or other of a finite number of possible conditions with different probabilities of being present (Fig. more often than not when we visit a doctor with a medical condition the doctor will recognize the problem and know the treatment that will best deal with it.1. We shall consider a hypothetical case of an uncertain diagnosis indicating three possible conditions. on average the programs suggested two additional diagnoses per case that the experts had not themselves proposed but which they thought were relevant. and C with probabilities A 0. since the symptoms are similar. There is a range of drugs available for treating these conditions and. 6. 6.20 C 0. By contrast. Fig. . A. the drugs will probably all have some efficacy for each of the possible conditions. B. But. Despite all the uncertainties of diagnosis.10 These probabilities will be based on a considerable body of information gathered by doctors and medical bodies and published in research journals.70 B 0.56 Everyday Probability and Statistics by the experts.
The outcomes. Drug a A 0. a.4 Drug b A 0. which have the following observed probabilities of dealing successfully with each of the conditions. and c.5 From the probabilities of the conditions it is clear that A is most likely to be present and from the effectiveness of the drugs it is clear that c is the best for treating condition A so should we conclude that.9 = 645. but numerically helpful. of the 200 patients with condition B.2 C 0. Hence the total number of recovered patients would be 420 + 200 + 40 = 660.0 = 200 will recover. a number 200 × 1.4 = 40 will recover. Of the 1. a number 700 × 0.75 B 0. C 100 .6 B 1.5 C 0.9 Drug c A 0.65 + 200 × 0. from the point of view of the patient the best treatment is to administer drug c? The answer is no — it is actually the worst decision to make! To see what the doctor should prescribe let us make the absurd. b. a number 100 × 0.000 patients.Probability and Health 57 We shall assume that there are three drugs available. and of the 100 patients with condition C. We now repeat this exercise for the other two drugs.6 = 420 will recover.65 B 0. assumption that he has 1. with the results for drug a repeated are For drug b the number recovering is 700 × 0.0 C 0. we would expect the following numbers with the three conditions A 700 B 200 If he uses drug a then of the 700 patients with condition A.5 + 100 × 0.000 patients with this uncertain condition and then work out the likely outcome of the three drug treatments.
75 + 200 × 0. there are many advantages in the use of synthetic drugs rather than natural products. branches of medicine have evolved over many centuries that involve the use of natural products. Again. it is necessary to consider the outcomes of all possible combinations of situation and reaction to optimize the chances of success. may depend on the weather conditions during the growing season. the dosage can be more accurately controlled.. conditions A. Testing Drugs In non-Western societies. The beneficial effect of many drugs is obtained by their action on particular proteins in the patient and drugs can be designed so that they . b. Another modern aspect of the pharmaceutical industry is that of drug design. B. Here. 6..58 Everyday Probability and Statistics For drug c the number recovering is 700 × 0. and c). turns out to be the one that is least effective for the most likely condition. like those described in Chap. and different possible reactions to those situations (i.5 = 615. 5 where the answer obtained from probability theory may be at variance with one's intuition. but it is far more effective for the other two conditions. applying drugs a. A. we have met another situation. origin.2 + 100 × 0. First. drug a. but sometimes of animal. The best drug to use. notably in China and India. the concentration of the active ingredient in a natural product. by extracting and analyzing the structure of the active ingredient in a plant. and C) with different probabilities. What we learn from this is that in cases where there are many possible mutually exclusive situations (i.2.e. However.e. with different probabilities of having a successful outcome. Many of these treatments are very effective and provide the basis for the synthetic drugs used in Western medicine. usually of vegetable. a drug company may be able to produce derivatives that are either more effective than the natural product or have less severe side effects — or both. say in the root of a plant.
a proper assessment should be made of the effectiveness of the new procedure. say a guinea-pig or a mouse. a harmless and inactive substance that in appearance resembles the drug. Tests on mice and primates had shown beneficial medical results without side effects. The patients themselves do not know whether they received the drug or the placebo so we can rule out any psychologically induced physiological effects.Probability and Health 59 bind themselves in specific ways to the protein and so either inhibit or increase its activity. although with much more stringent safeguards than hitherto. but sometimes it is not. We consider the hypothetical trial of a drug which is administered to 10 patients suffering from a particular disease. six men given the drug in a trial all became seriously ill. Occasionally. with failure of various organs and permanent damage to health. and fortunately rarely. However. so at the end of the day the only certain way to determine the effect on human patients is to test it on human patients. Of these patients 5 recover but 15 do not. Another group of 20 patients with the same disease are given a placebo. Of the 10 patients 7 of them recover but 3 get no benefit from the treatment. Many drugs are tested on animals where the assumption is that the reaction of the animal. to the drug will be similar to that of a human patient. It is important that when such trials are carried out. The theory behind these assessments is fairly difficult but we shall describe how the application of a few simple rules enables a decision to be made regarding the significance of the result of a clinical trial. as happened with the test of a drug in the United Kingdom in March 2006. Is there any evidence that the new drug is effective? To decide on this we are going to describe the steps in what statisticians call the χ2 -test (χ is a Greek letter written as "chi" but pronounced as "kye" so this is the chi-squared test). While this was a setback for testing drugs on humans it is certain that such tests will continue. Often it is. such tests can go badly wrong. .
for the number of patients given the placebo the expected number recovering and not recovering would be 8 and 12. Step 2 We make a hypothesis (called the null hypothesis) that the drug is ineffective. respectively.e. .4. a fraction 0. The totals given at the bottom of the columns are the total numbers of recovered and non-recovered patients. on the basis of the overall proportion of recoveries. Overall 12 out of 30 patients.. Based on the null hypothesis we now produce a table E (for expected) giving the expected numbers of recovered and unrecovered patients for those given the drug (now assumed ineffective) and those not given the drug. Similarly.60 Everyday Probability and Statistics Step 1 Represent the results of the trial in a table O (for observation) as shown below Recover Not recover Drug taken 7 3 10 Table O Placebo taken 5 15 20 12 18 The totals given at the right-hand side of the rows are the total numbers of patients given the drug and given the placebo. Notice that when the expected numbers are entered in the table the sums of rows and columns are the same as they were previously. recovered and if the null hypothesis is valid then this is the proportion of patients that recover spontaneously without any treatment. we would expect 4 to recover and 6 not to recover. of the 10 patients given the drug. i. respectively. Thus.
. part of which is shown in Table 6. was almost certainly untrue.1) where i = 1 to 4 for the four entries in the table and Oi and Ei are the entries in the O and E tables.625. even if the drug is completely ineffective the O an E tables would probably be different just because of random fluctuations. As an example. 4 6 8 12 (6. respectively.1.3). i=1 this value is χ2 = (7 − 4)2 (3 − 6)2 (5 − 8)2 (15 − 12)2 + + + = 5. if you obtained 90 heads and 10 tails you might suspect that your null hypothesis. (3. if you flipped a coin 100 times it is unlikely that there would be exactly 50 heads and 50 tails.Probability and Health Recover Not recover 10 61 Drug taken 4 6 Table E Placebo taken 8 12 20 12 18 Step 3 Now. Now. To do this we calculate the value of 4 χ = i=1 2 (Oi − Ei )2 Ei (6. that the coin was unbiased.2) Step 4 The value of χ2 is compared with the numbers in a standard table. However. Remembering the explanation of 4 given in relation to Eq. we see how to test the differences between the O and E tables to see whether they are so different that we might suspect that our null hypothesis was untrue.
Everyday Probability and Statistics
Table 6.1. Probabilities associated with the χ2 distribution for one degree of freedom. Probability χ2 0.20 1.642 0.10 2.706 0.05 3.841 0.025 5.024 0.02 5.412 0.01 6.635 0.005 7.879 0.001 10.827
Let us imagine that our value of χ2 had been 3.841. The table would tell us that if the null hypothesis is true, and the drug is ineffective, then the probability of getting this value of χ2 , or some greater value, is 0.05, which is 1 in 20. What significance you place on that value is a subjective choice that depends on the kind of test being made. If the test related to some commercial decision, e.g., testing consumer reaction to some new form of packaging, then possibly 0.05 would be regarded as sufficiently low to indicate that the null hypothesis, i.e., that the new packaging was ineffective, was probably untrue. In medicine the decision barrier is usually much more stringent and if the medical condition is a matter of life and death then it would be extremely stringent. The value of χ2 found in the example we have used indicates a probability just under 0.02 that the result could be obtained with the null hypothesis but this may not be small enough to pronounce that the treatment is effective and so to sanction its use. It is the usual practice to decide on the level of significance that would lead to acceptance before doing the test. If that is not done then there is always the danger that, with the result of the test already being known, a significance level will be set so as to achieve a desired outcome. Let us suppose that a significance level of 0.01 had previously been set as a criterion for judging whether or not the new drug should be adopted for general use. This means that the value of χ2 must be such that there is a probability less than 0.01 that the result of the test, or something even further from expectation, could have been obtained just by chance. The test described above would not have met the conditions for adoption but now let us suppose that
Probability and Health
the test had been carried out on a larger scale and that the observed and expected tables had been as shown below.
Recover Not Recover
Drug taken
Table O
Placebo taken
Not Recover
Drug taken
Table E
Placebo taken
These tables are just the ones we considered previously scaled up by a factor of 2, and now we have χ2 = (14 − 8)2 (6 − 12)2 (10 − 16)2 (30 − 24)2 + + + = 11.25. (6.3) 8 12 16 24
From Table 6.1, it is clear that there is less than 1 chance in 1,000 that the result of the test, or some more extreme result, could come about by chance if the null hypothesis was true. The test satisfies the criterion for adoption of the drug and its efficacy is confirmed beyond all reasonable doubt. This shows the importance of scale in carrying out significance tests. The χ2 test can be applied in many different contexts and we have just given one example here. In the legend to Table 6.1, there is a reference to "one degree of freedom". The concept of "degrees of freedom" is very important in statistics and is a measure of how
Everyday Probability and Statistics
many independent quantities there are in the system of interest. We noticed that although the O and E tables had different entries the totals of both rows and columns are the same. Thus, if we were told that 20 people had been given the drug and 40 given the placebo and that 24 patients had recovered but that 36 had not, then that would not enable us to construct a table. Both the O and E tables satisfy the information given and many other tables can be constructed that do so. In this context, one degree of freedom means that, given the totals of rows and columns, once a single entry is made in the table then all the other entries follow. We shall meet the concept of degrees of freedom again: there are other applications of the χ2 test, which will be described in Chap. 8, that involve many degrees of freedom. While it is a difficult concept to understand fully, fortunately, in the contexts in which it occurs, it is usually simple to find out how many degrees of freedom there are.
Problems 6
6.1. The symptoms for a patent indicate one of two conditions — A with a probability of 0.65 or B with a probability of 0.35. Two drugs are available, both of which have partial success in dealing with both conditions. Their effectiveness is summarized in the following table which gives the success rate in giving a cure.
Condition Drug a Drug b
To maximize the probability of obtaining a cure, which drug should the doctor prescribe?
Probability and Health
6.2. A chain store decides to test the effectiveness of a new form of packaging for one of its products. In two of its stores, in similar environments and with similar sales profiles, it arranges similar displays of the product. Of the first 100 customers that enter store A, 8 buy the product with the old packaging. Of the first 100 customers that enter store B, 20 buy the product with the new packaging. Calculate the value of χ2 . The null hypothesis is that the new packaging is ineffective. If the probability that the value of χ2 , or some larger value, could be obtained with the null hypothesis just by chance is greater than 0.05, then the store will not bother to change the packaging. Do they change it or not?
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5. If it goes into either of the two cups flanking the center one then the 10p is returned.1 Whenever the ball reaches a junction point. A Simple Probability Machine We consider a simple probability machine where a ball drops out of a hole and falls along a set of channels arranged in a pattern as shown in Fig. The Craps Game Revealed Un coup de dés jamais n'abolira le hazard. It can go left or right and if it goes left then it will reach the left-hand junction of the second row — the one containing three junctions.1. indicated by the larger circles. it has an equal probability of taking the left-hand or righthand path to the next line of junctions. 1842–1898) 7. a throw of a dice will never eliminate chance (Stéphane Mallarmé. are a set of cups. If it enters an outer cup then 30p is returned.Chapter 7 Combining Probabilities. 7. To start a ball on its journey a 10p coin must be inserted in the machine. Now consider a ball that reached the left-hand junction. shown by a small circle. For this to happen two independent events must take place — it must have gone left at two junctions with a probability 67 . into one of which the ball falls. If the ball enters the center cup the money is lost. At the bottom. What are the chances of winning or losing at this game? On leaving the first junction the ball has an equal probability of going to the left or right. so the probabilities of arriving at each of these junctions is 0.
7. Let us see what his chances really are.5. .25 0.25.5 = 0. each of which has a probability 0. To get to the middle junction of the row requires either of the combinations left–right or right–left.25 0.68 Everyday Probability and Statistics Falling ball 0. i. of getting his money returned is 0.0625 0. they add to unity since the ball must arrive somewhere on that row. Similar reasoning gives the probabilities marked along the third row of junctions and along the cups.125 0.25.5 = 1/2.375 0.625 of the occasions he either gets his money back or makes a handsome profit.1.125 0.25 = 0. 0. The pinball machine.375 0. These three probabilities are marked along the second row of junctions and. The probability of losing is 0.25+0. By symmetry this is also the probability of reaching the right-hand junction of that row.5 0.25 + 0.5 × 0.0625 Fig. He loses his money on 0. Since the combinations left–right and right–left are mutually exclusive the probability of reaching the middle junction is obtained by adding the two probabilities.375 0.375 ( = 3/8).25 0.5 that it will do so each time. It will be seen that the game is superficially attractive to the player. 0.5 × 0.375 ( = 3/8) of the occasions he plays but on the other 0.5 0. Hence the probability that it will get there is 0. of course.25 = 0.e.25 0. as marked in the figure.5 = 0.5 0..
5 = 0. is independent of the previous ones so the probability of each path is 0.0625 = 0.0625 but since the routes are mutually exclusive the total probability of reaching the central cup is 6 × 0. Of course. . This is an example where we have considered combinations of probabilities. some independent and some mutually exclusive. and 30p on one game.0625 = 0. if his luck is in and he quits when he is ahead then he can win something but if he plays the game a sufficiently long time he is almost bound to lose.5 × 0.Combining Probabilities. However.5 × 0. his return will be zero on three games. The Craps Game Revealed 69 and of getting three times his stake returned is 0.1. Probability machines that are much more complicated can be designed with many more rows of junctions and with the junctions. we can consider the ways the ball can get to the center cup in Fig. Representing a left deflection by L and a right deflection by R there are six mutually exclusive routes the ball can take LLRR LRLR LRRL RRLL RLRL RLLR Each step. 10p on four games. he inserted 80p to play the eight games so he is a net loser.125(= 1/8).0625 + 0. However.375. As an example. no matter how complicated the probability machine the chances of ending in each of the cups can be calculated using the principles explained above. so designed that the chances of going left or right are not equal.5 × 0. to left or right. So the expectation is that if he plays eight games. 7. which may be a simple pin off which the ball bounces. a return of 70p in all.
When all the players have received the cards they requested the dealer reveals his two cards and adds to them until he wishes to stop. according to which best serves his requirements. Another good combination is a "five-card trick" which is five cards totaling less than 21. The players then request further cards. one at a time. count for 10 points while an ace has a degree of flexibility and may be counted by the player as either 1 point or 11 points. If 21 points is obtained with two cards. When the player wishes to stop receiving more cards he says "stick" or "stand".. then that is called "pontoon" and will beat any other hand. up to a maximum of three further cards. 2 + 6 + 8.2. If he does not go bust then he pays out to all those with a higher number of points but takes the stakes of all those with less than or equal to his number of points. called pontoon in the United Kingdom as an extremely corrupted form of the French name vingt-et-un. The above is a basic description but there are extra aspects. A dealer initially deals two cards to each player and to himself. then he will lose that hand. If he goes bust then he pays all players who did not go bust an amount equal to that they staked. For example. is called blackjack in the United States and is a popular casino gambling game. The dealer's advantage . and then requests a further card that turns out to be a six or greater. Cards are associated with a number of points. Court cards.70 Everyday Probability and Statistics 7. a Jack. or King. giving a total of 16 points. he loses that hand and forfeits his stake money to the dealer. If his total exceeds 21 he goes "bust". which must be an ace plus a 10 or a court card. with the cards 2 to 10 giving the number they display. Thus there is a measure of judgement to be exercised — although luck also plays a major role. Queen. Pontoon — A Card Game This game. if the player has three cards. i. by saying "twist" or "hit" with the objective of getting as close as possible to 21 without exceeding 21.e. which beats all other hands except a pontoon. What has been given is not a complete description of the game but suffices for present purposes.
g.02413.00603 = 0. These four combinations are mutually exclusive so the total probability of getting 19 points with two cards is P(19) = 0.02413 = 0. Of course. e. 9) = 16 4 × = 0.02413 + 0. second.02413 + 0.Combining Probabilities.00603 The last two ways of getting 19 involve having an ace and 8 either way round. For example. One way is to get 10 points + 9 points. although not their total of points. The Craps Game Revealed 71 in this game is first that if he has a hand equal in value to that of another player then he wins and. A complete statistical analysis of pontoon is possible but complicated — just think of the number of different ways of achieving a five-card trick.02413 = 0.00603 = 0. 16 52 4 P(9. the number of cards he dealt them. 11) = 52 P(10. or about one chance in 16. that he knows something about what the other players have. 9) = 4 51 16 × 51 4 × 51 4 × 51 × = 0. Hence the probability of getting a 9 as a second card is 4/51.00603 + 0. we consider the probability of getting 19 with two cards. one may get 19 with three or four cards so the total probability of getting 19 in a hand of pontoon . There are some hands for which it is fairly simple to estimate the probabilities. 52 51 The probabilities of this and other selections are now listed.. Now there are 51 cards left in the pack of which 4 are 9s. The selections are independent so the chance of getting first 10 points and then 9 is P(10. 10) = 52 4 P(11. 8) = 52 4 P(8. The 10 may be obtained with a 10 or a court card and the probability of getting one of these from the pack is 16/52.06032.
to 3/32 = 0.2). Two ways of losing playing American craps! . 7. As a simple example. or 12 to lose.2. In the casino game of pontoon. Casino managers are always on the lookout for card counters and they are excluded from casinos when they are detected. 7.09375. slant the odds in his favor to the extent of wiping out the dealer's advantage and substituting a player's advantage.07692. The game can be won or lost in a single throw — getting a "natural" to win or by getting 2 (Fig. If 19 is obtained with five cards then that is a five-card trick and is a stronger hand than just getting 19 points. the cards are peeled one-by-one from a deck of cards placed face downwards. 3. The Throwers Chance of Winning at American Craps There are many different routes by which the thrower may win.72 Everyday Probability and Statistics is higher than this. there are more complicated Fig. the probability before any cards are dealt. in the hands of a clever gambler.3. or lose. or blackjack. if the first 20 cards dealt from the pack contained one ace then the probability that the next card will be an ace is increased from 4/52 = 0. However. There are professional gamblers who "count the cards" — that is they mentally record aspects of the cards they have already seen so as to reassess the probabilities for future cards. Such information can. 7. in the American craps dice game.
we have. the probability of obtaining 4 was found as 1/12. 2.3) . 12 3 36 (7.Combining Probabilities. Throwing a "natural" The probability of this was found in Sec. or getting 7 to lose with a probability of 1/6 on each throw. they are mutually exclusive so we need to identify all the routes. (7. The chance of getting a 4 before getting a 7 is probability of 4 = probability of 4 + probability of 7 1 12 1 12 2 9 = + 1 6 1 = . We shall find the combined probability of all the winning ways. Winning by this route has two independent components — first establishing a "point" of 4 and subsequently getting a winning throw of 4. 2. 3 (7.1) Similarly the probability of getting a 7 before getting a 4 is probability of 7 = probability of 4 + probability of 7 1 6 1 12 + 1 6 2 = . Setting a "point" of 4 (or 10) and then winning In Sec.2) Since the probabilities are the same for a total of 10. 3 but we are not directly interested in this probability because we are finding the total probability of winning. The only throws relevant to winning or losing after setting this point are getting 4 again to win.22222. The composite probability of this is P4 = 1 1 1 × = . P10 = P4 .5 as Pnatural = 0. Having obtained a 4 the thrower now embarks on a series of throws. find their probabilities and add them together. with a probability of 1/12 on each throw. The Craps Game Revealed 73 ways of winning and losing.5.
5 (7.5) Since the probabilities are the same for a total of 9. 45 9 5 (7. with a probability of 5/36 on each throw. with a probability of 1/9 on each throw.8) . or getting 7 to lose with a probability of 1/6 on each throw. the probability of obtaining 6 was found as 5/36.7) The composite probability of first establishing a "point" of 6 and subsequently getting a winning throw of 6 is P6 = 5 25 5 × = . Setting a "point" of 6 (or 8) and then winning In Sec. 2. or getting 7 to lose with a probability of 1/6 on each throw.5. For the relevant throws the chance of winning is probability of 6 = probability of 6 + probability of 7 5 36 5 36 (7. The only throws relevant to winning or losing after setting this point are getting 6 again to win. 36 11 396 (7. the probability of obtaining 5 was found as 1/9.5. we have. For the relevant throws the chance of winning is probability of 5 = probability of 5 + probability of 7 1 9 1 9 + 1 6 2 = . The only throws relevant to winning or losing after setting this point are getting 5 again to win. 11 (7. The composite probability of this is P5 = 2 1 2 × = . 2.4) Winning by this route has two independent components — first establishing a "point" of 5 and subsequently getting a winning throw of 5.74 Everyday Probability and Statistics Setting a "point" of 5 (or 9) and then winning In Sec. P9 = P5 .6) + 1 6 = 5 .
The Craps Game Revealed 75 Since the probabilities are the same for a total of 8. Problems 7 7. Thus on average the bank receives £7 profit for every £500 invested — a return of 1.9) Summing the probabilities of these mutually exclusive routes for winning.1. and the thrower stake an equal amount and the winner takes all.1. For every £500 staked by the bank another £500 will be staked by the thrower who will. is illustrated below.4% per game which is small but almost assured over a long period. + + + + + 9 36 45 396 396 45 36 Thus the bank has a slight advantage — which in any profitable casino is a necessary condition for running any game. A probability pinball machine.Combining Probabilities. on average receive back £493. similar in principle to that described in Sec. we have. P8 = P6 .492929. A B C D E The ball has a probability of 1/3 of going along each of the channels shown. 7. As the game is played the bank. (7. What is the probability of landing in each of . the overall probability of winning for the thrower is Pwin = Pnatural + P4 + P5 + P6 + P8 + P9 + P10 = 1 1 2 25 25 2 2 + = 0.
2. Analyze the profit or loss of using the machine if it costs 10p per ball. no return was made on cup C. In a game of American craps what is the probability that the thrower loses by (i) Setting a point of 4 and then subsequently losing by throwing a 7? (ii) Setting a point and then subsequently losing by throwing a 7? .76 Everyday Probability and Statistics the cups marked by a letter symbol.3. 7. What is the probability of obtaining a points total of 20 with two cards in a game of pontoon? 7. the 10p was returned on cups B and D. and 20p was returned on cups A and E.
To see how to answer this question. we defined logical probability which is based on reasoning and is heavily influenced by ideas of symmetry.1. which depended on the neutron energy. 3. 1819–1861) 8. When it's so lucrative to cheat. During the Second World War scientists in the United States were working on the development of a nuclear bomb. (Arthur Hugh Clough. 60 times out of 100? After all the coin may not have two heads but may be so unbalanced. were known but this knowledge could not be converted into an analytical solution of the problem. A neutron interacting with an atomic nucleus in the barrier could be unaffected. The problem was eventually solved by two of the scientists working 77 . we first consider a completely different problem concerning outcome and expectation. and Crooked Wheels Thou shalt not steal. The Need to Test for Fairness In Sec. We know that something is wrong when a coin comes down heads 100 times in 100 spins. or biased.Chapter 8 The UK National Lottery. that the mechanics of spinning it leads to a preponderance of heads.1. 70 times out of 100. One problem they faced was to determine how neutrons would penetrate a barrier — a problem that they needed to solve quickly but without any previous experience to guide them. or scattered. The probabilities of these possible interactions. Loaded Dice. But at what level should we be concerned — 90 times out of 100. an empty feat. absorbed.
if scattered. for example — so a procedure of this kind is called a Monte–Carlo method. With the advent of computers that enable large numbers of Monte–Carlo trials to be made. These bonds yield an average interest at a predetermined rate but the interest is not distributed as income to each individual bond holder but is used to provide prizes for a lottery based on the number of the bonds. They could estimate how far a neutron would travel on average before encountering a nucleus so they simulated the path of a neutron through a barrier.78 Everyday Probability and Statistics on the project. the method has been used in many areas. absorbed. Premium Bonds are a UK national institution that relies on the generation of random numbers. This produces truly random numbers since it is based on completely unpredictable physical events but for some scientific purposes it would be unsuitable. The numbers of the bonds are generated each month by ERNIE. by a novel numerical procedure. the output of which is interpreted as numerical digits and letters. By following a large number of such simulated paths they were able to determine the probability that a neutron would penetrate the barrier. The generation of a random number is similar to what happens in various gambling activities — throwing a die or spinning a roulette wheel. or scattered and. it is important for the success of the method that the random numbers generated should be truly random. an acronym for Electronic Random Number Indicator Equipment. The present version of ERNIE uses the random noise produced in transistors to control an oscillator. However. Ulam and von Neumann. At each interaction they decided what would happen by the generation of a random number that would determine whether the atom was unaffected. For many scientific calculations truly random numbers are required but it is desirable that they should be generated in a reproducible way so that the calculation that produces them can be rerun with the same numbers if necessary. In this way something of the order of one million random bond numbers can be generated per hour. both in science and in the social sciences. and also how far it would travel to the next interaction. in which direction. For this reason much effort has been .
To get started. That is what happens within the outer bracket in Eq. 8.1b) where we multiply xn+1 by 10 to give 4.40523013. that is intended to produce a string of random digits. x2 . To continue the series from xn+1 one can generate xn+2 and yn+2 and so on. . represented by Frac in the equation. which gives xn+1 . represented by Int in the equation. Testing Random Numbers To illustrate the methods of testing for the fairness of a random number generator. called the seed.1a). is to take the fractional part of that number. Now. (8.0523013 and then take the whole number part.40523013.The UK National Lottery. it is necessary to provide an initial fractional number x0 . Producing such generators is a surprisingly difficult thing to do and to be confident in the performance of a generator one must be able to test its output in some way. A series of fractional numbers is produced of the form: x1 . 0. This number is added to the special number π = 3. we move down to Eq. x3 . to give the digit 4. yn+1 = Int(10 × xn+1 ).14159265 to give 3. . The next thing that we do.1b) What this means will now be described. . Loaded Dice. we shall consider a generator.65382961. As long as the same computer is used then the same .2. . which is yn+1 . from 0 to 9. for the random digit generator. (8. The generator we test will be of the following form: xn+1 = Frac{(xn + π)2 }. and Crooked Wheels 79 expended in creating pseudo-random-number generators that produce a series of random numbers by some repeatable mathematical process.1a) (8. of which xn is the nth one. Let this fractional number be 0. This number is then squared to give 14.79542226. A computer can readily generate a long string of digits in this way. (8.
2 the expected table E.1) except that we now have ten boxes rather than four. we calculate χ2 as we did in Eq. Inevitably.80 Everyday Probability and Statistics Table 8. 6. (6. What we need is an objective numerical measure to see if what have obtained could have come about with reasonable probability with a good random-number generator.1. If a different set of random digits is required then it is only necessary to change the seed. O. To do this. 10 χ2 = i=1 (Oi − Ei )2 (2 − 5)2 (8 − 5)2 (10 − 5)2 + + = Ei 5 5 5 + + (4 − 5)2 (2 − 5)2 (1 − 5)2 (7 − 5)2 (7 − 5)2 + + + + 5 5 5 5 5 (6 − 5)2 (3 − 5)2 + = 16. Table 8. and Table 8. (8. the numbers of each digit found will vary and.1 represents the observed table with observations. The average times of occurrence of a digit is five but there is considerable variation about that average. A simple computer program was used to generate 50 digits in this way and the numbers of each digit are shown in Table 8. Observed table for 50 digits produced by the random-digit generator described in Eq.1. indeed. Can we deduce just from looking at the table that it is an unsatisfactory random-digit generator? Not necessarily so. just by chance.4. 5 5 .2 to test the efficacy of a drug. it would have been surprising — even suspicious — if we had found a row of ten 5s. Now. Digit Number 0 2 1 8 2 10 3 4 4 2 5 1 6 7 7 7 8 6 9 3 Total 50 string of digits will be generated by the same seed. we are going back to the χ2 test that was described in Sec.1).
50 0.2.005 χ2 8. Comparing the value of χ2 with the entries of the table shows that there is only about a 1 in 20 chance that the value found for χ2 . Published tables give the value of χ2 for various numbers of degrees of freedom.242 14. Using the same random-digit generator. given the sums in rows and columns. Digit Number 0 109 1 99 2 123 3 91 4 99 5 83 6 89 7 103 8 98 9 106 Total 1000 . Loaded Dice. The expected table E for 50 digits with a uniform distribution. Digit Number 0 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 5 Total 50 81 However.02 0.01 0. (8. Probability 0.919 19.1 for our probabilities because we have a different number of degrees of freedom in the present case. it is just the sum of the row.666 23. we cannot use Table 6.3. and we show an excerpt for 9 degrees of freedom in Table 8.05 0.30 0. that is relevant.4. There are ten digit boxes and until you have filled in nine of them the contents of the table are not defined — so in this case we have nine degrees of freedom.4. Analysis of 1000 digits produced by the random-digit generator described in Eq.656 12. You will recall that.023 19.20 0.679 21. the distribution for 1000 digits is given in Table 8.025 0.The UK National Lottery.589 Table 8.1).343 10. 50. could come about just by chance. with only one row. In the present case. Probabilities associated with the χ2 distribution for nine degrees of freedom. Table 8. and Crooked Wheels Table 8.2. This indicates that the random-digit generator might be suspect. or some greater value.3. 6. in Sec.684 16.10 0. the number of degrees of freedom was found as the number of arbitrary entries you could make in the table.
82 Everyday Probability and Statistics Table 8. (8.5. 100 100 100 This gives a better indication that the random-digit generator may be reliable since there is now about a one in four chance of getting this value of χ2 or greater. (8.1b).52. The step described in Eq. The general principles described here for testing random-digit generators can also be used for testing the output of the various pieces of equipment used for gambling. Digit Number 0 99 1 94 2 98 3 104 4 92 5 85 6 102 7 101 8 115 9 110 Total 1000 The expected table would have 100 in each box so for this distribution the value of χ2 is (−1)2 (23)2 (−9)2 (−1)2 (−17)2 (−11)2 (9)2 + + + + + + 100 100 100 100 100 100 100 + (3)2 (−2)2 (6)2 + + = 11. (8.76 and from Table 8. This distribution gives χ2 = 6.1b) gave the outcome shown in Table 8. 4.3. followed by Eq. Balls numbered from 1 to 49 are selected at random from .4.5. The UK National Lottery The way that the UK national lottery operates was described in Sec. The result of producing 1000 random digits by using one of the generators for a uniform distribution. Analysis of 1000 digits produced by a more sophisticated random-fraction generator followed by Eq.1a) generated a series of fractional numbers with a uniform distribution between 0 and 1 and there are much more complicated and effective processes for doing this. 8.2 we see that there is a somewhat greater than 50% chance that random fluctuations would give this value of χ2 or greater. One could have more confidence in this generator than the one first described. much used by Monte–Carlo modelers.
4)2 (138 − 138. There are 49 terms in the calculation of χ2 and we indicate the first two and last two of these in (8. given this total. we look at how frequently the numbers 1 to 49 have been picked and Table 8. Commonly available χ2 tables do not normally give probabilities for 48 degrees of freedom but there . (1) 134 (8) 133 (15) 125 (22) 137 (29) 137 (36) 127 (43) 158 (2) 138 (9) 144 (16) 123 (23) 152 (30) 146 (37) 127 (44) 159 (3) 133 (10) 142 (17) 132 (24) 126 (31) 149 (38) 167 (45) 141 (4) 130 (11) 151 (18) 136 (25) 155 (32) 145 (39) 129 (46) 134 (5) 131 (12) 141 (19) 146 (26) 138 (33) 147 (40) 148 (47) 156 (6) 145 (13) 115 (20) 108 (27) 143 (34) 130 (41) 111 (48) 153 (7) 147 (14) 132 (21) 125 (28) 141 (35) 141 (42) 137 (49) 135 a rotating drum that ejects them individually without human intervention. This might lead us to wonder whether this is an indication that the selection of numbers was flawed in some way. To see if this is so. (6.56. 138. for the purpose of working out the value of χ2 it will be taken as the expected value E to be inserted in Eq.4 138.The UK National Lottery.4 (8. Loaded Dice.4)2 (153 − 138.2): χ2 = (134 − 138. In this way it is expected that the numbers are produced without any bias.4 + (135 − 138. The frequency of UK lottery winning numbers up to 4 October 2006. The average number of times each number was selected is 138.4)2 + + ··· + 138.1). we need to do a χ2 test.4 138.2) The total number of numbers picked in the lottery up to 4 October 2006 was 6780 and.4.6 gives the relevant frequencies for the whole history of the lottery up to 4 October 2006. from 108 for number 20 to 167 for number 38. and hence not a possible frequency for selecting a number. The first thing we notice about this table is the large range of frequencies.6.4)2 = 51. and Crooked Wheels 83 Table 8. To check this. 48 of the frequencies must be given to define the table — meaning that there are 48 degrees of freedom in this significance test. while this average is fractional.
Now when the die rolls it will have a slight tendency to end up with the lead weight as low as possible. let us suppose that the dice are slightly loaded.50 47. American Craps with Loaded Dice In Sec.4.25 so we have no reason to suppose that the selection of national lottery numbers is biased in any way. the χ2 test confirms that this is not so and such variation is within the bounds of normal random fluctuation. Without doing such a test it might be concluded that the discrepancy between the number of times that 38 and 20 occurred indicated some failure of the number-selection system.1 on the line between the center of the 1-face and the 6-face (which are opposite to each other). but assured profit in the longer term. 8. Apartial table for 48 degrees of freedom is given in Table 8.7.91 0. closest to the table thus increasing the chance of getting a 1 and decreasing the chance of getting a 6. 8.2. However.56 or some greater value is greater than 0.02 0. Probability χ2 0. i. From the table. This example shows the importance of doing objective statistical tests before jumping to conclusions on the basis of just examining the numbers involved. it was shown that the throwers chance of winning for the American craps dice game is 0. Probabilities associated with the χ2 distribution for 48 degrees of freedom. Now.84 Everyday Probability and Statistics Table 8.10 60.01 73.68 are Web sources that enable such probabilities to be derived. we can see that the probability of getting χ2 = 51. we suppose that instead of the six numbers each .25 54.33 0.17 0.20 0. 7. However.7.5 so that the bank makes a small.025 69.e. just slightly under 0.05 65.492929. This could be done by inserting a small lead weight within each die as shown in Fig..
175000. 4× 1 1 × 6 6 = 0. Since the probability is not the same for all faces the analysis is a little more complicated but the principles involved are the same. We notice that this is less than 1 (= 0. A loaded die.1. Getting 7 6+1 1+6 5+2 4+3 3+4 2+5 2 × (0.2. In each case the probability of getting a particular combination of two numbers on the two dice is just the product of the individual probabilities for each of the numbers.158333) = 0. Total probability of 7 P7 = 0.158333. 6 and P(6) = 0. 8. How much difference would this make? To determine the new probabilities of winning and losing we must go through the analysis as was done in Sec. In all the following calculations of probability with . 7. the probability with 6 unloaded dice. Probability of a "natural" This requires one or other of the combinations shown with their probabilities as indicated. and Crooked Wheels The 1-face 85 Offset lead weight The 6-face Fig.1750 × 0.111111 = 0.The UK National Lottery. Loaded Dice. P(2) = P(3) = P(4) = P(5) = .111111.1666• ) the probabilities are: 1 P(1) = 0.166667).055416.166527.055416 + 0. having a probability of 1 6 = (0.
340847 = 0.058333 + 0.086111 + 0.1750 = 0.058333.111111).222222).113889 (0.166527 To win with a "point" of 4 requires the two steps — first.086111 (0.166527 + 0. Setting a "point" of 5 and then winning 4+1 2+3 1+4 3+2 2× 2× 1 × 0.029351 (0. Then. The probability of a "natural" is Pnatural = 0. Getting 11 6 + 5 5 + 6 2 × 0.086111 × 0.86 Everyday Probability and Statistics loaded dice we give the corresponding result with fair dice in parentheses.086111 = 0. Setting a "point" Setting a "point" of 4 and then winning 3+1 2+2 1+3 2× 1 × 0. Hence P5 = 0. 6 1 1 × = 0.055556. = P5 + P 7 0. = P4 + P 7 0.052778 = 0.027778.083333).406143 (0.113889 = 0.052778 (0. 6 6 Hence P4 = 0.219305 (0. the probability of getting 4 before 7 is P4 0.058333 + 0. Then. to establish the "point" and then to win with it.340847 (0.166527 .113889 + 0.158333 × 1 6 = 0. the probability of getting 5 before 7 is P5 0.058333.277778).055556 = 0.055556). 6 1 1 × 6 6 = 0.400000).333333).027778 = 0.1750 = 0. The probability of taking these two independent steps is Pw4 = 0.
136111 = 0.083333 = 0. = P6 + P 7 0. Setting a "point" of 6 and then winning 5+1 1+5 3+3 2× 3× 1 × 0.449749 (0.1666527 P8 + P 7 The probability of taking the two independent steps is Pw8 = 0. 6 1 1 × 6 6 = 0.136111 + 0.454545). and Crooked Wheels 87 The probability of taking the two independent steps is Pw5 = 0.138889).141666 = 0. the probability of getting 6 before 7 is P6 0.166527 The probability of taking the two independent steps is Pw6 = 0. Loaded Dice. Hence P9 = 0.449749 = 0.1750 = 0. Setting a "point" of 9 and then winning 6+3 5+4 3+6 4+5 2× 2× 1 × 0.113889 × 0. the probability of getting 8 before 7 is P8 0. Then.459667 = 0.058333.158333 = 0.058333 + 0. 4+2 2+4 Hence P6 = 0.141666 + 0.158333 = 0.046255 (0.061216 (0.083333 = 0.108334 (0.138889). = 0.454545). 6 1 1 × 6 6 = 0.083333.136111 × 0. Hence P8 = 0.459667 (0.044444).052778 + 0.065119 (0.063131).141666 (0.136111 (0. 6 1 1 × 6 6 = 0.052778 + 0.052778.083333.063131). .111111). Setting a "point" of 8 and then winning 6+2 2+6 5+3 3+5 4+4 2× 3× 1 × 0.141666 × 0.The UK National Lottery.406143 = 0.052778. Then.055556 = 0.055556.
6 1 1 × = 0. = P9 + P7 0.080556 × 0.108334 × 0.108334 = 0.108334 + 0.326028 (0.166527 The probability of taking the two independent steps is Pw9 = 0.400000).333333).166527 The probability of taking the two independent steps is Pw10 = 0.394141 (0. Then. While this may not seem very different from the probability of winning with the fair dice. A 43% increase in the rate of return is by no means trivial! . the probability of getting 9 before 7 is P9 0.326028 = 0.0% per game rather than the 1. The overall probability of the thrower winning with the loaded dice is thus Pw = Pnatural + Pw4 + Pw5 + Pw6 + Pw8 + Pw9 + Pw10 = 0.394141 = 0. Now for every £500 staked by the thrower he gets a return of £490 so that the banker makes a profit of £10 on a stake of £500 — a return of 2. 6 6 Hence P10 = 0. in fact.080556 (0.026264 (0.027778 = 0.490209.492929.080556 = 0.052778. 0.158333 = 0.4% received with the fair dice.080556 + 0.052778 + 0.88 Everyday Probability and Statistics Then. Setting a "point" of 10 and then winning 6+4 5+5 4+6 2× 1 × 0.0277778). = P10 + P7 0.027778. it is. a very important difference.042699 (0. the probability of getting 10 before 7 is P10 0.083333).044444).
The problem with this is that even with a fair die there would be some random fluctuations in the frequencies. However. To simulate what would happen if the loaded die was thrown many times. throw it many times.5.The UK National Lottery. Each number obtained is interpreted as a throw of the die in the following way: Between 0. we would have to carry out an objective test like the χ2 test. and Crooked Wheels 89 8.166667) is equivalent to throwing a 5 (probability 0.166667) is equivalent to throwing a 6 (probability 0.175000) is equivalent to throwing a 2 (probability 0.158333).166666) is equivalent to throwing a 4 (probability 0.341667 Between 0. we are going to use a high-quality random-number generator that gives a uniform distribution between 0 and 1.508333 and 0.175000 and 0. Testing for a Loaded Die The effect of slightly loading the dice in a crap game so that a 1 appearing is very slightly favored may not be very noticeable in the duration of a game involving one or two hundred throws. it can give a substantial proportional increase in the profit to the bank.675000 Between 0. find the frequencies of the numbers 1 to 6 and then check that the distribution found is not an improbable one for a fair die. .508333 Between 0.000000 and 0. as we have seen. as was seen in Table 8.175000 Between 0. To be sure that the die was loaded. We would start with the null hypothesis that the die was a fair one.5 that generated random digits with a perfectly fair random-digit generator.841667 Between 0.166667) is equivalent to throwing a 3 (probability 0.341667 and 0. Loaded Dice.000000 is equivalent to throwing a 1 (probability 0.841667 and 1.675000 and 0. If it were suspected that the dice were loaded then how could it be tested? The obvious way would be to throw the dice a large number of times to check that all the numbers appeared with equal frequency.
005 4.351 6.9.35 33. the result is as given in the first numerical row of Table 8.10 0. We can calculate the value of χ2 as we did for the random digits in Sec.90 Everyday Probability and Statistics Table 8.8.01 0.86 13.000 E 100 500 2500 7500 1 114 551 2628 7832 2 93 493 2471 7513 3 100 487 2505 7536 4 81 498 2510 7485 5 107 504 2512 7508 6 105 467 2374 7126 χ2 6. For the null hypothesis of a fair die.832 13.8.175000 of obtaining a 1.05 0.388 15. 6. A simulation of throwing a loaded die many times.30 0. the expected frequency of each number. the number of degrees of freedom is five.2 but now.750 χ2 . is 100. 8.000 15.80 7. and translated into equivalent throws of the die. Part of a probability table for five degrees of freedom is given in Table 8. a probability of 0.50 0. 100 100 Table 8.000 45.20 0.02 0.58 This gives a probability of 0.236 11. Probabilities associated with the χ2 distribution for five degrees of freedom.064 7.070 12. Total throws 600 3. The actual numbers vary between 81 and 114. with six faces of the die rather than ten digits.1583333 for a 6. When 600 random numbers were generated. shown in the column headed E.4.025 0. Probability 0.086 16.289 9. The value of χ2 for 600 throws is χ2 = (114 − 100)2 (93 − 100)2 (100 − 100)2 (81 − 100)2 + + + 100 100 100 100 + (107 − 100)2 (105 − 100)2 + = 6. and a probability of 1/6 (very nearly) for the other numbers — the loading of the die considered in Sec.9.80.
The value of χ2 . The slots are numbered from 0 to 36 and there are various ways of betting on the spin of the wheel.000 throws the proportion of throws for the various numbers are 1 0. he gets 36 time his stake returned. With 45.35. 7.2 of being found with a fair die. as found in European casinos. we see that there is about a 25% chance of getting this value of χ2 . which are very close to expectation.1668 6 0. with 45. Now suppose that we increase the number of test throws to 3. contains 37 numbered slots uniformly distributed around its rim within an annular valley.86. and Crooked Wheels 91 From Table 8. The Roulette Wheel A roulette wheel.000 throws the evidence for a loaded die becomes stronger. We may be fairly confident that the die is loaded. or some greater value. Loaded Dice. i.1675 4 0. One may pick one of the 36 non-zero numbers and if the ball lands on the number the punter gets odds of 35:1. there is no evidence that the die is biased for a test with this number of throws. However.1584.000 throws and a value of χ2 = 33.58 we may be completely certain that the die is loaded.The UK National Lottery.e.6. When we do a test with 15.1663 5 0.1740 2 0. The number 1 now appears clearly more frequently than any other but the value of χ2 . with a fair die. 8.1670 3 0. For a large-scale test it is possible to find not only that the die is loaded but also a good estimate of the way that it is loaded. or some greater value. has a probability of somewhat less than 0. 13.000.. has a 1 in 50 chance of happening with a fair die. Another way is to bet on either . We could not be confident that the die was unfair with this test.9. or some greater value. When the wheel is spinning quite quickly a steel ball is tossed into the valley after which it bounces around in haphazard fashion until eventually it lands up in one of the slots.
say. 8. From the point of view of fairness there are only two types of numbers — 0 and non-0. An American roulette wheel with 0 and 00. If the only numbers on the wheel were 1 to 36 then the casino and the punter would be playing on even terms. 8. or evens.2). 3. Suppose that a wheel gave a frequency of 0s 10% greater than it should.2. However. or on a group of six numbers where the odds are 5 to 1. the ball lands in the zero and then the casino takes all the stakes.700 spins of the Fig. A wheel which is slightly off-center might give an increased tendency for the ball to fall in the zero.92 Everyday Probability and Statistics even or odd numbers where the odds are 1 to 1. To test a roulette wheel for fairness it is necessary to spin it many times. The ball is usually made of steel so a bias toward zero may also be made with a magnet but this form of cheating is so easily detected that it is never used. this is its profit. The standard roulette wheel used in the United States is somewhat more profitable to the casino because it not only has a 0 slot but also a 00 slot when the casino takes all (Fig. . on average once every 37 spins of the wheel. Could this be detected with.
100 times in 37. Observed 0 non-0 110 3. either locally or nationally.1) would not indicate bias with any certainty. and Crooked Wheels 93 wheel? With the null assumption that the wheel is fair the observed and expected tables are as shown below. Problems 8 8.600 The value of χ2 for these tables is 1.28. if 0 turned up 1.000 spins the value of χ2 would be 10. they are also monitored and the punter may be reasonably sure that the equipment used is fair and without bias. When a die is thrown 600 times the occurrence of the various sides is as follows: 1 2 3 4 5 101 6 84 112 113 81 109 On the basis of this test is there a greater than 90% chance that the die is unfair? . However.590 Expected 0 non-0 100 3. Loaded Dice. Where casinos are licensed to operate.028 which for one degree of freedom (Table 6. the chances of getting this value of χ2 or a greater value is only one in a thousand so there would then be clear evidence for a biased wheel.1.The UK National Lottery.
Would you be justified in suspecting that the wheel was biased? . or some greater value.3. In 370 spins of a European roulette wheel a 0 occurs 20 times. with a fair die is less than 10%.94 Everyday Probability and Statistics Note: The wording of this question implies that the probability of getting the associated value of χ2 . 8.2. A heavily biased die has the following probabilities of obtaining the various sides: 1 1 4 2 1 6 3 1 6 4 1 6 5 1 6 6 1 12 What is the probability that the thrower will get a natural? 8.
Non-physical characteristics can usually be expressed in a qualitative way but may be difficult or impossible to quantify. another excitable. Joe may 95 .Chapter 9 Block Diagrams Variety's the very spice of life. nevertheless.1. but if he has an inadequate diet then he will never achieve the capability given by his genes. In particular. An individual may be genetically inclined to be tall. The fact is that we are all the products both of nature and nurture and. this variety is all part of the rich tapestry of life. weight or blemishes on their surfaces. people have wide differences of temperament. it is a matter of everyday experience that all human beings are different. even the most similar identical twins differ in some small characteristics. one may be of happy disposition and another morose. Differences between people can be both physical and non-physical so that. There is the saying "They are as alike as two peas in a pod" but when you examine the peas in a pod carefully they are all different — in size. One person is placid. whatever our genetic inheritance. for example. the environment in which we develop and live and our life experiences will influence our characteristics. That gives it all its flavour (William Cowper. Variation in Almost Everything Most natural classes of objects show variations. 1731–1800) 9. As William Cowper stated in the quotation that begins this chapter. Identical twins come from the same divided ovum and are thus identical in their DNA composition but.
This example gives us a warning. and in geographical features such as the length of rivers — in all these there is variation. however similar they may seem superficially.2. Only scientists deal with objects. Thus. Jim may be reflective and deliberate in his actions and John is somewhere in-between. Comparative terms such as tall or short only have meaning when the context in which they are applied is known. to describe in a systematic way. In other animals. but how can these characteristics be numerically described? By contrast physical characteristics are easily measured. A short man in Europe may be average. sell them to a retailer . a man who is 1. in plants.1 m (nearly 7 ) or more. In the things that we can see no two are alike. We should stick to numbers — these mean the same thing everywhere. There is no way that two electrons can be distinguished — if there were then the whole of physics as a subject would be radically different and the physical world would be very difficult. A Shoe Manufacturer The objective of a shoe manufacturer is to make shoes to be sold at a profit to a shoe wholesaler who will. short or of average height but we can be more specific by giving their actual measured heights. or even tall. they can be of greater importance and may even be of commercial significance. many of whom grow to a height of 2. Usually variations are just matters of note but. We may generally describe people as tall.96 Everyday Probability and Statistics be impulsive. which by their very nature are identical. 9. But atomic particles are things we cannot see although we can infer their existence with complete certainty. at some level of refinement they will be found to be different.91 m (6 3 1/4") tall would be regarded as tall in a European context although he would be considered of normal height by the Watutsi people of Ruanda. by the standards of the pigmy people of the Congo and tall in Europe may not mean the same as tall in Ruanda. All atomic particles of the same kind are precisely the same. particularly in the human context. or even impossible. Variation occurs in entities other than human beings. in turn.
Such a sampling exercise would probably find that the great majority of men's shoes fell within the UK sizes 6–13 with a tiny fraction outside that range. sell them to a customer at a profit.1. If it were possible to check the shoe size of every man in the community then the proportions of shoes required of each size would be known.1 for shoe sizes 6–13 in steps of 1/2. If a customer comes into a shop and cannot find the size of shoe to fit him or her then a sale is lost. are shown in Table 9. It is of interest to all that the number of shoes of a particular made size should approximately match the number the customers needing that size. However. Notional proportions. Another adverse factor in the profit-making enterprise would be if a large surplus of some particular sizes of shoe were to build up at any stage of the chain because this would correspond to an unproductive tie-up of capital. at least in the upward direction. Since values. Let us consider men's shoes. within the normal commercial range the obviously efficient thing to do would be to manufacture numbers of men's shoes with sizes in the same proportions as occur in the adult male population.1 is very common in naturally occurring quantities. somewhat idealized. The last link in this chain is the customer and satisfying the needs of customers is the key to the success of the whole enterprise. 9. The symmetrical bell-shaped distribution shown in Fig. in which the length of a block gives the proportion of shoes required of the indicated size. either .1 gives the essential information about the proportions of shoes of each size required but the detailed numbers must be studied at some length to appreciate the variation of demand over the size range.Block Diagrams 97 at a profit who will. or histogram. A better visual picture giving the overall pattern of demand is given in Fig. that is described more fully in Chap. by which a close approximation to the relative numbers required of each size can be found. in turn. Extreme sizes. 14. where the same information is presented in the form of a block diagram. would fall in the province of shops dealing with unusual clothing requirements — and the cost to customers of providing such a need is usually quite high. there is a process called sampling. Table 9. Although that would not be a feasible exercise. 9.
1.1210 0. at the extreme smaller end of the scale or at the extreme larger end. .0270 0.2 0.98 Everyday Probability and Statistics Table 9. Notional proportions of men's shoe sizes between 6 and 13. are usually rare.1210 0.1 0. Shoe size 6 61/2 7 71/2 8 81/2 9 91/2 Proportion 0. The distribution of men's shoe sizes shown as a histogram.0022 0. there are exceptions to the conditions that give bell-shaped distributions and other types of distribution are possible.1. a bell-shape is very common.1761 0.1761 0.0270 0.0004 0.0088 0. 9.0648 0.15 Proportion 0.1995 Shoe size 10 101/2 11 111/2 12 121/2 13 Proportion 0.0004 Distribution of shoe sizes between 6 and 13 0. However.0088 0.0022 0.0648 0. although it may not necessarily be symmetrical.05 0 6 7 8 9 Shoe size 10 11 12 13 Fig.
i. there is some similarity.Block Diagrams 99 9. Average cloud cover on a monthly basis at Trenton.3.e. .2 gives the average night-time cloud cover. we can interpret the cloud-cover histogram as a bimodal distribution.2. in fact. one with two peaks.2 starting with August as the first month then the cloud cover steadily rises each month up to January then falls steadily until April. occurring in January and June. The bimodal nature of the cloud-cover distribution is not too pronounced but there are situations where much stronger bimodal Fig. 9. Although this seems very different from the histogram representing shoe sizes. If we re-plotted Fig. 9. Enhanced cloud-cover in May and June spoils the bell-shaped pattern. The months of the year run cyclically and deciding to begin the year with the month of January is an arbitrary choice. Fig. Histogram Shapes As an example of a non-bell-shaped histogram. In fact. month-by-month. averaged over a five year period at Trenton in Ontario — something of interest to astronomers. 9. Ontario.
. 9. Most eruptions last either about 2 mins or 4 mins with 3 mins eruptions. in Yellowstone Park in the USA. Figure 9. One of the great wonders of nature.000 to 32. with durations divided into intervals of 1/3 min. sending anything from 14. The duration of each eruption is also very variable.000 liters of boiling water to a height between 30 and 55 m. is the Old Faithful geyser (Fig.100 Everyday Probability and Statistics Fig.4 shows a histogram of the relative number of eruptions of various durations. mostly at the longer end. Yellowstone Park.833 and 2.4 the bimodal nature of the frequency is clearly seen. This geyser may be faithful but it is somewhat variable in its behavior. being between 1. 9. In Fig. This means that the first interval is between 1.833 mins. It erupts at intervals between 65 and 92 mins. being comparatively rare. the second between 1.5 and 5 mins. The Old Faithful Geyser.3). USA (USGS). or thereabouts. 9.167 mins and so on. distributions can occur.3.5 and 1.
9. a . shoe size. can only occur with particular values.Block Diagrams Eruptions of Old Faithful geyser in Yellowstone Park 1 101 0.4 0. However. Other shapes of histogram can occur but only rarely. there are quantities that are continuously variable — height. No shoes of size 9. The bellshaped and bimodal histograms account for most of the distributions that occur naturally.1 a bell-shaped histogram comes about from a quantity. especially in environments such as an army barracks.4.2 0 0 1 2 3 duration in minutes 4 5 6 Fig.6 0. The distribution of eruption durations for Old Faithful.. 9. We have already introduced this idea for the eruption durations of Old Faithful where the durations were divided into 1/3 min intervals. Lofty and Shorty In Fig.e.35 are manufactured and if you require shoes that are not one of the standard sizes then you must go to the expense of having them made to measure.4. weight etc. It is common to give people nicknames dependent on their heights. 9. which is discrete — i. and we shall now see how to represent these by means of histograms.8 relative freaquency 0.
in a strictly scientific sense.765 m. chosen at random.62 m man (5 4 ) is called "shorty. 9. The question only makes sense if we ask for the probability that the man will be in a certain range of height — for example between 1.7601 m tall." Sometimes. If someone says that he is 1. the 1.5. will have a height of 1. perversely and as a form of irony.755 m and 1. . is meaningless since there may be nobody who is exactly 1. for example.76 m 1.76 m tall although there may be several men between 1. If we could explore the whole population of British adult males and look at the proportions in various heights in the range of 1 cm.76 m tall what he really means is that his height is closer to 1.76 m than to either 1.75 m 1. the short man is called "lofty" and the tall man "shorty" — but the main point here is that in some way or other the extremes of height are being highlighted in some way.102 Everyday Probability and Statistics school or a workplace. we could produce a histogram looking something like Fig.755 m 1.765 m 1. 1. 9.77 m Implied range Fig. We now consider the question "What is the probability that an adult British male.77 m. Thus.91 m (6 31/4 ) man becomes known as "lofty" and the 1.7599 m and 1. If a man is asked for his height he will usually quote it to the nearest centimeter or quarter-inch but it is only an approximation.76 m?" The question.75 m or 1.755 m and 1. could be applied to most men.765 m.76 m. The implied range of possible heights for a quoted height of 1. Figure 9. In fact in giving his height as he does he is implying that his height is somewhere between 1.5 shows the basis of this implication.6. The majority of men fall in the middle range of height and so they are not granted a distinguishing nickname relating to height — "middley".
7 1.01 0 1.02 0. A histogram showing proportion of UK males in 0.7.1 and this smoothness is dependent on the size of the interval we choose for the blocks.8 1.5 1.025 Proportion in 0. in Fig.9 2 2.03 0.4 1.6. 9.8 1.7 1.6 1.7 we show the same distribution of heights but .1 height (meters) Fig.9 2 2. We notice that the histogram shown in Fig. 9.05 proportion in 1 cm intervals 0.01 0.5 1.4 1. 9. A histogram showing proportion of UK males in 1 cm height intervals.04 0.6 1.6 is somewhat smoother in appearance than the one in Fig.5 cm intervals 0.1 height (meters) Fig.03 0.015 0.5 cm height intervals. 9. 0.Block Diagrams 0. 9. For example.005 0 1.06 103 0.02 0.
75 m will obviously be one-half of the proportion in a 1 cm interval.13 18 20 22 24 26 0.5 cm interval centered on 1.04 0. Women's off-the-peg dresses in the UK are normally manufactured in sizes ranging from 6 to 26 in steps of 2. We notice that the lengths of the blocks are less in Fig. In the following chapter we will see what happens when we take this to the limit of very tiny (theoretically zero) intervals. are: 6 8 10 12 14 16 0. 9. 9.20 0.1.02 0. excluding the very few outside that range. In the interests of its shareholders should it reduce the range of sizes it sells? . 9.104 Everyday Probability and Statistics this time plotted in half-centimeter blocks and the result is smoother than that shown in Fig.15 0.01 Draw a histogram to represent this table.7 than in Fig.09 0.6. The shop has a policy of charging the same price for all sizes of a particular style. The proportions of women in each size category.16 0. If it only stocked sizes 8 to 18 its overall costs would be reduced and its profit per garment sold would increase by 20%. Problem 9 9.6 since the blocks are narrower and the proportion of men with heights in a 0. As we make the intervals smaller and smaller so the saw-tooth serrations along the boundary get finer and finer.10 0.03 0.07 0.
10.057 to about 0. Eventually we reach a distribution that is completely smooth. 1843–1916) 10. You may well ask why it is that when we represent the situation of having infinitely narrow blocks the maximum shoots up to nearly 7. as illustrated in Figs. Probability Distributions Let us imagine that we take the distribution of men's heights. How does this come about? We can best understand this by going back to Fig. (Henry James.7 the lengths of the blocks went down.6 and 9. looking somewhat like the curve shown in Fig.1 and interpreting what we see there.028.6 to Fig. Another way of thinking about this is to say that if we pick a pair of shoes at random from all the available pairs then there is a probability of 0. with the maximum length going from about 0. Similarly. When we went from wider to a narrower blocks in going from Fig. 9. the length of the block corresponding to size 10 (and also to size 9) is 0. 9. 9.6 and 9. you may have noticed an important difference between this distribution and the ones shown in Figs. However.7 and we make the intervals narrower and narrower so that the appearance of the curve becomes smoother and smoother. The length of the block corresponding to size 91/2 is 0.1.Chapter 10 The Normal (or Gaussian) Distribution …that of the strange and sinister embroidered on the very type of the normal and easy. 9.7.1995 that it will be size 91/2.1995 corresponding to the proportion of shoes required of that size.1.1761 and this is the probability that a pair of shoes chosen at random will be 105 . 9.
size 10 (or size 9).4 1. With this interpretation of the lengths of the histogram blocks we now consider Figs.. 9.e. . Extending this idea we now ask what the probability is of choosing a pair of shoes at random and having a size somewhere in the range 6–13.8 1.6 1.106 8 7 6 5 4 3 2 1 0 Everyday Probability and Statistics Proportion per unit interval 1.1. 10. Since the choices are mutually exclusive — if one size is chosen then the other two sizes are excluded — then the answer is obtained by adding the lengths of the blocks corresponding to the three sizes — i.1.5 1.1 Height (meters) Fig. on average.7 1.1995 + 0. all the shoes are between sizes 6 and 13 so it is completely certain that the pair chosen will be one of these sizes.9 2 2.6. A probability of 1 means "certainty".7 as there are in Fig.1 and the answer is found to be 1 — which can be checked from the figures in Table 9. size 91/2 or size 10?".6 the lengths of the blocks in Fig. 9.1761 = 0. 0.5517. A distribution of heights for infinitely small intervals. This is obtained by adding the lengths of all the blocks in Fig.7 are.6 and 9. The sum of the lengths of the blocks has to be 1 in each of the figures but since there are twice as many blocks in Fig.7 which portray the notional height distribution of UK males.1761 + 0. 9. We now ask the question "What is the probability that a pair of shoes chosen at random will be one of size 9. one half of the lengths of those in Fig. 9. 9. 9.
has to be considered in a different way.5 1.10 m is the total area under the curve and must be 1. of thickness 0. is the total area under the curve between those two limits.7 107 Probability per unit interval 1. also shown in Fig.3 shows the curve covered by a mesh of blocks.2 we show a narrow strip of the curve. 10.1.6 1. In Fig. The continuous curve in Fig.40 m and 2. 10. Some blocks fall completely under the curve and some partially so and within each block is given the estimated fraction of the block in the region below the curve to the nearest 0.01 centered on 1. because all. The strip covers the range of heights from 1.The Normal (or Gaussian) Distribution 8 7 6 5 4 3 2 1 0 1. each of area 0. Fig.01 m centered on the height 1.70. fall inside this range.1). 10. or nearly so. except a very few individuals.8 h1 h2 1.2. Adding all these . To check this.70 m.1.1 (length 1 × breadth 0. 10.4 1. A narrow strip of the distribution of Fig.1. Extending this idea the probability that a man chosen at random would have a height between 1. which contains no blocks of defined width with lengths to represent probabilities.9 2 2. 10.705 m and the probability that a man chosen at random would have a height between these limits is now taken as being given by the area of the strip — not the length of the strip as for the previous histograms.1 Height (meters) Fig.695 m to 1.2. 10. of width 0. It follows from this that the probability that the randomly chosen man would have a height somewhere between h1 and h2 .
10. To illustrate this.0 1. 10.3 0.1 is the one that occurs frequently in nature.1.0 1.0 1. a French mathematician. Because of its symmetrical form it is clear that the peak of the curve represents the average.4 1. gives the area under the curve as 1 — as it should be.8 0. or mean. .4).6 1 0.3 0.0 which. The fractions of blocks under the curve judged to the nearest 0. This distribution is now called a normal distribution or.8 0 1.9 0. It was first used in a mathematical context by Abraham de Moivre (1667–1754). numbers gives 10.0 1.2 0. a Gaussian distribution.0 Proportion per unit interval 4 0.3.8 1. especially by physicists. 10.1 2 0. The Normal Distribution The symmetrical bell-shaped distribution shown in Fig.1 Height (meters) Fig.7 1. 10.6 2 2.3 1.6 0. value. the area of each block.1.2 0.2. in 1733 but its widespread applications were not realized at that time.5 1.1 3 0.108 8 7 6 5 Everyday Probability and Statistics 0. and its properties are of great interest to the statistician.7 1. What is also of interest is the spread of the curve. multiplied by 0. Its mathematical form was first described in detail and made well known by Johann Carl Friedrich Gauss (1777–1855) — who is usually known as Carl Gauss since for some reason he discarded the Johann part of his given name (Fig.
05 0 0 5 10 x 15 20 Fig.35 0. Since the area under all the curves equals 1 this means that the greater the lateral spread the less is the height of the curve — as is . Three normal distributions with the same mean but different spreads.15 0. 0. 10.1 0. Carl Friedrich Gauss. in Fig.4 0. 10.4. 10.45 0.5 there are shown three normal curves with the same mean but with different spreads.2 0.5.The Normal (or Gaussian) Distribution 109 Fig.5 0.25 0.3 P(x) 0.
7 This quantity V that describes the spread of the set of numbers is known as the variance and taking its square root gives the standard deviation of the set of numbers. . so the average of each set is the same. To see what this could be we consider the following two sets of seven numbers: Set A Set B 7 1 8 6 9 9 10 10 11 11 12 14 13 19 The sum of each set of numbers is the same.5 is some numerical quantity related to their spread of values.5. usually represented by the Greek letter σ. We now have to find out how it is that different curves.110 Everyday Probability and Statistics seen in the figure.3. as shown in Fig. 10. The Variance and Standard Deviation Clearly. equal to 10 (70/7). 10. Set A Difference from average (10) Square of difference 7 −3 9 8 −2 4 9 −1 1 10 0 0 11 1 1 12 2 4 13 3 9 We notice here that the square of −3 is the same as the square of +3. can all be described as normal curves. 10. The sum of the squares of the differences is 28 and now we find the average square of difference for the seven numbers as V= 28 = 4. equal to 70. what is needed to describe the difference between the curves in Fig. Here we have √ √ σA = V = 4 = 2. We now illustrate this with set A. What we now do is to find the sum of the squares of the differences of each number in a set from the average for that set.
perhaps many millions. no matter . and 3σ from the mean.29. 10. we consider Fig. of quantities and for these quantities. 10. 2. Now a normal distribution may come from a large number. there will be a standard deviation. on both sides of the mean. they all have the same shape. 10. and hence for the distribution function they form. 10.5 and wonder how that can be so — surely the three curves shown have different shapes! To understand what is meant in saying that all normal curves essentially have the same shape. Properties of Normal Distributions An important characteristic of normal distributions is that. respectively.The Normal (or Gaussian) Distribution 111 The same process carried out for set B gives: Set B Difference from average (10) Square of difference 1 −9 81 6 −4 16 9 −1 1 10 0 0 11 1 1 14 4 16 19 9 81 The variance is V = (81 + 16 + 1 + 0 + 1 + 16 + 81)/7 = 196/7 = 28 and hence σB = √ √ V = 28 = 5. are measures of the spread of the two sets of numbers and clearly indicate that set B has a much larger spread than does set A. and 3 and hence standard deviations of 1. indicated by vertical lines. You may look at Fig. and √ 3. the three distributions shown in Fig. In fact. 2.6 that shows a normal curve with distances 1σ. The values of the standard deviations. σA and σB . 2σ.4.5 have √ variances of 1. The characteristic of all normal distributions. The total area under the curve is 1 — a necessary condition for a sensible distribution curve. essentially.
6826 = 0. is 0. 10.45 0. By the same reasoning we can also see that the probability of having a quantity more than 3σ from the mean is only 0. ±2σ and ±3.05 0 0 1 2 3 4 x 5 6 7 8 -2 σ -3σ 2σ 3σ Fig. roughly one-third.3174. or beyond.25 0.6. It follows from this that since the total area under the curve is 1. by 1σ or less. In the language of statisticians a part of the curve beyond a particular specified limit is referred to as a tail of the distribution. is that: The area between −σ and + σ is 0.2 0.35 0. we can say that the probability that the quantity differs in magnitude from the mean by more than 1σ is 1 − 0. Table 10. Finding the areas from the mean either out to.6826. either in the positive or negative direction.1 gives the area from . A normal curve with lines at ±σ.9973 We now discuss what this means? If a particular quantity has values with a normal distribution then the probability that one of the values chosen at random differs in magnitude from the mean.6826 The area between −2 σ and +2σ is 0. about 1 chance in 400.4 0.9545 The area between −3 σ and +3 σ is 0. a particular number of standard deviations for a normal curve is frequently required in statistical work.0027.5 0.1 0.15 0.112 Everyday Probability and Statistics 0. what are their standard deviations.3 P(x) -σ +σ 0.
49984 0.49989 0.4713 2.49379.4452 0.7 standard deviations from the mean then we go down the column headed 1.49379 0.4554.48214 0.5σ from the sixth entry down in the column headed 2. The distribution of men's heights shown in Fig. In the other direction the height 1.5 − 0.0398 0. Area from the mean to n standard deviations from the mean for a normal distribution.47725 0. 000×0.0793 0.49653 0. This is 0.00621.3643 0.1 0.49865 0.1915 0. or 2.48610 0.5σ less than the mean height so 186 of the townsmen should be shorter than this height. Of these 30.5σ from the mean.3413 0.0 0. Thus.06 m (6 cm). We can now find the area under the curve from the mean out to 2. if we wish to find the area from the mean out to 1.4 0. 0.4032 0.1 σ. The total area in one half of the curve is 0.90 m is 15 cm.90 m.48928 0.0000 0. To illustrate the use of this table.49180 0.3849 0.1554 0. N 0.8 0.2257 0.60 m (5 3 ) is 2.49997 the mean out to four standard deviations (just in one direction) in steps of 0.000 adult males. which is the probability that a man chosen at random has a height greater than 1.2 0.00621 = 186 of them should be taller than 1. To see what this means we consider a town with 30.49993 0.7 and read off the area.90 m.9 0. .4192 0.0 down to the row labeled 0.2580 0.0 0.1 has a mean of 1.6 0.1. 10.75 m and a standard deviation of 0.The Normal (or Gaussian) Distribution 113 Table 10.49995 4.4332 0.49966 0.3 0.0 0.1.4554 0.1179 0.0 0. the probability that a man chosen at random is taller than 1.2881 0.49977 0.0 in Table 10.7 0.5 0.49744 0.49813 3.90 m (equivalent 3 to 6 2 4 ) can be found by noting that 1.49903 0.5σ is 0.0 0.4641 0.5 so that the area beyond 2.0 0.49952 0.3159 1.49534 0.49379 = 0.49931 0.
716924 2.488320 2.000.005)200 (1 + 0.000.704814 2.000 = = = = = = = = = = = ε 2.5. starting with n = 2 and increasing n to the final value of 1. Here we show the value of ε.718281828. Some Special Numbers A number that occurs a great deal in mathematics is the exponential e given by e = 2.000.715569 2.02)50 (1 + 0. One way of defining e is to take the equation 1 ε= 1+ n n (10. A Little Necessary Mathematics 10.718146 2.001)1000 (1 + 0.711517 2.002)500 (1 + 0.1)10 (1 + 0. .691588 2.250000 2.2)5 (1 + 0.0001)10.000 (1 + 0.000 (1 + 0.593742 2.1) and to make n very large — in fact infinity.5.000.000001)1.1. It seems a fairly arbitrary number but it crops up over and over again in mathematics and the sciences in a very natural way.114 Everyday Probability and Statistics 10.01)100 (1 + 0.718280. n 2 5 10 20 50 100 200 500 1000 10.05)20 (1 + 0.5)2 (1 + 0.653298 2. It is not easy to explain to a non-mathematician how and why it occurs so naturally — but it does.000 1.
for example. mathematically we can have 22·3 and this cannot be written as 2 multiplied by itself a finite number of times. We see there the notation 22 for the square of 2 and 23 for the cube of 2. (8.5.The Normal (or Gaussian) Distribution 115 As we make n larger and larger so ε gets closer and closer to e.2.1) to find the successive approximations to e.141592654 that is easily described as the circumference of a circle divided by its diameter. The first step in understanding what this means is to consider the following: 22 × 23 = (2 × 2) × (2 × 2 × 2) = 32 = 25 x ×x =x a b a+b or.1). However. 10.2) . Thus. represented by the Greek letter pi. a = 2 and b = 3. (10. This was used in the simple random-number generator described in Eq. which we can write mathematically as 22 = 2 × 2 = 4 and 23 = 2 × 2 × 2 = 8. given by π = 3. for example. this idea can be extended to the product of many powers of x so that. we know that the square of 2 is 4 and the cube of 2 is eight. the same is not true for another frequently occurring number. While e cannot be justified as a special number very easily. In fact. in general. The general equation can be related to the original specific example by taking x = 2. (10. we come . Powers of Numbers Another piece of mathematics that is fairly familiar consists of taking powers. a process that was used in the calculations using Eq.
(10. In fact. the number such that the product of n of them gives x. We finish with a last little oddity.. it can be found with a scientific hand-calculator in the form 22·3 and is 4.e. With the above information we are now in a position to understand what 22·3 means. For example. Using Rule 1 we find: xa × x0 = xa+0 = xa from which we come to the conclusion that x0 = 1.9246. 1 1 1 1 1 (10. the number such that the product of three of them gives 2. According to Rule 2 we can write it as 22·3 = (223 ) 10 which is the tenth-root of 223 . (10.3) Now. This generalizes to the second rule Rule 2 xa b = xa×b . illustrates this (22 )3 = (2 × 2) × (2 × 2) × (2 × 2) = 22×3 = 26 .116 Everyday Probability and Statistics to our first rule for handling products of powers Rule 1 xa × xb × xc × xd = xa+b+c+d .4) We can now understand what is meant by fractional powers of numbers.e.5) i. This is absolutely true — any finite number raised to the power zero equals 1! 1 . The following example. taking the square of 2 cubed. i. from Rule 1 2 3 × 2 3 × 2 3 = 21 = 2 so that 2 3 is the cube-root of 2.. we consider what happens when we take a power of a power of a number. By generalization we can see that x n is the nth root of x.
10. (10. Random and Systematic Errors When a scientist carries out an experiment. he will make various measurements and then combine them to get the quantity he wants. measuring the speed of light may involve a measurement of distance and a measurement of time and both these measurements will have some associated error.7) ¯ With x representing height.1. x is itself a complicated function and there is obviously a problem in showing ex clearly if x is a bulky function to express.The Normal (or Gaussian) Distribution 117 What we have established is that any positive quantity can be raised to any power and that this power can be fractional or even zero. To help with this problem there is a convention by which one can write ex = exp (x).7.06 m this produces the curve shown in Fig. The Form of the Normal Distribution The formula describing the normal distribution involves the exponential e raised to some power — something of the form ex . The factor before "exp" ensures that the curve is properly scaled so that the area under the curve is 1. 10.6.75 m and the standard deviation σ = 0. (10. We do not have to worry about interpreting such calculations when we need them — they can easily be found by calculators or computers. 10.6) With this notation we can represent the normal (Gaussian) distri¯ bution for a quantity x that has a mean value x and a standard deviation σ as ¯ (x − x)2 1 P(x) = √ exp − 2 2σ 2πσ2 . the mean height x = 1. For example. . say to find the speed of light. However.
The errors in the different measurements — and in some experiments there can be many of them — will combine together to give an error in the estimated value of the quantity of interest. perhaps several kilometers long. The distribution of the values of these estimates would be Gaussian with a mean close to the true value of the quantity and a standard deviation that would depend on the precision of the individual measurements. Baselines of length 7–10 miles were measured with an accuracy of about 1 in — approximately two parts in a million. 100 feet long. Considering the complication of this process the measurements were made with surprising precision. carried out by the British army at the beginning of the 19th century. If the measurements were all made very precisely then the standard deviation of the distribution would be small. which was supported horizontally by wooden coffers each 20 feet long. Even earlier a standard chain was used. The tape was suspended above the ground attached to supports adjusted to be at the same level at each end. Another kind of procedure that would involve errors is the measurement of a baseline in surveying.118 Everyday Probability and Statistics The error may be large. To create a map it is necessary to measure a baseline. and completed by Colonel George Everest whose name was given to the world's highest mountain. that will create the correct scale for the mapping procedure. of which . These were used in the great survey of India. A classical way of measuring a baseline was with a steel tape. Every measurement of a single chain length. The length of the tape was marked off repeatedly over the whole span of the baseline with each length indicated by score lines on a metal plate attached firmly to the ground. Because of its weight the tape was not straight but bowed downwards in a shape known as a catenary and a correction had to be made for this effect. Now. small or even minute but no matter how much care the scientist takes there will be some error. if they individually had large errors then the standard deviation would be large. we imagine that the scientist repeats the experiment many times (he would not do so in practice) and ends up with a large number of estimates.
Let us suppose that the steel tape. produced in large numbers by mass-production processes. Electric Light Bulbs Hot-filament electric light bulbs are a common and cheap commodity. just as indicated by a normal distribution.8.9545. 10. positive and negative errors would compensate to some extent. the assumed length of the chain. we have assumed that the errors in the measurements were random and equally likely to be negative as positive. if the quantity is measured as X with an estimated standard deviation (usually called a standard error in this context) of σ then the probability that the true value is in the range X + σ and X − σ is 0. there is another kind of error for which this supposition is not true. used for the baseline measurement had stretched — which sometimes they had.1.6826 and in the range X + 2σ and X − 2σ is 0. if measured many times. This error of the mean is called a systematic error. would give a normal distribution of estimated distances with a standard deviation dependent on the magnitude of the random errors but with a mean shorter than the true distance. Assuming that only random errors are present. is actually slightly longer so the total distance is going to be underestimated independent of any random errors. clearly.8. However. The total baseline measured with a stretched chain. Now a distance estimated as 100 feet. sometimes positive and sometimes negative. Some Examples of the Normal Distribution 10.The Normal (or Gaussian) Distribution 119 there would be several hundred along the baseline. In the above discussion of errors. if the baselines had been measured many times the estimates would have had a normal distribution centered on a value close to the correct value and with a standard deviation of the order of 1 in. or the chain. Inevitably there is some variation in their characteristics and the . Nevertheless. The sum of these errors gave the total error in the measurement of the baseline and. would involve a small error.
no matter what the relative cost per hour. devoted to the work of the inventor Thomas Edison (1847–1913) who invented the first practical filament light bulb in 1879. Assuming that the lifetimes have a normal distribution then what is the probability that a particular bulb will have a lifetime of more than 1. From Table 10.5 − 0. Clearly. if the effort of changing a bulb is a primary consideration. People on Trolleys and Under-Used Resources Occasionally a story hits the news media in the UK concerning an individual left on a hospital trolley in a corridor for several hours.1 the area between the mean and 2σ from the mean is 0. In the museum in Fort Myers. We would not be rushing to pay the shop extra money if we had a bulb of unusually long lifetime.47725 so the area in the tail more than 2σ from the mean in the positive direction is 0. Hence the proportion of bulbs with a lifetime of more than 1. 10. one of the original bulbs is still burning. If we bought a light bulb and it only lasted 700 h would we be justified in complaining? Not really — in the sale of millions of light bulbs some would inevitably fail early. We now consider a particular brand of light bulb that is found to have a mean life of 1.200 h is about 1 in 45.000 h and a standard deviation of 100 h. Again the pattern of use must be taken into account since constantly turning a bulb on and off subjects it to thermal shocks that shorten its lifetime.120 Everyday Probability and Statistics one that most people are concerned with is their lifetime. A more expensive light bulb that lasts longer may be preferred to a cheaper one with a shorter lifetime. inadequate and failing — or so the tabloid press would have you believe! . It has never been switched off. We should also note that this is also the probability of a lifetime of less than 800 h.47725 = 0. Florida.2.8.200 h is 2σ from the mean we are interested in the area of that part of the normal distribution that is more than 2σ from the mean in the positive direction.200 h? Since 1.02275. the National Health Service is under-resourced.
Let us see what the implications of this argument are by considering a hypothetical hospital with 1. estimated over a long period of time. From the light-bulb example we know that the area of a single tail more than 2σ from the mean is 0. then clearly on 50% of days. on average. (b) On how many days per year. It could be argued that although the times of occurrence of these eventualities are unpredictable it is known that they occur from time to time so provision should be made for them. staff cannot be engaged and dismissed to match demands fluctuating on a timescale of a few days then some inefficiency must be present in the system. is the hospital working at less than 90% capacity? Since 90% capacity is 900 beds. This is the problem of providing any resource for which there is a fluctuating demand. If the demand were constant then the resource could be made to exactly match the demand. every demand would . The demand for beds. on average. Now.02275 = 8 days to the nearest whole number. the average demand. the hospital resources are being underused.000 beds were required then the demand would be more than 2σ from the average. or 188 days per year. A cold snap or an influenza epidemic may mean a sudden large increase in the admission of elderly patients. Sometimes the excess demand can be moved to a neighboring hospital but if all local hospitals are under pressure then the hospital-trolley expedient becomes necessary. suggests that the average demand on any day is 900 beds with a standard deviation of 50.000 beds.02275 so the number of days per year that demand cannot be met is 365 × 0. is the hospital unable to provide all the beds required? If more than 1. on the (dubious) assumption that the daily demand follows a normal distribution we consider the following questions: (a) On how many days per year.The Normal (or Gaussian) Distribution 121 The provision of a health service is made complicated by the fact that the demands on it are spasmodic and unpredictable. Since. under any reasonable employment regime.
should be aware of the problems being faced by hospitals by the very nature of the service they provide.122 Everyday Probability and Statistics be met and the resource could be provided in the most economical way. . but not the right kind of beds. for example. Again. a train crash or a terrorist attack. not of good and bad. and should therefore be able to take an objective view. and to the person affected. By spending enough money a system can be set up in which hospitals can meet every demand and. intensive-care provision is expensive and the number of beds devoted to it is limited. The detailed structure of the hospital provision also plays a role so that. hospitals can be run economically — but not both at once. and his or her relatives. by ensuring that excess capacity is kept to a minimum. In some circumstances the hospital may have spare beds. such stories are often driven by political agendas. By increasing the number of beds to 1050 the incidence of being unable to meet demand falls to 1 day per two years. the statistics of the situation may not seem relevant. On the other hand there are then 246 days per year of underused resources — including staff. such as an epidemic. it is a question of choice. that we have used to illustrate the hospital-capacity problem. The normal distribution. However. those who are not directly involved. The stories of patients on trolleys are certainly worthy of report. The actual situation that hospitals have to face will actually be much worse than that described above — demands can increase dramatically due to some emergency. alternatively. will certainly not be an adequate description of the extreme and rapid fluctuations of demand that can occur. Employing agency staff on a short-term basis as needed is another possibility — but agency staff are considerably more expensive than permanent staff. Managers have various ways of trying to optimize the efficiency of their hospitals — moving patients between hospitals to try to balance over-demand in one hospital by spare capacity in another has already been mentioned. This is all a question of priorities.
Write in the form 3n the following: (i) 32 × 34 × 35 4 (ii) 33 (iii) 313 1/10 10. The average daily sale of a daily paper is 52.The Normal (or Gaussian) Distribution 123 Problems 10 10.2. On how many days a year will its sales be: (i) below 47. and the standard deviation of the following set of numbers: 2 4 6 7 8 10 12 10. the variance.3.000 with a standard deviation of 2.000 (ii) above 55.000.1. Find the average.000? .
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at birth. that most babies weigh about 3 kg. 2. 3. etc. 10.984 kg. or variance (Sec. or a little more. which would be best expressed by the standard deviation.591 kg. if any. which no man could number (Revelation. of the babies' weights.3). Again. Another quantity of interest would be the spread of the weights. there was. Regional health offices record the birth weights of all the babies in its area and each sends a complete list of weights to the ministry. New Testament) 11. or some other electronic medium for computer input. She flicks over the pages and sees the list of weights 3.1. 125 .228 kg.Chapter 11 Statistics — The Collection and Analysis of Numerical Data A great multitude. then the average could be found quickly. and she just receives confirmation of what she knew anyway. The Minister is now presented with a document about the size of a large telephone directory containing about three-quarters of a million numbers. A useful number is the average of the weights and if the information were provided by each region on a CD. Too Much Information We consider a Minister of Health who wishes to keep a check on the birth weights of babies born in each year so as to be able to see if there are any significant trends in the country as a whole. The average from year to year could then be compared to see what trend. if the information was provided in an electronic form this could readily be found.
While that is a perfectly correct definition of the variance. The bar over the top means "average of" so x2 means the average of x2 where the individual numbers of set A are the x s. it does challenge the imagination somewhat. We also found previously that the average of set A. and the way that we worked it out for the distributions A and B. 11. If each region had submitted the number of babies born. The weights both in the individual regions and in the country as a whole would have a normal distribution.3 we described the variance of a set of numbers as: the average of the square of the difference between each number and the average of the numbers. We now apply this way of calculating variance to the sets A and B given in Sec.2. that . With a little mathematical manipulation the variance can be expressed in an alternative way as: the average of the square minus the square of the average. To see how to combine the regional estimates into a national estimate we have to do some calculations.3 to check that it gives the same results as we found previously. their average weight and the standard deviation (or variance) of their weights that would be a complete description of the situation in their area. 10. 7 (11. Set A Square 7 49 8 64 9 81 10 100 11 121 12 144 13 169 The average of the square is x2 = 49 + 64 + 81 + 100 + 121 + 144 + 169 = 104. 10.1) Notice the terminology x2 .126 Everyday Probability and Statistics Actually there was no need for the local regional health offices to have sent all the individual weights in the first place. Another Way of Finding the Variance In Sec.
Since. . From Regional to National Statistics We now consider one of the regions. for which the number of babies born was Nj .4) as found previously. the average weight was wj .2) The average of the square is x2 = 1 + 36 + 81 + 100 + 121 + 196 + 361 = 128 7 (11.6) That means for each of the regions.5) We find by rearranging 2 wj = Vj + wj 2 . 2 Vj = wj − wj 2 . Similarly. designated by the letter j. is 10 so the variance is V = x2 − x2 = 104 − 102 = 4.3) Again the average is 10 so the variance is V = x2 − x2 = 128 − 102 = 28 (11. We can now use these results to see how to convert the regional birth weight data into that of the nation as a whole. (11. since we are given Vj and wj . 11.3.Statistics — The Collection and Analysis of Numerical Data 127 ¯ we indicate as x. for set B Set B Square 1 1 6 36 9 81 10 100 11 121 14 196 19 361 (11. (11. by our latest definition of variance. the same as we found previously. 2 we can find wj . and the variance of the weights was Vj .
8) is the sum of the squared weights of all the babies in the country and dividing by the total number of babies gives the average squared weight. That is: 2 wall = M 2 j=1 Nj wj . The divisor in (11.7) is the total number of babies born so dividing the total weight of the babies by the number of babies gives the average weight of the babies for the whole country. The average baby weight is wall = M j=1 Nj wj . and the variance of their weights — and all that each region had to send in was three numbers.128 Everyday Probability and Statistics We are now in a position to find the average baby weight and the average of the square of the babies' weight for all the babies combined from all the regions. The fourth column shows the standard deviation (square root of the variance) of the weights in each region. M j=1 Nj (11.1 we show the results of such an exercise for a hypothetical set of data. M j=1 Nj (11.8) The top of Eq. In Table 11. (3. their mean weight. The term Nj wj is the total weight of all the babies born in region j and summing this over all the regions gives the total weight of all the babies born that in the year for the whole nation (about 2.7) The summation symbol was explained in relation to Eq.000 tonnes for the UK!). By an exactly similar process we can find the average squared weight of the babies. (11. 2 Now by combining wall and wall we can find the variance for all the babies as 2 Vall = wall − wall 2 . (11.9) The Minister has now combined the information from the regions to find the total number of babies born.2). .
05987 σ (kg) 0.Statistics — The Collection and Analysis of Numerical Data 129 Table 11.101 3.657162 9.251 0.267 0.148084 9.178 3.231 0.220 w2 (kg2 ) 9.997 3.077850 10.1.225 93. 3.515 64.05987)2 = 0. the economy.250 0. and their average weight. (11. demography. Region South South-east South-west West Central East North-east North-west Far-north Outer Isles Average (sum) N 66. The UK Office of National Statistics collects and analyses statistical data over a wide range of national activities concerned with health.8).062180 9. The variance of the weights is 2 V = wall − wall 2 = 9. The Minister now has available all the information she needs.219 (700.362 26.212 82.215514 9.002 2.641.621 76.424592 − (3.358 37.0617875 kg2 corresponding to a standard deviation σ= √ V = 0.996 3.062 2. respectively.237 0. 700. are found from Eqs. and business. Many of its analyses are based on the assumption of a normal distribution but there are other kinds of distribution that we shall now consider.219 0. Data for the birth weights of babies born in different regions. employment.7) and (11.226 0.099 3.296 108.117197 10.424592 The averages. at the bottom of columns three and five.496 41.239 0. .337 104.038516 9.438845 9.039130 10.011 3.224 0.167 3.185 3. The table directly gives the total number of babies born.641) w (kg) 3.05987 kg.2486 kg.672370 9.
089 m 0.086 m 0. (11.130 Everyday Probability and Statistics Problems 11 11. Use Eq.372 m Standard deviation 0.267 m 1.1. Four schools in a particular district send the following information concerning the heights of their 11 year old boy pupils.1. 11.2) to find the variance of the set of numbers given in problem 10.091 m 0.090 m What is the mean height and standard deviation of heights of 11 year old boys in the district? .352 m 1. School 1 2 3 4 Number of pupils 62 47 54 50 Mean height 1.411 m 1.2.
We decide to divide time up into one-minute intervals and in each interval the number of cars passing is recorded. he wrote a paper with a title that translates into English as Research on the Probability of Criminal and Civil Verdicts in which he developed and described what is now known as the Poisson distribution. of considerable importance. 12. 1781–1840) 12. To illustrate an application of the Poisson distribution let us consider the problem of taking a census of traffic flow on a quiet country road.Chapter 12 The Poisson Distribution and Death by Horse Kicks Life is good for only two things. was a French mathematician and physicist whose contributions covered an enormous range of topics and who must be considered as one of the scientific giants of the 19th century.1. 131 . In 1 h.1). Siméon-Denis Poisson. discovering mathematics and teaching mathematics. we now describe. (Siméon-Denis Poisson. and one of these. 1781–1840 (Fig. Although it never again featured in any of his mathematical publications this distribution turns out to be of great importance in dealing with the statistics of many real-life situations. Rare Events Although the normal distribution is by far the most important naturally occurring distribution there are others. In 1837.
132 Everyday Probability and Statistics Fig. 12. The numbers of cars per one-minute interval were small — for many intervals none. . which shows the number of intervals for each number of cars passing. Fig. We can illustrate the mathematical nature of this distribution by taking another situation where the number trials (minute intervals in the case we have just taken) is large but the numbers of events per trial (the passing of cars) are small.2. for less intervals two. It is to explain the variation in frequency in situations like this that Poisson developed the theory of what is now called the Poisson distribution. for somewhat fewer intervals one. and for even fewer intervals three. e the 60 numbers recorded are 0 1 0 0 0 0 1 0 0 1 1 0 0 1 1 0 2 1 2 2 1 0 1 1 0 2 0 2 1 0 1 0 1 0 1 0 1 0 0 0 0 3 1 0 1 1 1 1 0 2 1 0 0 0 0 0 1 0 3 1 These results can be represented by the histogram. 12.1. Sim´ on-Denis Poisson.
The first thing we can find is the average number of errors per page.2. 12. the typist makes occasional typing errors although. When the pages are proof-read the numberof-errors-per-page are found.2. Histogram for numbers of cars passing per minute on a quiet country road.The Poisson Distribution and Death by Horse Kicks 30 133 25 Number of 1 minute intervals 20 15 10 5 0 0 1 2 Number of cars passing in 1 minute 3 Fig. Typing a Manuscript We consider the task of typing a manuscript of 100 pages. The total number of errors is (30 × 0) + (36 × 1) + (22 × 2) + (9 × 3) + (2 × 4) + (1 × 5) = 120 and. Number of errors Number of pages 0 30 1 36 2 22 3 9 4 2 5 1 More than 5 0 This distribution is illustrated as a block diagram in Fig. Being human. 12. for a good typist. 100 . the average number of errors per page is a= 120 = 1.3. hence. 12. the average number of errors per page would be small.2. and hence prone to error. with the following results.
The numbers of pages with particular numbers of typing errors. the number of pages with two errors. we multiply that by a to obtain 30 × a = 36. a. The nearest whole number is 9.134 Everyday Probability and Statistics 40 35 30 number of pages 25 20 15 10 5 0 0 1 2 3 4 5 number of errors per page Fig. . The Poisson distribution is completely defined in terms of the average.3. Now.8. which is the number of pages with one error. we multiply the number of pages with two errors by a and divide by 3 22 × a ÷ 3 = 8. we multiply the number of pages with one error by a and divide by 2 36 × a ÷ 2 = 21. Starting with the number of pages with no errors. 12. The nearest whole number is 22.6. Now. the number of pages with three errors.
10. The term at the bottom is factorial r that was described in Sec. cars passing in one minute) where the average a is F(r) = e−a ar . 12.2. the number of pages with five errors. Now we multiply the number of pages with four errors by a and divide by 5 3 × a ÷ 5 = 0. 4. The nearest whole number is 1. The nearest whole number is 3.The Poisson Distribution and Death by Horse Kicks 135 Now.1) The two terms at the top of this formula have been explained in Sec. but the actual number of pages with four errors is 2. The Poisson Distribution as a Formula Expressing a Poisson distribution in a mathematical form. we multiply the number of pages with three errors by a and divide by 4 9 × a ÷ 4 = 2. These calculated and observed numbers of errors do not exactly agree in this case. the fraction of cases with r events (error on a page. r! (12. Although this gives the numbers of pages with different numbers of errors related by a sequence of factors it is better to have an explicit formula to give the number of pages with a specified number of errors.7. Let us see how the formula works for the typist's errors where the average number per page is 1.1.7. which is partly due to taking nearest whole numbers in the calculated values. With a calculator we find that .3.5.
to the nearest whole number. 4! e−1.2 = 0. .23 = 0.3012 0! since. 2! e−1.2 × 1. we can write F(0) + F(1) + F(2) + F(3) + F(4) + F(5) + F(6) + · · · = 1.3614. Like any sensible distribution giving probabilities the sum of the probabilities for all the possible outcomes must be 1 — that is to say that it is completely certain that there must be some outcome.20 = 0.2 × 1. 1.0260.9984 since the probabilities from F(6) onwards are missing.14 pages with one error.20 = 1 and 0! = 1.2) where. The sum of the probabilities for F(0) to F(5) for the typing errors is 0.2 × 1.0062. 3! e−1.25 = 0. The fraction of pages with one error is F(1) = Similarly F(2) = F(3) = F(4) = F(5) = e−1. 5! giving 22 pages with two errors giving 9 pages with three errors giving 3 pages with four errors giving 1 page with five errors.0867.136 Everyday Probability and Statistics e−a = e−1. Once again there is a discrepancy for the number of pages with four errors but this is due to the rounding-off of the numbers — there cannot be 36.2 × 1. on the left-hand side.2 × 1. 30 of them will be error-free.2 × 1.2169. Because there are 100 pages in all. (12. 1! giving 36 pages with one error. e−1.22 = 0. Thus. The fraction of pages with zero errors is thus F(0) = e−1. as we have previously seen. we have an infinite number of terms.3012.2 = 0.24 = 0.
5 and (b) average 3.4 the distributions for averages 0.2 0.7 0.15 p(x) 0.3 0. Poisson distributions for (a) average 0.4 0.1 0.4.4.5). 12.The Poisson Distribution and Death by Horse Kicks (a) 0. 12.5 and 3. Death by Horse Kicks One of the most famous applications of the Poisson distribution was made by a Russian statistician Ladislaus Bortkiewicz.25 p(x) Poisson distribution with mean 3.0.2 0. To get a feeling for Poisson distributions we now show as histograms in Fig.6 0.5 137 (b) 0.0. 1868–1931 (Figure 12. .5 0.05 0 0 2 4 x 6 8 10 Fig.1 0 0 1 2 3 x 4 5 6 Poisson distribution with mean 0.0 0. 12.
138 Everyday Probability and Statistics Fig. from 1875 to 1894. and analyzed them statistically. Statistics of deaths by horse kicks for 200 cavalry-unit-years. .5.1. From his figures the total number of deaths was 65 × 1 + 22 × 2 + 3 × 3 + 1 × 4 = 122. 12.1. Bortkiewicz took the record of such deaths for 10 cavalry units of the Prussian army over a period of 20 years. Number of deaths per unit per year 0 1 2 3 4 More than 4 Number of unit years 109 65 22 3 1 0 At the end of the 19th century cavalry units formed a component of most armies and once in a while someone in a cavalry unit would be killed by a horse kick. Bortkiewicz showed that the number of deaths per cavalryunit-year followed a Poisson distribution. Ladislaus Bortkiewicz. Table 12. For the 200 cavalry-unit-years the information he derived is shown in Table 12.
1.61 × 0.61 × 0. is N(0) = 200 × e−0.7 0! or 109. 3! e−0.0 (4).1.6 (1) 4! and the similarity of these figures to those given in Table 12.611 = 66. Equation (12. 12.1 will be evident.61 × 0. rounded off to the nearest whole number. with the rounded-off figure in brackets. N(1) = 200 × N(2) = 200 × N(3) = 200 × N(4) = 200 × e−0. given the average.61. This is exactly the figure found in Table 12. mainly for bombarding London and the South-east of .3 (66).613 = 4. Flying Bomb Attacks on London During the final period of World War II the Germans designed a weapon. Indeed.5.612 = 20.61 × 0. a χ2 test shows no significant difference between the expected numbers we have just calculated and the observed numbers found by Bortkiewicz.5.2 (20).The Poisson Distribution and Death by Horse Kicks 139 so the average number of deaths per cavalry-unit-year was 122 200 = 0. The Poisson distribution is important in many areas of life and we now give two further examples. Similarly. 2! e−0.610 = 108.614 = 0.1) gives the proportion of cases with r events (deaths) so the number of cavalry-unit years that no deaths would be expected. Some Other Examples of the Poisson Distribution 12. 1! e−0.61 × 0.
and side winds so that their falls were distributed all over the London area.1 1. which could hardly be missed. The number flying bombs landing within 1/4 km2 squares in London.140 Everyday Probability and Statistics Table 12. they were a very imprecise weapon and depended for their success on the large size of the target. London was divided up into 576 squares of side 0.5 211. There was a suspicion that the fall sites of the bombs tended to occur in clusters so it was decided to apply a statistical test on the fall of the bombs in London up to that time. The final column of Table 12. known as the flying bomb. So the average number of bombs per square equaled 535/576 = 0. Number of bombs 0 1 2 3 4 5 Number of squares 229 211 93 35 7 1 Poisson distribution 227.2. They were at the mercy of unpredictable tail winds.3 98. Of the 8600 that were launched roughly one-half penetrated the defenses.2.3 England. driven by a simple ram-jet motor that traveled at 575 km hour−1 mph and had a range of 240 km.1 30. It was a small pilot-less aircraft. Clearly.2 shows the numbers expected from the Poisson distribution. They were launched from a ramp pointing toward London and when the fuel ran out they fell to the ground as a bomb containing 1 ton of explosive. the rest being shot down by anti-aircraft fire and fighter planes.5 km and the number of bombs that had fallen in each square was recorded.9288. The similarity between the . The results are shown in Table 12. head winds.4 7. The total number of bombs falling in this period was 211 × 1 + 93 × 2 + 35 × 3 + 7 × 4 + 1 × 5 = 535.
1 + + = 0..15 1. This gives: P( ≥ 8) = 1 − F(0) − F(1) − F(2) − F(3) − F(4) − F(5) − F(6) − F(7) or P( ≥ 8) = 1 − e−1. the combined probabilities of 8.e. 11 etc. up to infinity.The Poisson Distribution and Death by Horse Kicks 141 observations and the Poisson distribution is evident and a χ2 test shows that there is no significant difference between the two. say testicular cancer.1 1 + 1.16 1. In testing for the significance of an outcome greater than the expectation we find the total probabilities of the actual result and all results which are more extreme — i. 10. In a city with 50.12 1. 12. 8 of them are found to have the condition. Is there any evidence that this is a significant cluster of incidence? The expected number for 50. A condition affecting young men is known to occur with a frequency of 2. is higher that would be expected in a particular region then one might look for contributory causes of a local nature — particular industrial processes or the use of pesticides.14 + + 2! 3! 4! 1. The idea of clustering was not supported by this analysis. Clustering of a Disease The medical authorities are often interested in what is known as clusters of particular diseases.17 + + 5! 6! 7! .13 1. Let us consider a hypothetical example.2.000020124.2 per 100.5. 1.2) the sum of all the probabilities of a Poisson distribution must equal unity we can find the probability of 8 or more cases by taking away from unity the sum of the probabilities of 7 or less cases. 9.000 men. If the incidence of a condition.000 young men.1 so we are clearly interested in a Poisson distribution with that average. Since from (12.000 young men is 1.
000. This may be genetic. . a compromise number of operators can be found that will be acceptably economic while.000.000 men so. one of those sets of 50.. dietary or of some other origin and would take further investigation to find out which. Suppose the answer had been 0. (i) The number of calls at a call center per minute. One aspect of servicing a large business premises is to replace failing light bulbs. leading to excessive customer dissatisfaction. Statistics must be treated with care — they can be misleading but also.2 per 100. at the same time. which will be in terms of Poisson statistics. There might be an average of 7 failing per day but the actual number per day. will give a Poisson distribution with average 7. i. those with bad intent can attempt to use them to mislead! 12. This would come about just by the vagaries of a random distribution of cases so one could not then conclude that there was a significant cluster. (ii) The number of light bulbs that fail per day in a large building. By analyzing the demand. sometimes. In judging this significance we have to beware to not jump to conclusions. cultural. having a sufficient number of operators to meet the greatest peak demand will mean that operators are often idle.142 Everyday Probability and Statistics Now. Then there are 100 sets of 50. taken over a long period of time. on average. giving an acceptable service to the customer. leading to inefficiency in the running of the call center.5. this is a very small probability so we may conclude that this is a significant cluster so that some local factor is operating to produce the condition. one chance in 100 — could we conclude then that the cluster was significant? Let us say that there are 5. At the other extreme.e.3.01.000 young men would give the outcome we found. Some Further Examples To finish our description of the Poisson distribution we give here a sample list of other situations in which the distribution plays a role.000 young men in the region that gave the incidence rate of 2. By having too few operators the customer may have a long wait.
e. What is the probability that at least one will fall on the Earth within the next 1. DNA is the material within the animal cell that controls the genetic make-up of the individual. On how many days per year would the number failing be 15? .000. Large asteroids fall onto the Earth at the rate of 1 per 10. The number of faulty components in a manufacturing process follows a Poisson distribution with a mean of 1 per 100 components. in a statistical sense.3.000. When exposed to radiation DNA can become damaged and will then reconstruct itself. sometimes but not always giving modification of the original genetic information it contains — i. The average number of electric light bulbs failing per day in a factory is 10.The Poisson Distribution and Death by Horse Kicks 143 (iii) The number of mutations in a given stretch of DNA for a given dose of radiation.000 years. (iv) 5 faults? 12. a mutant has been created. (ii) 1 fault. The number of mutations for a given radiation dose will be a matter of chance but. Problems 12 12.. the numbers will follow a Poisson distribution. What is the probability that a random selection of 100 components will have: (i) No faults. It consists of a long chain of many thousands of units of which there are four basic types.000 years? 12.1. (iii) 3 faults.2.
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If the sample is taken just from those entering an expensive jewellery store in the West End of London the results of the poll would be as unlikely to be representative of the whole population as sampling just those entering a fish-and-chip shop in a run-down northern inner-city area. Politics is not an exact science (Bismark. will contest national and local elections. polling organizations collect the opinions of samples of the population and then give the estimates of the voting intensions of the whole population based on the declared intentions of those sampled. religion or region of the country. To obtain a good sample is a complicated exercise.1. age. The clientele of both those establishments should be a part of the polled sample. by social class.Chapter 13 Predicting Voting Patterns Die Politik ist keine exakte Wissenschaft. for example. The major polling organizations show great skill in getting good samples and in general 145 . Ideally. As an alternative. There are various ways of categorizing members of the population. gender. but not its totality. To start with one must be sure that those sampled are a good representative selection of the population as a whole. from time to time. 1815–1898) 13. one would like to know the voting intention of everyone who intends to vote but this is tantamount to actually holding an election. income. ethnicity. Election Polls In democratic countries there is always a natural interest in knowing the support given to the various parties who.
000 representative members of the public is taken and indicates the support as follows: National Democrats 522 People's Party 478 Translated into proportions. The Chicago Daily Tribune newspaper. It turned out that the sampling of public opinion had been carried out by telephone. They have learned from many early mistakes. Dewey. pressured by a printing deadline before the actual result was known. 13.2. that the choice is basically between two alternatives. Thomas E.4% advantage over his opponent. Polling Statistics We consider a situation where just two parties.478 On the face of it the National Democrats have a clear lead and must be considered as likely to win a majority of the popular vote. brought out an edition with a huge headline "DEWEY DEFEATS TRUMAN". In the event. the National Democrats and the People's Party. Truman won the election with a 4. initially. are offering themselves to the electorate. For the purposes of our discussion of polling we shall assume that the sample of the population is a good representative one and.146 Everyday Probability and Statistics their results are now reliable. However. had a comfortable lead of between 5% and 15% over his Democratic opponent. can it be said that they are certain to do so? Clearly not — the . Truman. Harry S. In the 1948 presidential election in the United States all the polls showed that the Republican. Dewey certainly had majority support from those that owned telephones in 1948 but hardly any support from the many without a telephone at that time. the voting intentions of the sample become National Democrats 0. A poll of 1. Later we shall consider the situation when there are three or more choices of party or individual.522 People's Party 0.
in this case 1.5 of the votes and this value is also shown in the figure.1) Let us see what this means with our hypothetical example. the actual proportion of the electorate supporting the National Democrats could be different from p. We have r = p = 0.1. We shall now find what these probabilities are. Let the size of the poll be N. in this case 0. Getting a majority of the popular vote means getting more than 0.478 = 0.522 × 0. but small.5 and the area to the left is the probability that the People's Party will come out on top. the poll sample included a majority of National Democrats supporters. probability that the People's Party will win and we now see how to estimate this mathematically.522. The probability that the National Democrats will get more of the popular vote is the area to the right of the line indicating r = 0.0158. .000. However. 13. say r. and the proportion of the poll sample supporting the National Democrats be p. as illustrated in this figure. The reality of the situation is that there is a finite. The mathematics tells us that the most likely outcome of the election is that a proportion p will vote for the National Democrats.2) (13.522 and hence σr = 0. has unit area under the curve. just by chance. N (13. Theory shows that on the basis of the polling evidence the unknown value of r has an approximately-normal probability distribution with mean r=p and with a standard deviation σr = p(1 − p) . 1000 We show this probability distribution in Fig.Predicting Voting Patterns 147 possibility must exist that there is actually a majority in favor of the People's Party but that. The form of the normal distribution.
.0158 from the mean of the distribution.55 0.4 0.65 Proportion voting for National Democrats Fig.081.5 0.919.522 − 0.5 (0. Hence the probability that the National Democrats will get most of the popular vote is the sum of the area to the right of the mean (0.5 would still be 1.4 standard deviations from the mean is 0.148 Everyday Probability and Statistics 30 25 Relative probability 20 15 10 5 0 0.081 — unlikely but not impossible.4σr 0. If the analysis had been done from the aspect of looking at the probability distribution of support for the People's Party then the distribution would have been centered on 0.45 0.4σp ) giving a total probability 0. but in the other direction.1.478 but the standard deviation would have been the same. From Table 10. 13.6 0.1 the area from the mean out to 1.5 is 0.5 = 1. and the probability deduced for the likelihood that the People's Party would come out on top would be found as before — 0. The line r = 0. The probability distribution of those voting for the National Democrats.419 which is the area out to 1.4192.0 − 0. The line r = 0.5 because it is one half of the area under the curve) and the area to the left of the mean out to r = 0.919 = 0. This means that the probability on the basis of the poll that the People's Party will win is 1.4σp from the mean of the distribution.
this is not necessarily so. 0.522.000 and. However. as prevails in the UK. to 0.081.478 = 0.0112. The size of the poll has a considerable effect on the significance of the results.0112 from the mean of the distribution.0 standard deviations from the mean is. In a system giving proportional representation. differing slightly in their polling techniques but all giving worthwhile estimates of voting intentions. party support may vary through a pre-election period so the data for .1. from this.3. 2000 13. the proportion of those voting for the National Democrats was 0. The chance that the People's Party would win has reduced from 0. care should be taken in making such combinations — for example.023. Using (13.97725. about one chance in 43.2) the standard deviation now becomes σr = The line r = 0. Combining Polling Samples There are a number of different polling organizations.Predicting Voting Patterns 149 We have stressed all the time the concept "proportion of the popular vote" because how that would turn out in terms of seats in the parliament would depend on the electoral system. In a first-past-the-post system.522 − 0.5 is now 0. the probability of the National Democrats getting more of the popular vote is 0. or about one chance in 12. the party with a majority of the popular vote would also have the most seats. Assuming that they are all equally valid then it is possible to combine them to give an estimate of voting intentions more precise than any individual one of them.5 = 2. again.522 × 0. The area from the mean to 2. from the same proportions in a larger poll.0σr 0. 0.47725 and. from Table 10. We suppose that the poll size was 2.
0158.521) to the critical 0. The area from the mean of the probability distribution (0. .00913 σXYZ = 1.000 National Democrats 521 National Democrats 517 National Democrats 495 People's Party 479 People's Party 483 People's Party 505 For poll X alone: σX = 0. three polling organizations in a two-party election: Poll X Poll Y Poll Z Poll size 1.511 on a poll of size 3. 1000 0. as with all human activities. When the three polls are combined we find that the indicated fractional support for the National Democrats is 0. equal to 0.5 level of support is.1.150 Everyday Probability and Statistics the separate polls should not be widely separated in time.33σX from the level 0. making the probability of a National Party victory 0.511 × 0. the probability of victory by the National Democrats is 0.000 Poll size 1.376 from poll Z alone (it predicts a People's Party victory). Similarly.1 this gives a probability of 0. from Table 10.521 × 0. Polls taken at different times during an election campaign can be quite informative of trends in the voting intensions of the electorate.011 and support for the National Democrats is 0.489 = 0. We consider the following information from.000.885 for a National Democrat victory.0158 σX = 1.860 from poll Y alone and 0. However. From Table 10.5. For this combined data σ'XYZ = 0.479 = 0. there is always some unpredictability even when the polls are properly conducted.00913 3000 0.000 Poll size 1.20σXYZ from the critical 0.5 level.908.021 Support for the National Democrats is thus 0. beyond which they are the majority party.408.
0097. 2000 σA = 0. B.25 × 0. pB = 0.0110. and C. and pC = 0. and pc for party C.4) Swapping around A. Polling with More than Two Parties Once there are more than two choices of party or individual. (13. and C gives the means and standard deviations for the support of the other parties.75 = 0.35. Consider a case where a poll of size 2.6 = 0.4.000 gives pA = 0.25.0107.4 × 0.3) To find out what the range of possible support is for party A. the simplest approach is to consider a two-choice scenario where the choice is party A or not-party-A.35 Party C ¯ pC = 0. We now consider a three-party system.25 σC = 0. the analysis of polls is somewhat more difficult. 2000 σB = 0. pB for party B.4 Party B ¯ pB = 0.4. N (13. We have the condition that pA + pB + pC = 1. B.Predicting Voting Patterns 151 13. Then the expected means and standard deviations for the parties are: Party A ¯ pA = 0. with mean pA and standard deviation σA = pA (1 − pA ) = N pA (pB + pC ) . with parties A.65 = 0. The probability distribution of the proportion voting for party A can be regarded as a normal distribution.35 × 0. for which a poll of size N gave a proportion pA for party A. 2000 .
37 × 0.55σLab 0. with the actual voting percentage in parentheses: Labour Conservative Liberal Democrat 37.152 Everyday Probability and Statistics A precise mathematical analysis of this situation is quite complicated. from the point of view of getting a general impression of where the parties stand.2 standard deviations below and the Liberal Democrat vote was 2.5% 24. 0. the probability estimates for getting a majority of the vote were the same whether one considered the distribution for one party or the other. when there were two parties. and by how much the actual result is likely to differ from the mean.3%) (32.3%) (22. Similarly the Conservative vote was 0. the minor parties did better than the polls suggested. A gain by one party relative to the poll estimate completely defined the loss by the other and the probabilities of gain and loss were consistent. with a poll size of 2. As an illustration a poll taken just before the 2005 election in the UK. On the basis that votes were either Labour or non-Labour.011 below the poll prediction.1%) The minor parties accounted for the remaining few percent.011 2000 . a particular gain by one party involves a net loss by the other two but the pattern of loss between the other two parties is not defined. gave the following percentages.017 = 1. Here. the standard deviation for the Labour vote was σLab = so that the actual vote was 0. However. the individual means and standard deviations as given above serve their purpose well.0% (35.000. with three parties.0% 32.0 standard deviations below their respective poll predictions. Clearly.63 = 0.
Predicting Voting Patterns
13.5. Factors Affecting Polls and Voting
Apart from the problem of selecting a good representation of the population for polling there are other factors that can distort the poll results. For various reasons — perhaps out of sheer mischief — people may say that they are voting for a party other than the one they intend to vote for. Another factor is that if the polls are indicating a strong lead for a party then its supporters may not vote on the grounds that their votes will not make any difference. Many countries forbid the taking of sample polls in the week or so before the election is due to take place to prevent the polls themselves from influencing the voting pattern of the electorate. By-and-large polls give a good indication of how people intend to vote although it is not clear that they serve any important democratic function. Some individuals may, indeed, be influenced by the polls in the way they vote. If a small party with extreme views is indicated as having significant support then those generally sympathetic to those views may be encouraged to vote for that party since they would not then feel they were part of a freakish minority in doing so. However, whether or not we would agree with their choice, we must accept that, that is what democracy is all about! There are other influences, to some extent unpredictable, which can affect the results of a General Election. In the 1992 UK General Election the Labour Party, headed by Neil Kinnock, was ahead in the opinion polls and seemed to be heading for a comfortable victory. Just before the election, the Labour Party held a rally in the Sheffield Arena that was conducted in a triumphal rather noisy American style that was jarring to the UK public. In the opinion of many commentators this lost the Labour party some of its support. In addition, the influential and widely read tabloid newspaper, the Sun, brought out a pre-election edition with the front page showing a large picture of Kinnock with a headline requesting that if Kinnock won the election that day then the last person to leave Britain should turn out the lights. The Sun is also known for its scantily-clad young ladies
Everyday Probability and Statistics
who appear on page 3. In that particular issue the lady on page 3 was not young and very overweight with the caption "Here's how page 3 will look under Kinnock". It was thought that this intervention by the Sun, intellectually irrelevant as it was, influenced enough votes to give a Conservative victory. In the event, John Major became Prime Minister with his party achieving the highest proportion of the popular vote ever obtained by a British political party, albeit that he had a reduced and rather small majority in the House of Commons. The Sun certainly thought that its influence had been decisive. On the day following the election its headline, in typical Sun style, was "It's the Sun wot won it". Subsequently the Sun switched its support to the Labour Party (relabelled 'New Labour') which won the three subsequent elections. Such are the factors that influence democracy!
Problems 13
13.1. A large scale poll of 3,000 electors gives 1,560 supporting party A and the remainder supporting party B. What is the probability that party A will get more than 50% of the popular vote? 13.2. In a random sample of 200 baby snakes 92 were found to be male. What is the probability that for the whole population of snakes more than 50% are male?
Chapter 14
Taking Samples — How Many Fish in the Pond?
…few are chosen (St Matthew)
14.1. Why do we Sample?
Consider that we have some large number of entities — they could be human beings, animals, plants, or material objects — and we wish to determine some characteristic that they possess. Some examples of this are: (i) (ii) (iii) (iv) (v) (vi) The average height of 14-year-old boys within a country. The distribution of hat sizes of males in the United Kingdom. The average house income of those living in social housing. The average weight of codfish living in the North Sea. The number of people watching a particular television program. The proportion of faulty components made in a particular factory.
What is clearly impracticable, if not impossible, is to examine every individual entity to determine the required characteristic. It would be possible, although extremely uneconomic, to measure every 14-year-old boy or to check every component for its perfection but it is certainly not possible to catch every codfish in the North Sea. The technique that is used to estimate the required characteristic of a complete set of the entities is to take a sample — that is, after all,
Everyday Probability and Statistics
what was described in the previous chapter to estimate the voting intentions of the entire electorate.
14.2. Finding out from Samples
In sampling theory, the complete set of entities, whatever it is, be it humans, fish or brake pads, is referred to as the population. This population has some property of which we wish to know the average ¯ value, ap , and the standard deviation, σp . The technique that we use to estimate these quantities is to measure them for a sample of size n — that is to say that we take a random selection of n entities for our measurements. The individual values of the quantity a in the sample are: a1 , a2 , a3 , . . . , an ¯ and from these we can obtain the sample average, as , and the sample variance, Vs , found by the method described in Sec. 11.2. Without any other information available, the best estimate that can be made ¯ for the average of the whole population, ap , is the average for the sample, so that ¯ ¯ ap = a s (14.1)
Notice the notation here; the brackets mean "the estimate for" the quantity they enclose. Now it might be thought that the best estimate of the variance of the population, Vp , is the sample variance, Vs , but a detailed mathematical analysis shows that it is actually given by
2 Vp = σp =
n Vs . n−1
n The factor n−1 is called the Bessel correction and is especially important in its effect for small samples. For a sample size of 1,000, the Bessel correction is 1.001 but for n = 9 it is 1.125. If the sample size
then one could have less confidence that the mean of the single sample was a good estimate of the mean of the whole population. To see how to do this we first imagine the following scenario — there is a large population of millions of entities and we repeatedly take different ¯ random samples of n of them and find the sample means. or nearly so. is not available. each ¯ s themselves would have a distribution.4) . if the variance of the sample means is large. Another matter of interest when taking samples is to know how much uncertainty there is in estimating the mean of the population from that of the sample. as . if the variance of the distribution ¯ of the values of as is small. Conversely. at least for larger values of n. the reality is that when you take a sample. Theory shows that if the variance of the whole population is Vp . Now. we find Vas = ¯ Vp n Vs 1 Vs = . then the estimate of the population variance would be infinity. implying that it cannot be estimated at all — which is sensible since there cannot be any variation in a single quantity.Taking Samples — How Many Fish in the Pond? 157 were 1. Instinctively we know that the larger the sample the more reliable is the estimate of the population mean. = × n n−1 n−1 n (14. and time. The values of a statistical theory shows that. ap . n (14. then it is likely that the estimate of the population mean taken from the single sample would be a good one.3) and using (14. Taking estimated values on both sides of (14. Vp . and it is from this single sample. that you must estimate the variance of the sample mean. that is all the information you have. centered on the true average ¯ for the whole population. then the variance of the means of samples of size n is Vas = ¯ Vp . but we would like to be able to quantify that reliability.3) However. it would be a normal distribution. the true value of the variance of the whole population.2) as the best estimate of Vp .
A quantity quoted by a grower would be the minimum average weight of the batch of apples he is providing.203 0.252 0.146 0. This is the basis of the following example. buying apples from a grower.152 0. However.436 = 0. In selling apples an important factor in setting the price is the size. The individual weights of the apples.150 0.137 0. while trees do not provide apples of uniform size they can be separated by size and sold accordingly.211 0. in kilograms.192 0. the larger the sample the more likely it is to give an average close to the population average — the effect of n − 1 in the divisor of (14. A wholesaler.139 0.158 Everyday Probability and Statistics Apart from being found by analysis. An Illustrative Example We will better understand the subject matter of Sec.139 0.1718 kg. 14. selects 20 of the apples at random.0305542 kg2 .3. If Vs is small.123 0.167 (i) What is the mean weight and standard deviation of the weights for this sample? Adding the weights and dividing by 20 gives w= 3. or weight. this is an indication that there is not much variation in the values of a so that the sample mean is likely to be close to the population mean.174 0. 20 . this is a sensible result with the right sort of characteristics.611084 = 0.162 0.155 0.2 by applying the results to a numerical example.224 0. whatever the variance of the individual values in the sample.156 0.176 0.198 0. Again.180 0.4). of the individual apples. are as follows: 0. and a wholesaler would be interested in checking the grower's claim. 20 Adding the squares of the weights and dividing by 20 gives w2 = 0. 14.
1718 kg.000054682 kg2 . which is at least 0. If the claim is true.03223 kg. The grower's claim is almost certainly untrue.007395 kg. ¯ (iii) The grower claims that the average weight of the apples he supplies is more than 0.2 kg. then taking many samples of 20 from his apples would give an average sample mean of 0. and the standard deviation for the sample is σs = Vs = 0.1718 = 0.0282/0.4).0282 kg from the claim. the variance of the weights for the sample is Vs = w2 − w2 = 0.00103896 kg2 . n−1 19 This gives the estimate of the standard deviation of the sample mean as σas = ¯ Vas = 0.17182 = 0.007395 = 3.8 standard deviations or more from the mean is the area in the tail of a normal distribution and is found from Table 10. so that Vas = ¯ 0.00103896 Vs = = 0.0305542 − 0.007395 kg.Taking Samples — How Many Fish in the Pond? 159 Hence.00007 — so very unlikely. then the probability that the mean weight of the sample is 0.1 to be 0. (ii) Find an estimate for the standard deviation of the sample mean? The estimate of the variance of the sample mean is given by (14.2 − 0.2 kg or more with a standard deviation estimated to be 0. The one sample we take has a mean weight of 0. Can this claim be rejected on the basis of the sample results? If the grower's claim was correct. .
for smaller samples. we assumed that if we took many sample means they would form a normal distribution. For the manufacturer there is a fine balance to be struck. this would ensure that there were no returns but the expense of doing this would have to be passed on to the customer. make his product less competitive. General Comments on Sampling The kind of analysis that was used to test the grower's claims about his apples could equally be applied to claims about the lifetime of electric light bulbs or in other situations. instead of using a table for the normal distribution we should use a table for what is known as the Student t-distribution.5. 14. For a sample of size 20. samples. the error is not negligible but neither is it large.160 Everyday Probability and Statistics 14. a machine tool becoming defective then he might not detect this for some time and then he would suffer the double loss of needing to replace many articles and also of customer confidence in his product. S. so using the normal distribution gives simplicity at the expense of some accuracy. This would only be strictly true if we took very large. . At one extreme he could check every article before it leaves the factory.4. However. theoretically infinite. Gossett who derived the t-distribution in 1908. It also represents a cost to the manufacturer of the product who has made an article that had yielded no income. At the other extreme he could check nothing but then if the manufacturing process developed a flaw through. He needs to implement a Student was a pseudonym used by the statistician W. In examining the claim about apples. which would adversely affect future sales. and so reduce his sales. he finds it to be faulty and has to return it to have it replaced. Otherwise.a a distribution that depends on the sample size. the error would be significant. say. Quality Control It is extremely annoying to a customer if. having bought some product. for smaller samples.
01 − f from the mean.000 tested bicycles for his statistical analysis. 1000 (14. so the manufacturer must monitor his production to keep faulty products below that level.6) Assuming a normal distribution. He decides on a strategy of testing every tenth bicycle and using as a sample the last 1. 13. We will deal with this problem in a similar way to that in Sec.5) Following the pattern described in Sec.01 will be the probability of being more than 0.2.Taking Samples — How Many Fish in the Pond? 161 something between these two extremes — something called quality control.01 − x/1000 . where voting intentions were considered.000 bicycles such that there is a probability of 0.01 − f 0. He also wants to be 99% sure that the 1% fault level is not exceeded.7) The condition that there should be a probability less than 0.01 or less that 1% or more of the bicycles are faulty?" It is worth reading that last sentence again so that you properly understand the nature of the exercise. or a number of standard deviations t= 0. To illustrate the basic concept of quality control we consider a factory turning out 1. there will be a probability distribution of failure rate centered on f with a standard deviation σf = f (1 − f ) . Experience in the industry suggests that it is acceptable if less than 1% of the bicycles have a flaw.01 of a 1% failure rate is equivalent to having an area in the tail of the . 1000 (14.000 is x then the estimated proportion of failures is f= x . = σf x/1000(1−x/1000) 1000 (14.000 bicycles a day. the probability of an actual failure rate greater than 0. The question that we now ask is "What is the maximum number of failures in the current sample of 1. 13.2. If the measured number of failures in the sample of 1.
t = 2. x = 5. Fewer checks would give less reliable statistics and more would give a greater interval between the time of realizing that something was wrong and the time from the beginning of the sample when things started going wrong. t = 3. The owner of a large fish pond wishes to know how many fish it contains — at least to an accuracy of . What has been described here is much simpler than any system that would be used by a large-scale manufacturer. 14. How Many Fish in the Pond? Sampling theory can be used in various ways and here we describe an interesting application. but it illustrates the general principles involved.000 bicycles sampled.8) Finding a value of x that inserted in (14.000 checks on which to base the statistics and to be 99% sure of not exceeding the target fault rate.242. overhaul either his manufacturing equipment or his assembly process. There is some arbitrariness in the quality control system described here. x = 4. 4. if it was every 20 bicycles that were checked this would also increase the time between the occurrence and detection of a manufacturing problem. Similarly.047. this is found to require t > 2. If the manufacturer finds more than 4 faults in the last 1.006.33. It was decided to use the most recent 1.162 Everyday Probability and Statistics normal distribution beyond t standard deviations less than 0.7) will give condition (14. The highest value of x satisfying the condition is 4. Getting a good balance is part of the skill of quality control. if necessary. (14. t = 4.1.8) is best achieved by trial and error — just trying values of x until you find the maximum value satisfying the condition. and 5 are shown below: x = 3. The values of t found for x = 3.6. then he should check and. From Table 10.01.
2) the standard deviation of that estimate is σp = p(1 − p) = 0. On that basis the best estimate of the proportion of tagged fish in the pond is p= 60 = 0.1538 and phigh = 0.0231 = 0.2 − 2 × 0. so a process for finding the number that takes about a month will not be bedevilled by excessive changes of the total number by births and deaths. In this case.000. The fish are then released back into the pond. from (13.2462. 300 However.0231 = 0. nets a few fish and tags them with a light plastic tag attached to a fin that causes the fish no distress. He suspects that the number is of order 3. he finds that of the 300 fish he has caught. 60 are tagged. the limits of p up to two standard deviations from the mean are plow = 0. It is not possible to count the fish individually. On a daily basis he goes to different parts of the pond.1.2 + 2 × 0. we find that. . When he carries out his final netting.95. Once he has tagged about 400 fish he then starts a process of again going to different parts of the pond and netting fish.2. so what should he do? The average lifetime of the type of fish in the pond is about 3 years. but it could be half as much or even twice as much as that. Let us see how he finds the number of fish in the pond from this procedure. but now he just counts the number of fish caught and the number of those that are tagged. We take a numerical example to illustrate the process.0231.Taking Samples — How Many Fish in the Pond? 163 about ±50%. 300 From Table 10. the probability of being within 2σp of the mean is about 0. for a normal distribution that we assume here.
06 1.000.85 1.99 1. Problems 14 14. for the number of fish in the pond are 400 = 0.93 2. A sample of 20 Watutsi adult males gave the following heights.164 Everyday Probability and Statistics If the number of fish in the pond is N. then the actual value of p is 400/N.73 2.97 .2462 or N2 = 1.10 2. with p = 0.92 1.97 1. (ii) The standard deviation of the sample.2. 625. 601.88 2.14 1.08 Find (i) The mean height of the sample.2 is N = 2.87 1.03 1.87 1.00 m.95. (iv) An estimate of the standard deviation of the sample mean.89 2.1.1538 or N1 = 2. (iii) An estimate of the standard deviation for the whole population. with a probability 0.04 1.02 1. The owner of a fish pond catches and tags 100 fish. in meters: 1. (v) The probability that the mean height of Watutsi adult males is greater than 2. He then catches 100 fish and finds that 20 of them are tagged. and 400 fish were tagged. N2 The most probable number.92 2. the limits found.97 2. Hence. because we know that 400 fish were tagged. 14. N1 and 400 = 0. What is the most likely number of fish in the pond? 1.
The Significance of Differences It is believed that there are as many brown rats as there are people in the United Kingdom. although in some less-well developed societies they can be seen roaming the streets at night. We consider a hypothetical example of comparing the weights of male rats from London and Manchester. while a sample of the same size from Manchester had a mean weight of 587 g and a standard deviation of 38 g. moving no more than 100 m from their nests if the food supply is adequate. at least in Western societies. then we could say that there is evidence that the populations in London and Manchester are significantly different. or more. 1645–1696) 15. Once ensconced within an urban environment they are not great travelers. If that probability is very low. 165 . The people have little intelligence (Jean de la Bruyère. The question we consider is the likelihood of having that difference.Chapter 15 Differences — Rats and IQs Le people n'a guère d'esprit…. otherwise there would be no such evidence. in the mean weights if the samples were taken from the same population. By and large they keep clear of the human population. A sample of 400 rats from London had a mean weight of 595 g and a standard deviation of 41 g.1. mostly living within city environments. and so it is quite feasible that the gene pools of rats in well separated places may differ significantly.
98 g and 95% of them will be less than 3.916 g2 .0 standard deviations. if we take samples of 400 from the same population. From Table 10.1) where n1 and n2 are the sample sizes.96 g.4. In considering the significance of the difference we must consider the possibility of having a difference of four standard deviations in either direction. then from Table 10. in this case that London and Manchester rats are all part of the same population so that there is a single population from which we have drawn the two samples. 14. To spell out what this signifies.00006. but for the large samples we are using here.00003. For the present example.2). the area in one tail more than 4σ from the mean is 0. Another theoretical result is that the variance of the difference will be Vdiff = 2 2 σ1 + σ2 .166 Everyday Probability and Statistics As is usual in problems of this kind we make a null hypothesis (Sec. n1 + n 2 − 2 (15.e. i. 6. if we compare the sample means from a very large number of samples taken from the same population then the difference from one sample mean to another is sometimes positive and sometimes negative. so in the two tails it is 0. is appropriate. 68% of the magnitudes of differences of sample means will be less than 1. that is.1. 400 + 400 − 2 giving the standard deviation for the difference of sample means. Theory shows that the expected difference of the sample means is zero. the Student t-distribution. This is a very small probability so that the difference in the means of the two samples is unlikely to have occurred by chance and we may conclude that the rat populations . for this kind of problem. we must find the area in the two tails of the distribution more than four standard deviations from their means. or just over 4. The actual difference in the sample means from the two cities is 8 g.. Really. as mentioned in Sec.98 g. σdiff = 1. but averages to zero. assuming that the normal distribution applies will give little error. this is Vdiff = 412 + 382 g2 = 3.1.
002 0 450 500 550 600 weight (grams) 650 700 750 Fig. 0.012 London Manchester 0. 15. 15.1. 15.Differences — Rats and IQs 167 in the two cities are significantly different. Whether or not one can say that they are genetically different will depend on a judgment about environmental factors. are shown in Fig.1. The probability distributions for the weights of rats in London and Manchester. and that London rats have a somewhat higher mean weight. However.2." Faced with an average London rat and an average Manchester rat you would be hard-pressed to detect any difference. as indicated by the samples.004 0.008 Probability 0. Significantly Different — So What! The probability curves for the rats' weights. what is also clear is that the weights are very similar and the statistical term "significant difference" must not be confused with the judgmental term "important difference.006 0. is clear from this form of presentation.01 0. If there is some prevailing condition in Manchester that leads to a smaller food supply. . The fact that the distributions are different. then the difference in mean weights may be due to that factor alone and not due to genetic variation.
by ethnicity. how could the anthropologist with his western city background know one way or the other? However. An anthropologist who specialized in the study of tribes in New Guinea recounted a story where he accompanied some members of a tribe on a hunting trip. These factors are usually popularly referred to as. respectively.168 Everyday Probability and Statistics We have already indicated that a statistical difference may be the result of environmental factors. a large animal had passed through.. the headman stated that. and physique. When asked how he knew this he replied that some of the vegetation had been disturbed — but the anthropologist looking round the clearing could see nothing of any significance. Athletic prowess may well depend on genetic factors. While ascribing differences of athletic ability to genetic factors is largely non-controversial as soon as there is any discussion of differences of ability related to the workings of the brain there is much controversy. Such a deception would be safe enough. height. or it may be due to genetic factors. a better food supply for rats in London than in Manchester.g. in general. by religion etc. nurture and nature and. If nurture was the sole cause of the difference in weight between Manchester and London rats. people of West African origin seem to do well in track sprint events. An Ethiopian sprint champion would be regarded as an oddity but no one is surprised when an Ethiopian or a Kenyan wins an international marathon race. He decided that the headman was trying to impress him and that really there was no evidence of the passage of a large animal. e. they occur together. Some of these divisions are related to genetic differences that manifest themselves through visible characteristics such as skin color. while endurance events on the track tend to be the speciality of those from East and North Africa. On reaching a clearing in the jungle. although cultural and environmental factors will also play a role. hair color. Thus. the anthropologist . since they had last been along that path a few days previously. The human population of the world is subdivided in various ways — by nationalities. then a group of Manchester rats transported to London would have progeny with the same mean weight as London rats.
there are often different patterns of child development dependent . The intellectual capacity of an individual and the main skills developed will depend to a great extent on nurture — the society in which he or she lives and his or her early experiences — but also on nature. The pattern of matches set up was not perfect. The anthropologist concluded his description of this remarkable feat by noting. Then. On returning to the village he spilled a box of matches into a heap on a table and asked the headman to examine it and commit the arrangement to memory. It has a few basic instincts — to cry when hungry or uncomfortable and to suck when hungry and the opportunity presents itself — but otherwise its brain is almost like a blank sheet of paper on which anything may be written. these people would appear to be subnormal." When a baby is born. by any intelligence test that they would devise we would appear to be subnormal. A child that gets plenty of stimulation in its early years is likely to be intellectually well developed in later life. The New Guinea tribesmen needed and acquired different skills from those of the anthropologist but there is no evidence that the intellectual capacity of their society was in any sense inferior to that of his. Using a Polaroid camera he then photographed the pile from several different directions to preserve an accurate record of how it had been. in the sense that genetic factors will also come into play. homogeneous in a genetic sense. its brain has little ability to recognize the world in which it lives. contacts are established by synaptic connections in the brain that condition the child for involvement in the society in which it will live and enable it to develop the skills necessary to flourish in that environment. after 3 days. As the child grows so. one that is deprived of such stimulation will be handicapped thereafter. "By any intelligence test we have devised. Most of this process of brain development takes place in the first 3 years of a child's life. Within a particular society. but it was topologically correct in that the system of matches being under and over others was exactly as in the original pattern.Differences — Rats and IQs 169 decided to conduct a test. he gave the headman the same box of matches and asked him to arrange them as he had seen them previously.
Testing different social or racial groups may reveal significant differences. However.1. Parents lacking parenting skills will not enable the full potential of their children to be realized and these children. with the right upbringing. that potential could be realized. Trying to create those circumstances is a difficult. given the right circumstances. For chickens fed on the new regime.170 Everyday Probability and Statistics on social class. does not affect the potential bound up in the genes of the affected individuals and. sad though it is. For the chickens fed on the original regime.7 g with a standard deviation of 2. Such differences as are found are often statistically significant but may be of no importance. one can learn little about the genetic quality of individuals from such differences. Problem 15 15. with the right upbringing. The child. perhaps impossible. might.2 g with a standard deviation of 2. in a statistical sense. could. This cumulative neglect. in their ability to deal with intelligence tests or their ability to track animals but will tell you nothing of interest about the genetic characteristics of the groups being tested. Statistics is a useful tool for understanding society — but it is important that we should first understand statistics. the mean weight of 100 eggs is 57. have become a theoretical physicist. the mean weight of 100 eggs is 57. in their turn. task that many governments have tried to tackle. A farmer decides to test the effect of a new feeding regime on his chickens.3 g. What is the probability that there could be that difference or more in the means of the two samples if the new feeding regime was ineffective? . The child of inadequate parents.1 g. who would usually perform badly with intelligence tests. will often become inadequate parents. influenced by the teaching of a parent who was a rabid racist. have become a tolerant high dignitary of the Church of England. otherwise we may draw the wrong conclusions from its applications.
whereas prosecution and conviction may do no more than create another citizen with a criminal record. When a young offender is caught committing a minor crime. For one thing the police exercise a certain amount of discretion in applying the law. Actually this is not so. A minor act of pilfering. then there would be millions of criminal acts and prosecutions per day. Mostly. O liberty! O liberty! What crimes are committed in your name! (Mme Roland. for example. 171 . the police will either not take action because the violation is not at a gross level or. a formal warning may be enough to steer him away from further crime. such "crimes" are not detected.1. Crime and the Reporting of Crime Committing a crime involves breaking the law and since what is unlawful is defined in great detail it would seem that the presentation of crime statistics would be a completely objective process. they will just issue a warning.Chapter 16 Crime is Increasing and Decreasing O liberté! O liberté! que de crimes on commet en ton nom!. perhaps. The effort of reporting the crime may be more troublesome than the loss itself. Another uncertainty in crime statistics is that not all crimes are reported. coupled with the knowledge that the police are unlikely to devote much effort to solving such a trivial crime when they have more serious matters under consideration. 1754–1793) 16. if exceeding the legal speed limit on roads by a modest amount was always detected and prosecuted. but even when they are. might not be reported. say involving the loss of money or goods just worth a few pounds.
family life. or he. will be accused of having been a consensual partner in the act.2. change the direction of their criminality.172 Everyday Probability and Statistics Even serious crime is sometimes unreported. Here. As a result of various government and police initiatives. should we be more or less fearful about crime or is the incidence of crime fairly stable? If we take the politics out of crime we may see it as it really is. usually her. in which statistics are bandied to-and-fro. The Trend for Overall Crime in England and Wales The British Crime Survey is issued by the Home Office and reports on various types of crime in England and Wales over the recent past. might well conclude that crime is simultaneously increasing and decreasing. All major types of crime are included in the survey. there are changes over time in both the reporting and recording of crime. Based on the facts. because the victim is too ashamed to reveal the sad state of. A criminal seeking to fund a drug habit may therefore be deflected away from car crime toward some other criminal activity. Rape seems to be more reported than once it was as procedures have changed and victims are less fearful that they will be exposed to humiliation both in the initial police investigation and in the courts. Acts of domestic violence may be unreported. Proving rape beyond reasonable doubt is often very difficult. particularly in England and Wales. except fraud . Another factor is that criminals may. rape. from time to time. In recent years the security of motor vehicles has been improved so that car crime has become more difficult to carry out. we shall look at the official crime statistics in the United Kingdom. and this is one of the factors that confuse any consideration of trends in crime statistics. over the past few years to see what really has been happening. because the victim fears that she. Another serious crime. 16. Crime is a rich source of point-scoring for politicians and an observer of a political debate on crime. is notoriously under-reported.
On the other hand. headed for most of the time by Margaret Thatcher. Since 1995. 16.Crime is Increasing and Decreasing 173 Fig. Labour would claim that since it took power. crime rose steadily from 1981 to 1995.1. the last years before the Labour election victory of 1997. 16. Such is politics! The graph shows the facts of the situation but it is not the whole story. so the Conservatives can justly claim to have launched the improvement which lasted for over a decade. The figures for overall crime. the total number of offences fell steadily and in the year 2004–2005 fell below the 1981 figure.1. Some kinds of crime are of more concern to society than others. a point made much of by the opposition political parties. reported by the survey. The rise in crime took place during the Conservative administration. are given in the form of a graph in Fig. The numbers of reported crimes of all types in England and Wales from 1981 to 2005. It will be seen that in the period under report. overall crime has fallen and might neglect to mention that it inherited an improving situation. and forgery. there was a marked decline between 1995 and 1997. Vehicle crime is not trivial — especially to those affected by it — but . increasing by approximately 80% in that period.
16. 16.2. more than doubled. but that murder had greatly increased. starting in the Conservative administration and continuing under the Labour government." i. It mirrors to some extent the pattern shown in Fig. many now have immobilizers which means that they cannot be started by "hot-wiring. by joining wires behind the Fig. then an average citizen would feel that everyday life was much more threatening. Vehicle crime in England and Wales from 1981 to 2005.3. Part of the decrease is probably due to improved security on motor vehicles. . Vehicle Crime. 16. If it turned out that burglary and car theft had greatly reduced. between 1981 and 1995 and thereafter a fall.1 for overall crime — a steep rise.e. as given by the British Crime Survey.2. A consideration to bear in mind in judging the significance of this graph is a substantial increase in car ownership during the period — by 16% in the decade 1990–2000 alone. is shown for England and Wales in Fig. 16.. and Violent Crime The incidence of vehicle crime over the period 1981–2005. Burglary.174 Everyday Probability and Statistics everyone in society would regard murder as much more serious.
There are no such feelings about vehicles. which were once an optional extra.Crime is Increasing and Decreasing 175 dashboard. in 1993. 16. Another factor is that radios. A burglary is a violation of that security and some victims say that they never again quite feel the same about their homes once they have been burgled. The crime of burglary. a pattern that by and large resembles those for overall crime and vehicle crime except that the decline began a little earlier.3. Burglary in England and Wales from 1981 to 2005. but require the key to get them started.3 shows the incidence for burglary in England and Wales from 1981 to 2005. for example — in view within cars that have been left unattended. are now standard on all cars so that there is little market for stolen car radios. A home is a special place where an individual or a family spends most of time and in which they normally feel secure. stealing by breaking into homes. Even the police cannot enter a home without a warrant or some other authorization. a country at war exhorts its citizens to defend their homes — not their motor cars! Figure 16. Finally. is less common than vehicle crime but far more traumatic for the victim. publicity campaigns have persuaded people not to leave valuables — handbags and mobile phones. . Fig.
vehicle crimes and burglary rarely. but at least it shows the same decline from 1995 as the other categories of crime so far described. However. A high proportion of the assaults are the regular Saturday night "punch-ups" that occur in town and city centers after too much alcohol has been .4. The scale of violent crime. In the year 2004–2005. 347. and mugging.000 cases of mugging. Violent crime in England and Wales from 1981 to 2005. The category of violent crime includes domestic violence. several million per year. involve violence and it is violent crime that is of the greatest concern to most people. and the remainder were assaults of one sort or another.000 cases of domestic violence. 16. The pattern as presented so far is reassuring in that the present trend is downward.176 Everyday Probability and Statistics Fig. is quite surprising. there were 401. To have the alloy wheels of your car stolen while you sleep may be annoying but to be in hospital after being violently assaulted is an experience in a different league altogether. but does not include homicide. The occurrence of violent crime in England and Wales from 1981 to 2005 is given in Fig. common assault. 16.4. although sometimes.
However. The numbers per year are measured in hundreds. Homicide Homicide is the category of crime that includes murder and manslaughter. a celebrity. multiple deaths. 16. as displayed in Fig. A murder is always reported in the local press and on local radio and television news programs. The incidence of homicide in England and Wales from 1993 to 2005. on the contrary. say. 16. alas. a citizen who leads a quiet domestic life and has friendly neighbors is not very likely to become a victim of violence — although.4. Homicides in England and Wales. This concentration of attention on homicide in the news media leads to the impression that it is a common occurrence in society. .5. 16. involving. and is the one that most impinges on the public's general awareness of crime. some do. whereas the numbers per year of the other types of crime to which Fig. it is a very rare crime. A homicide with special characteristics. 1993–2005. a child or the use of a firearm will usually become a national news item.Crime is Increasing and Decreasing 177 consumed.5 shows that.
As might be expected the rates tend to be higher in major cities: the figures for three major cities in the United Kingdom are London Edinburgh Belfast 2. in that year. but accompanied by a huge suicide rate).6 per 100. Taking these special factors into account.000 of the population per year. only Japan has a lower homicide rate (0. some quoted figures for major . because the 2000–2001 figure includes the deaths of 58 Chinese illegal immigrants who suffocated in a lorry that brought them into the United Kingdom and the 2002–2003 figure is greatly distorted by including all the victims of the mass murderer Dr Harold Shipman. Given the population of England and Wales as about 53 million. The later entries in the graph are a little misleading. However. However. has more than twice the UK rate. To put the UK city figures in perspective. Of all major countries. it is probably true to say that the underlying rate has flattened out at a level less than 10% above what it was in the late 1990s — to something in the region of 830 per year. this gives a rate of 1. even Switzerland. estimated to be between 150 and 250. the graph shows a different evolutionary pattern to those of the other types of crime.4 The Belfast figure is affected by a component of sectarian homicide which is not present to any extent in the remainder of the United Kingdom.9 per 100. It is customary when reporting homicide rates in any society to express it as homicides per 100.1 2.000 per year. it turns out that the United Kingdom is one of the safest countries in the world in terms of homicide.178 Everyday Probability and Statistics we have referred are measured in millions. while any homicide is one too many. although their deaths occurred over several years.4 4. generally regarded as a haven of harmony and tranquility.000.
Crime is Increasing and Decreasing
cities in Europe and the United States are as follows: New York Amsterdam Lisbon Chicago Baltimore Washington, DC 7.3 7.7 9.7 22.2 38.3 45.8
Despite all the impressions to the contrary the United Kingdom is a safe place to live — at least as far as homicide is concerned. A UK citizen should be concerned about homicide, but it should not overshadow his life — he is four times more likely to be killed in a road accident than to be the victim of a homicide.
16.5. Crime and Politicians
There are many issues that affect the lives of citizens of any country — education, law and order, health, employment, and taxes to mention just a few. The facts relating to any of these issues will be complex and interact with other facts in unexpected ways. In this chapter, a numerical account of the incidence of crime in the United Kingdom has been given. But even this picture, objective as it seems, is open to criticism. The British Crime Survey finds out about crimes that have occurred by asking victims of crime (with the obvious exclusion of homicide victims) about their experiences, and their survey includes both crimes reported to the police and those unreported, Their claim is that this form of collecting statistics is more comprehensive than just relying on police figures and also reveals more about the public's attitudes toward crime. However, some question the validity of this way of collecting crime statistics. There may be a number of reasons why crime statistics vary from year to year. There will be a statistical fluctuation based on the random pattern of the incidence of crime. If over a long period of time
Everyday Probability and Statistics
it is assessed that the average rate for a particular crime is N per year, then the standard deviation in that number due to random fluctua√ √ tions is N per year. For N = 2,000,000, we have N = 1414 which is 0.071% of the number of crimes committed — a very small fractional fluctuation. On the other hand, if 900 homicides are expected √ in a year, then N = 30, which is 3.3% of the expected number, a much larger random fluctuation. Because of the numbers involved, with the exception of the homicide figure, the graphs shown in this chapter are little affected by random fluctuations — the trends are real. Another important reason put forward for changes in the crime figures is changes in the unemployment rate. During the period that crime rates have fallen in the United Kingdom, so has the unemployment rate, and there is an assumption that crime rates will be higher among the unemployed. This was not evident during the slump years of the 1920s and 1930s, but during that earlier period there was no television, and so much less exposure to advertising, that stresses the attractiveness of acquiring ever-more worldly possessions. No individual government can be blamed for a worldwide change in advertising pressures or for a worldwide depression. So, perhaps it is unfair to blame a government for an increase in crime during its tenure of office. The argument just made as an excuse for the increase in UK crime from 1981 to 1995 may or may not be sustainable. The point really being made is that a simplistic interpretation of any set of statistics, whether it relates to crime or anything else, may be flawed. We live in a complex world. Politicians for their own ends will try to put forward a simple picture that either presents their own party in a favorable light or the opposition in an unfavorable light. The rule is to try to think for yourself; a useful exercise is to be a devil's advocate and to see if you can make a case for the other side, because then you might see the inconsistencies in the argument being presented to you. One should beware of confidence tricksters, whether they are after your money or your vote.
Crime is Increasing and Decreasing
Problem 16
16.1. The following graph shows the number of road users killed in road accidents in the United Kingdom from 1964 to 2005. Estimate from the graph: (i) The ratio of the number killed in 2005 to the number killed in 1964. (ii) The fraction of those killed who were pedestrians in 1964, 1982, and 2005. Suggest some reasons for the general decline in the figures over the period of the graph.
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their eyes become red and inflamed. 1577–1640) 17. labeled as BRCA1. hay-fever sufferers may have their futures blighted by performing below their true capability. Genetics and Disease At the beginning of summer many plants enter their period of reproduction and release vast quantities of pollen. Breast cancer in women is now recognized to have a strong genetic association and. the ruin and overthrow of body and soul. These are the sufferers from hay fever. unless some allowance is made for their condition.1. Hay fever is recognized as being of genetic origin and its incidence occurs in families. Unfortunately the onset of the hay-fever season coincides with the period when many young people take important examinations and. Some relief is obtained by taking antihistamine tablets. their nasal passages become blocked with mucous. They sneeze violently and continuously. (Robert Burton. BRCA2. It is a prototype of a genetic disease of which there are many other examples. but for an unfortunate minority it presages a time of great misery and discomfort. For most people this is a matter of no consequence. and 183 .Chapter 17 My Uncle Joe Smoked 60 a Day damned tobacco. often associated with the occurrence of eczema and asthma. the genes associated with the condition. but they have the unpleasant and potentially dangerous side effect of causing drowsiness. indeed. and they have difficulty with breathing.
but who is little exposed to the agencies that cause it. Cancer may be caused by various agents. someone without any genetic disposition toward cancer could well contract the disease under any level of exposure but with the risk increasing with exposure. may never contract the disease. There is an element of chance in determining whether or not a particular individual is affected. It can happen that someone who has a genetic tendency toward some form of cancer. This could happen by exposure to radiation — for example. if the tumor cells migrate through the body affecting other tissues. then the condition is malignant and much more difficult to handle. Nevertheless only 5% of cancers are associated with the presence of the aforementioned genes. The probability that it will do so would depend both on the degree of exposure to it and on the propensity of the individual to suffer such damage. If the condition remains localized.184 Everyday Probability and Statistics BRCA3. Cancer is a condition in which some cells of the body run out of control and repeatedly divide. It is against this background that we will discuss the relationship between smoking and smoking-related diseases. so there are clearly other factors at work. but basically all these have the effect of damaging some of the protein molecules in the cell. forming a tumor. By contrast. Researchers have stated that 60% of women with the BRCA3 gene will develop cancer before the age of 70. However. . then the tumor is benign and removing it by surgery or locally-applied high radiation doses essentially cures the condition. by excessive sunbathing — or by the action of carcinogenic chemicals of which there are many. The breast cancer example shows that not all those with a genetic predisposition toward a particular disease will suffer from that disease and also that those without the predisposition sometimes will. It is even possible that such a person could be heavily exposed to a carcinogenic agency and not contract the disease — but he would have to be lucky. A potentially cancer-inducing agency may or may not damage a cell protein to cause a cancerous cell. have now been detected.
in the latter half of the twentieth century evidence began to emerge of a connection between smoking and various diseases. During the early part of the twentieth century. Often they would use irrational arguments to justify continuing to indulge in their harmful drug habit. in about 1564 but was popularized by Sir Walter Raleigh. of the kind "My Uncle Joe smoked 60 a day for the whole of his life and lived to be 101". Eventually the cigarette companies had to give ground as responsible governments all over the world took steps to encourage people to give up smoking.My Uncle Joe Smoked 60 a Day 185 17. It probably came to England through Sir John Hawkins. The companies were to a great extent initially supported by smokers themselves who were in a state of denial. beneficial to health — perhaps because the smell of tobacco was preferable to the prevailing smells of the time.2. or his crew. At first the major cigarette companies fiercely disputed this evidence. saying that it was not well founded. The Incidence of Smoking in the United Kingdom The smoking of tobacco originated with the aboriginal inhabitants of North America and was introduced to Europe by various explorers during the sixteenth century. Anti-smoking organizations even suggested that the companies were withholding information connecting smoking and lung cancer that their own research had revealed. notably lung cancer and also many others. Sometimes the same Uncle Joe would be cited by other nephews and nieces as a reason to drink alcoholic beverages to excess because that is what he did. The advertising of tobacco products . During the eighteenth and nineteenth centuries the idea was prevalent that smoking was. a rakish adventurer and a favorite of Elizabeth I — but somewhat less favored by her successor James I who ordered his execution. However. in some way. smoking cigarettes was regarded as a rather sophisticated thing to do and films from the pre-World-War-II period show people smoking (and drinking) quite heavily with languid wealthy young ladies doing so with the aid of long cigarette holders.
All this pressure against smoking gradually had an effect and the proportion of smokers of both sexes has steadily decreased in the United Kingdom — from 45% of all adults in 1974 to 25% in 2005. In 2004. in terms of the health hazard that it imposes. By definition. In addition. for example. Nevertheless Uncle Joes do exist so we will now consider why it is that Uncle Joe got away with his reckless lifestyle while other clean-living. The various component countries of the United Kingdom followed suit in 2006 and 2007. those continuing to smoke either accept the risk involved. For them the "Uncle Joe" example is the only one they can understand. reducing the overall proportion of those smoking will reduce the incidence of smoking-related illness. the tobacco companies had to put warnings on the packets of their products. the Republic of Ireland banned smoking in what had been. non-smoking individuals die young of lung cancer or some other disease associated with smoking. the smokers' paradize — the public house. perhaps because they are so addicted that they cannot stop. has been won. The UK government has a target to reduce the proportion of smokers down to 21% or less by 2010. . in theatres and cinemas.186 Everyday Probability and Statistics was banned on television and in other media. and many airlines prohibited smoking even on long-haul flights. Figure 17. By and large the argument against smoking.1 shows the variation in the proportions of men and women who smoked from 1974 to 2004/2005 published by the UK National Statistics Office. so that one has the bizarre spectacle of individuals giving large sums of money in return for packets bearing the words "Warning — cigarettes can kill". until then. Many of the early films in which smoking was portrayed in such a glamorous way had to be viewed through a thick haze of tobacco smoke. Smoking was banned in many public places. or because they are completely unable to understand the arguments that have been accepted by the majority. Restaurants designated smoking and non-smoking areas.
there is the hazard of secondary exposure when in the presence of those smoking.1. The incidence of cigarette smoking in the United Kingdom. There is no way of avoiding some exposure to potential carcinogenic hazards. This is especially true in the home if some. Background radiation is everywhere and is particularly high in areas where granite exists in the rocks. of the family smoke and young children are particularly put at risk by the smoking of one or both parents. While smoking is still a legal activity. to any other carcinogen.My Uncle Joe Smoked 60 a Day 187 Fig. the exhaust systems of cars emit some carcinogens and occasionally one sees badly-maintained vehicles illegally emitting black exhaust fumes that are both unpleasant and highly toxic. 17. or genetic screening of some kind. as in Aberdeen. an individual cannot know what his or her sensitivity is to tobacco smoke or. or where granite is heavily used as a building material. indeed.3. The Smoking Lottery Unless there is a family history of smoking-related illness. 17. Source: UK Office of National Statistics. but not all. . such as in parts of Cornwall. Even with the best of filters.
although exactly quantifying the risk is not easily done. A notional representation of the probability of acquiring a carcinogenrelated illness as a function of exposure and sensitivity. one's sensitivity to such hazards is unknown the prudent course of behavior is to avoid any other additional exposure.188 Everyday Probability and Statistics Given that there is no escape in the modern world from some exposure to carcinogenic hazards and also that. . such as smoking.2. The sensitivity is given on a scale of 1–5 on the basis that nobody can have so little sensitivity that they are immune from such illnesses. in general. Fig. 17. Just as an illustration of what the situation could be there is given in Fig. The probability of carcinogen-related illness increases with sensitivity and increases with exposure. 17. and exposure is also given on a scale of 1–5 on the basis that there is a minimum exposure that cannot be avoided.2 a notional representation of the dependence of the probability of acquiring a carcinogen-related illness related to exposure and sensitivity.
Uncle Joe may well have had maximum exposure but if he had minimum sensitivity then the likelihood that he would acquire a carcinogen-related illness would only be 0. both in sensitivity and exposure. a nerdish shopkeeper is awakened from a cryonic state after 200 years into a world where many beliefs and practices had changed. exposure level 3.0 the probability of getting a carcinogen-related illness goes from 0. However. and which ones would not. The story about smoking and illness is one that is now well understood and. In the comedy film The Sleeper. (ii) Sensitivity level 5.2 find the probability of an individual acquiring a carcinogen-related illness with: (i) Sensitivity level 1.0. Problem 17 17.0. will not change with time even though the exact nature of the risks involved may be quantifiable with greater accuracy as the results of further research are made available. From the graph in Fig.0.5.My Uncle Joe Smoked 60 a Day 189 From the figure it will be seen that.1. for someone with a sensitivity rating of 2. if there were 20 people like Uncle Joe.45 with maximum exposure — again an increase factor of 3. Which one would succumb. a reality.0. (iii) Sensitivity level 3.5.16 at maximum exposure — an increase factor of more than 3 in risk. That is the smoking lottery.15 with minimum exposure to about 0. essentially. starring Woody Allen. exposure level 5. . In one scene he is being examined by hospital doctors who prescribe a course of smoking cigarettes to improve his health. Smoking to promote health may make good cinema humor but is not. and never will be. 17.05 at minimum exposure to about 0. could not be determined ab initio even if the facts about sensitivity and exposure were known.05 — about a one in twenty chance. exposure level 1. then the chances are that one of them would become ill. Someone who is maximally sensitive goes from a probability of 0.
0. living with a background exposure level of 2. the increase in exposure level due to smoking 20 cigarettes a day is 2. By what factor has he increased his risk of acquiring a carcinogen-related disease by deciding to smoke? .0.190 Everyday Probability and Statistics For someone with a sensitivity level of 3.0.
1. for example. Luck.Chapter 18 Chance. In primitive societies this idea of being helpless. The idea arose that there were all-powerful agencies. and would act accordingly. This verse summarizes the beliefs of many people about the nature of life — a series of chance happenings that toss you this way and that. then this was a sign that a god was angry and was venting its feelings by killing some of those that had offended it in some way. sometimes for good and sometimes for ill. And one by one back in the closet lays. 1769–1821) 18. coupled with ignorance about the way that nature works. and Making Decisions Has he luck? (Napoleon I. In most early societies the idea was established that there was a multiplicity of gods dealing with different aspects of human affairs 191 . interpreted as gods. and mates. and slays. led to superstitious beliefs in the supernatural. The Winds of Chance In Edward Fitzgerald's famous translation of The Rubáiyát of Omar Khayyám there is a verse that runs: 'Tis all a chequer-board of nights and days Where destiny with men for pieces plays Hither and thither moves. who shared many of the emotions and characteristics of humankind — anger and generosity. If a volcano erupted and life was lost.
warns Caesar to "Beware the Ides of March" but one is not told how he derived his prediction of danger. In the more advanced societies. There were two main reactions to this belief in a panoply of gods. and offered guidance to those who came. and the carrying out of sacrifices. visionary. temples were built and a priesthood was established. from Greek culture. love. fertility. tea leaves. etc. In Shakespeare's Julius Caesar. The second reaction was to try to predict what the gods had in store. palmistry. Belief in powers of prediction has declined greatly in the modern world. a human priestess. but there are still those who use tarot cards. They would arrange and supervise the recital of prayers. or any part of it. was the Oracle of Apollo at Delphi. One of the great scientific discoveries of the last century. astrology.192 Everyday Probability and Statistics such as war. the offerings of goods. A famous example of this. so that one could take some kind of remedial action. by prayer and supplication and by making offerings of one sort or another. The pronouncements about the future were usually a bit vague (a characteristic of many religious texts) and required further interpretation. The Romans seemed to favor an investigation of chicken entrails to predict the future. referred to as the Heisenberg Uncertainty Principle. a temple in which Apollo spoke through a Pythia. and would therefore react. is that there is no way of precisely defining the future state of the universe. The first was to try to modify the way that the gods felt. then the future development . at least in a philosophical sense. prosperity. or a crystal ball to attempt to find out what the future has in store for them. in addition to soothsayer there are the terms seer. This is in contrast with the deterministic view that existed at the end of the nineteenth century when it was believed that if the position and motion of every particle in the universe were known at any instant. usually of animals. a soothsayer. The idea that some people can predict the future has given rise to a host of terms for describing such people — for example. and also of human beings in some societies. a person who can foretell the future. which included experts who could act as intermediaries between the layman and the all-powerful gods. and fortune teller.
it will cause it to move in an unpredictable way. Caesar would not have gone to the Senate on the Ides of March. and in its interaction with the body. Basically. As the Rubááiyát says. the uncertainty principle applies to bodies of any mass but for large bodies of the type we meet in everyday life.Chance. then this is at the expense of having no idea at all of how it is moving and if you try to find its velocity exactly then you have no idea at all of where it is. If one were to accept Shakespeare's version of Roman history. The uncertainty in behavior most strongly manifests itself in the behavior of particles of very small mass. but we cannot control our lives completely. if you try to locate its position exactly. Theoretically. i. Finding the position of a body requires it to emit or reflect radiation for some detector to record. Similarly. and Making Decisions 193 of the universe would be completely defined. we are moved hither and thither. Snooker players do not have to worry about the uncertainty principle in planning their next shot. The uncertainty principle indicates that it is impossible to know exactly where a particle is and how it is moving. its velocity. had had her way. It seems that the flow of events that dictates both the life of an individual and the destiny of a society are unpredictable and uncertain. it has no practical importance. The most precise definition of position is obtained if the radiation has a very high frequency. like electrons. but then the radiation has a great deal of energy associated with it.e. By making sensible choices we can minimize the harm that we experience. The basis of the uncertainty principle is that in trying to determine the position and velocity of a body you have to disturb it in some way.. The Heisenberg uncertainty principle actually quantifies the relationship between the uncertainty of position and the uncertainty of velocity. If Calphurnia. Luck. Caesar's wife. then you influence that body and change its characteristics. . any attempts to find out about the future will influence and change that future. then Caesar would not have been assassinated and human history would have been changed. the principle says that if you go through the process of finding the characteristics of a body. in relation to human affairs.
Everyday Probability and Statistics
18.2. Choices
Every day people make decisions of one sort or another. They are mostly trivial and sometimes involve a choice of alternative actions, i.e., to go shopping or not to go shopping, and sometimes a choice of several different possibilities — for example, deciding on which resort to choose for a holiday. In Chap. 6, we examined another kind of decision-making when a doctor decides on a course of treatment based on some diagnostic information that leaves the patient's condition not uniquely defined. In that case, we were able to assign numerical values to the probabilities of particular conditions and to the probabilities of the effectiveness of various drugs and so to identify the course of action most likely to lead to a favorable outcome. Nevertheless, in a decision based on the calculation of probabilities there is the possibility, or even certainty, that from time to time the wrong decision will be made. In such circumstances, if it turned out that the treatment prescribed by the doctor was not the best for the patient's actual condition, it would be unfair to criticize the doctor or to say that he or she had been professionally incompetent. Any decision, no matter how trivial it seems, can have unexpected, and sometimes tragic, consequences. The trip to the supermarket by car may be marred by a traffic accident. How many times have those involved in such incidents uttered the words "If only I had decided to do it tomorrow!"? People who chose to have a holiday in Phuket, Thailand, over the Christmas period in 2004 could not have anticipated or imagined the tragedy of the Asian tsunami that claimed more than a quarter of a million lives. How many relatives and friends of victims have said "If only they had decided to go somewhere else!"? We see from the above that in making decisions there are two types of outcome, those that are predictable and those that are not. The latter category is almost infinite in its range of possibilities. When going to the supermarket, almost anything can happen from meeting an old school friend, spraining an ankle, having a traffic
Chance, Luck, and Making Decisions
accident, or winning a prize for being the store's one-millionth customer. If you let your imagination flow you can fill several pages with possible outcomes of widely varying likelihood. By contrast, by their very nature, predictable consequences are usually bounded. However, one cannot accuse of foolishness, either a doctor making a wrong decision based on an estimate of probabilities or a holidaymaker who travels to the scene of a tsunami. That is what life is like and it cannot be any other way.
18.3. I Want a Lucky General
There are those who move majestically through their lives with everything going in their favor and others who experience one mishap after another. From a statistical point of view, what we are seeing here are the extremes of a distribution of what we might call good fortune or luck. It was Napoleon who, when confronted with the curriculum vitae of candidates for a vacant general-officer post, pushed them aside and said that what he wanted was a lucky general. Rationally that statement did not make sense. A winning streak at the roulette table would not have any influence on the outcome of future spins of the wheel and, equally, someone who had been lucky in the past was not more likely than anyone else to be lucky in the future. Nevertheless, where decision-making does not depend on mechanical devices like a roulette wheel, there may be people who appear to be lucky in that they make many more good decisions than bad. Actually such "luck" may just be a subtle manifestation of more skill. The golfer Arnold Palmer is reported as having said "It's a funny thing, the more I practice the luckier I get". So, it is that a lucky general may just be one who has a better appreciation of the lie of the land and so disposes his forces more effectively. Sporting fixtures very often give unexpected results with the "underdog" individual or team winning, and sometimes winning well. That is just one aspect of the variation that occurs in all human affairs. In 1985, in the fourth round of the FA Cup Competition,
Everyday Probability and Statistics
the third division club, York City, beat one of the giants of English football, the first division side Arsenal, 1–0. Not only that, but on the day Arsenal were outplayed. The game was played at York, but nevertheless, the outcome was almost unbelievable. If York City and Arsenal had played 100 games of football, it is likely that Arsenal would have won 99 times, but on that day it was the 100th game. Was York City lucky that day? In a sense they were; presumably there was a concentration of factors operating in their favor and against their opponents. The Arsenal team was in no sense at fault — it was a day when they were at the low end of their performance distribution and their opponents were at the high end of theirs. What we learn from all this is that in all human affairs there is an element of chance that may affect either performance in achieving certain objectives or the outcome of decision-making. Members of a mature and thoughtful society will recognize this and accept the ups and downs of life for what they are, the manifestation of entirely unpredictable statistical fluctuation of outcomes in particular situations. If rainfall had been low, so that the levels of reservoirs had substantially fallen, then a prudent government minister would order restrictions on the use of water — to the great inconvenience of those affected. If 1 month after the restrictions were imposed there was a period of heavy rain that quickly replenished the reservoirs, then it may be possible to show that the restrictions had been unnecessary. In the nature of the shallow thinking that accompanies most politically motivated judgment, the minister will then undoubtedly be accused of making an unnecessary premature decision. With the benefit of hindsight decision-making is much easier, and is extremely easy for those who do not have to make the decisions in the first place. Actually, a government minister who ignored the drought by imposing no restrictions and whose decision was rescued by the subsequent rain would be worthy of dismissal, despite the fact that his course of action had less inconvenienced the public. He might be a lucky minister on that occasion, but it would be unwise to put one's trust on the future luck of such a reckless minister. In the next
Chance, Luck, and Making Decisions
decision he makes, his luck might run out and then a disaster could follow his lack of judgment.
18.4. To Fight or Not to Fight — That is the Question
The world is a very dangerous place because of the nature of its inhabitants. There seems to be an inbuilt tendency in human being toward violence based on a tribal instinct. However, unlike the smallscale conflicts that went on between tribes in primitive societies, the conflicts of the modern world encompass a huge range of scales. When the football teams Glasgow Celtic and Glasgow Rangers meet, there is an antagonism between the supporters based on a religious and political divide that has its origins on the island of Ireland. The skirmishes that sometimes accompany such games are unpleasant, sometimes tragic, but are not a threat to mankind in general. Curiously, when Scotland plays England at football, the erstwhile rival fans of the Glasgow teams are united in their opposition to the English tribe. The tribes are not static but can reorganize themselves in various ways based on a football team, a country, color of skin, religion, school, university, etc. Any individual may be a member of several tribes and so may be an ally of another individual in one context and an opponent of the same individual in another context. The most important decision that any country can make is whether or not to go to war. If the country is attacked and wishes to defend itself, then there is no decision that needs to be made — someone else has made it already. There are many reasons for going to war as an action rather than as a reaction. Some local wars are caused by border disputes, often between ex-colonial countries where the colonial powers had created artificial frontiers that did not respect traditional boundaries. In the past, many wars were based on imperialism, the wish to extend the territorial possessions of a country. That was the motivation of the many wars that established the colonial empires of, for example, the United Kingdom and France. Such empire building is now outmoded, a thing of the past. Another
was that it did. the combination of unstable regions of the world plus the presence of oil in those regions has proved to be a very combustible mixture. or was developing. The most significant conflict at the beginning of the 21st century has been the invasion of Iraq and the overthrow of Saddam Hussein. The United Kingdom and France declared war on Germany on 3 September 1939. There are many overtones associated with this conflict. headed by Hans Blix. There may have been other background reasons. and by the United Nations itself. The Falklands conflict of 1982 might also be claimed to have been a war based on principle since the economic benefit to the United Kingdom was far exceeded by the cost of reoccupying the islands and the potential dangers that such a military expedition involved. if so. but the general belief at the time. France and Germany. There was good reason for such a belief. often complained . based on intelligence sources. there are wars based on attempts to obtain economic benefits. for the Iraqis it was more based on a border dispute somewhat fueled by traditional antagonism between Persians and Arabs. Iraq certainly had once possessed chemical weapons and had used them against its own population and against Iranian soldiers in the Iran–Iraq war that raged from 1980 to 1988. because Germany had invaded Poland and the United Kingdom and France had treaty obligations with Poland to defend its integrity. Iraq never fully accounted to the United Nations for the stock of chemical weapons it was known to have had in 1991 although at some time they must have been destroyed by the Iraqis themselves. chemical or biological weapons? Was the public deliberately misled about the perceived nature of the threat posed by Iraq? Of course we now know that Iraq did not have available stocks of chemical or biological weapons. Was the invasion of Iraq legal or illegal? Were there good reasons. The UN weapons inspection teams. to believe that Iraq had. In this war the Iraqis were largely supported by the United States that saw the Iranian revolutionary government as a threat to its oil supplies. they were not the ones advanced at the time. but.198 Everyday Probability and Statistics reason for declaring war could be of a question of principle. even in those countries that opposed the war such as Russia. Finally.
Germany alone could be seen to be taking a principled stand without any discernable ulterior motive. and Germany. The UN position on Iraq was basically dominated by two blocks. but of even greater significance. which urged that the UN inspection teams should be given more time to hunt for weapons. The United States and the United Kingdom clearly believed that the General Assembly would prevaricate endlessly and that the time had come to offer those serious consequences as a real possibility and so assembled forces on the borders of Iraq. leading to a large loss of life. but the long-term future of Iraq became uncertain. Russia. and Making Decisions 199 of obstruction by the Iraqi authorities and one must wonder why they were being obstructive since there was nothing to hide. Given the character of Saddam Hussein. passed by the United Nations had concluded in section 13 that "… the Council has repeatedly warned Iraq that it will face serious consequences as a result of its continued violation of its obligations". Resolution 1441. but the follow up to the invasion by the coalition forces was pathetically inept and the situation developed into a state of chaos with an influx of foreign terrorists. which wanted strong and immediate action. based on antagonistic branches of Islam. That we now know — but it was not known for certain in 2003 and there were good reasons to think otherwise. and the second by France. armed resistance to the coalition forces. which owed both of them a great deal of money. Iraq was invaded successfully. Luck. The general view is that the US position was mainly dominated by a concern about oil supplies coupled with a deep distrust of Iraq after that country's invasion of Kuwait in 1990. his history of violent behavior.Chance. Of the major players in this crisis. it seems remarkable that Iraq was free of chemical weapons in 2003 — but so it was. . the first headed by the United States and the United Kingdom. conflict between different Iraqi groups. An elected government was formed after a successful general election. and his reputation for deviousness. France and Russia were the countries with the strongest commercial links with Iraq. The rest is history as they say. On the other hand.
200 Everyday Probability and Statistics Both in the United States and the United Kingdom. and the fact that the justification given for invading Iraq turned out to be invalid. as we have previously stated. the tide of public opinion which. to pursue its own path. They would argue that. 18. There can be no doubt that the outcome of the war was not as had been expected — but.5. However. there was a possibility of a catastrophic outcome if nothing at all had been done by the international community and that the international community was far too divided to decide on a course of action. The Mathematics of War For many people the rights and wrongs of the Iraq war are simply matters of legitimacy of the action. there is a counterargument that. We shall consider the question of whether to invade or not to invade by looking at the probabilities of postulated favorable or . There is strength in that argument. it created a dangerous precedent in which the one superpower now on Earth overrode the primary agency for international order. Here we shall sidestep all these questions of legitimacy and morality — important though they are. even if the war had produced only benign outcomes. the United Nations. we now try to construct a theoretical framework to assess whether or not there was any rational justification for the path taken by the United States and the United Kingdom. then it should have been taken later with the approval of the whole international community — if that were possible. Many countries in the United Nations believed that if action was to be taken. hindsight is a wonderful guide to policy. at first supported the war in Iraq — more so in the United States than in the United Kingdom — later swung decisively in the opposite direction with strong and vituperative criticism of both the President and the Prime Minister. with the state of knowledge that existed at the time the invasion was carried out. Putting aside all questions of the veracity of the leaders or the quality of their intelligence sources. taken without specific UN sanction.
the nature of the good and bad outcomes will also depend on individual assessments of what the reactions of participants — the Iraqi people. Thus. p5 +p6 = 1. p9 +p10 = 1. since the outcomes are mutually exclusive. As has been previously mentioned. p1 +p2 = 1. A decision tree for deciding whether or not to invade Iraq.1. and Making Decisions Do not invade P2 WMD do not exist P4 WMD exist P1 P2 WMD do not exist P8 P9 Good outcome E P10 Bad outcome F 201 Invade P1 WMD exist P3 Good outcome A P5 P6 Good outcome C P7 Bad outcome D Bad outcome B Fig. where alternative possible outcomes deviate from the single point. 18. p3 +p4 = 1. and the Iraqi government — might be. Firstly. p7 +p8 = 1. This is a notional exercise to illustrate the principles involved in decision-making.Chance. . The decision tree for this exercise is presented diagrammatically in Fig. foreign terrorists.1. the UN term for what it was seeking in Iraq. inserting different probabilities. Lastly. will come to different conclusions. 18. Several things must be said about this decision tree before we contemplate using it. the reader should insert his own ideas to see what the decision tree suggests. The actual values of the probabilities will be a matter of judgment and different individuals. where WMD stands for weapons of mass destruction. Here we suggest some probabilities and outcomes. there are an infinite number of unpredictable outcomes. as we discussed in relation to medical treatments in Chap. the sum of the probabilities must be 1. It is not intended as either justification for invading Iraq or for not doing so. 6. Luck. unfavorable outcomes.
confident that the UN will not physically intervene.8 (p2 = 0. p1 = 0. in considering outcomes A.2). As of A except that there are no WMD to destroy. Restarts nuclear weapons research and uses threat of chemical and biological attack on Israel. Iraq continues to threaten its neighbors. This scenario might not appear for a few years and would be difficult to counter. Emboldened by lack of UN action the Iraqi regime used WMD and intermediate-range missiles to threaten or bully its neighbors. and foreign terrorists exploit the chaotic situation. and a democratic government is established. It also stops persecuting opposition groups within Iraq. There is local resistance to the coalition forces. Iraq becomes peaceful. Bad outcome B Good outcome C Bad outcome D Good outcome E Bad outcome F We now consider some notional probabilities to attach to these outcomes. Good outcome A The WMD are destroyed. B. delivered by rockets. In view of the rather tempestuous nature of Middle-east societies. where political. religious.202 Everyday Probability and Statistics Based on general expectations by the United Nations and by individual countries. There is a possible threat of Middle-east war involving several countries and consequent world economic chaos when oil supplies are disrupted. to deter intervention by western powers. sectarian strife ensues. or other differences often lead to armed conflict. a tyrannical government is removed. Despite having no WMD. The Iraqi regime does not attempt either to begin or to expand WMD production and settles into peaceful coexistence with its neighbors. and C a bad rather .
Without WMD. However. as it must be. The sum of these probabilities is 1. Invading Probability of a good outcome is p1 p3 + p2 p6 = 0.6.2 × 0.34.5 = 0.6 + 0.Chance. bad outcome D.2 × 0.4 = 0.4.4 (p4 = 0. Iraqis are.6 = 0. Almost every reader would disagree with some aspect of this analysis or even reject the whole concept of looking at the problem this way. However.4 + 0. We take p7 = 0.8 × 0. Iraq would be less likely to be aggressive.7 + 0.8 × 0. or some variant of it.6) and p5 = 0.2 × 0.66. what should not be rejected is that it is possible .5 (p10 = 0. We are now in a position to estimate the total probabilities of good or bad outcomes for both the decision to invade and the decision not to invade. seems more likely than the alternative outcome E.8 × 0. Probability of a bad outcome is p1 p4 + p2 p5 = 0.8 × 0. since good and bad outcomes are mutually exclusive.6 (p6 = 0.7 (p8 = 0.4). Not invading Probability of a good outcome is p1 p8 + p2 p9 = 0. and Making Decisions 203 than a good outcome would be expected. Probability of a bad outcome is p1 p7 + p2 p10 = 0.5). With the assumed possession of WMD and taking into account Iraq's previous wars with its neighbors. on the whole.3). This gives the judgment p3 = 0. so the expectation of a good outcome might be better than is generally expected in the region.2 × 0. Luck.3 + 0.5 = 0. so we take p9 = 0. intelligent and well educated and are the inheritors of an ancient civilization.
but it is not a sensible thing to do as the potential consequences of a small-probability accident are so bad. Crossing a quiet country road with your eyes shut might be a fairly safe thing to do. There are many possible scenarios that could be envisaged. Such decisions have to be made on the basis of perceived probabilities of this or that happening. What is the outcome of a good outcome in this case if no invasion was carried out? Could this have led to a different decision about whether or not to invade? . eliminating it as a possibility might be considered to be of paramount importance.204 Everyday Probability and Statistics that both options — to invade or not to invade — could more probably give a bad outcome than a good one. Problem 18 18. often they are between what is bad and what is worse. Decisions often have to be made without complete knowledge — knowledge that. one will be open to criticism — the critics will only see the bad outcome and not the potentially worse outcome of the alternative choice. It could be argued that without the availability of WMD.1. even if its probability was very low. What has not been considered in the above analysis is the relative quality of the bad outcomes.1 should be 0. just as one bets on a horse in the light of a probability based on its previous performances. Whatever one chooses in such a situation. the eventual critics will have. Bad outcome D could be so catastrophic that. but the only one we know the consequences of is the one that actually happened — and that turned out to be very bad indeed. 18. with hindsight. Backed horses do not always win and logically-taken decisions do not always turn out well. There is a tendency to look for scapegoats when things go wrong. Choices in real life are not always between good and bad. That is something an educated citizen should appreciate. and with the previous experience of the war in which Kuwait was liberated. Iraq would have followed a peaceful path so that the probability p10 in Fig.
along with some periods of strong winds too. This is the probability that a given patient with the disease will die. Since all numbers are equally likely to occur the probability 1 is 12 . The fraction of patients who died in the epidemic is 4205 = 0. 1.2. with warm weather in any sunshine in the east. rather warm conditions are expected.1. Low pressure is expected to affect northern and western parts of the United Kingdom throughout the period. Most likely Least likely 123 1. There is a risk of some showery rain over south-eastern parts over the first weekend but otherwise much of eastern England and possibly eastern Scotland should be fine. 205 . However. More central and western parts of the United Kingdom are likely to be rather unsettled with showers and some spells of rain at times.4. 1. with a southerly airflow dominating.029. The words showing lack of certainty are underlined.3.Solutions to Problems Problems 1 1.
and 5 + 1. Since getting the 5s are independent 6 events the probability of getting the two 5s is P5+5 = 1 1 1 × = . the probability of getting a particular combination of 1 numbers is 72 . There are four aces and getting them are mutually exclusive events. 12 6 72 2. For each coin the probability of getting a head is 1/2. From 220.127.116.11. 52 52 52 52 13 2. The probabilities that the dodecahedron and die will give a 5 1 are 12 and 1 respectively. Hence the probability of getting a number less than 6 is P1 to 5 = 1 1 1 1 1 5 + + + + = . The probability of picking a particular ace is 52 .1.5. 3 + 3.206 Everyday Probability and Statistics Problems 2 2. These combinations are . 12 12 12 12 12 12 1 2. 13 2 2 2 2 104 2. 4 + 2. There are five sides less than 6 and getting these are mutually exclusive events. 2 + 4.2 the probability of picking a jack 1 is 13 . Since all the events are independent the total probability is PJ+4H = 1 1 1 1 1 1 × × × × = smaller than that found in 2.3. The combinations that can give a total of 6 are: 1 + 5. Hence the probability of getting an ace is Pace = 1 1 1 1 1 + + + = . From the argument given in 2.3.
2 5 7 9 13 19 . 41 41 41 41 1681 Problem 3 3. We calculate the summation on the right-hand side of (3.0835. f ). 41 Note that the sum of the two probabilities is 1 (the gene has to be one or other of the mutually exclusive possibilities) and that the ratio is 1:40 as given.1.6. 72 72 2.0476. the combinations being mutually exclusive. Hence the probability of being a carrier is Pcarrier = 40 40 1 80 1 × + × = = 0. The probability that a particular gene will be f is Pf = 1 41 and that it will be F is PF = 40 . F) or (F. (i) To have the disease each gene of the pair possessed by the individual must be f . 41 41 1681 (ii) To be a carrier the gene pair must be (f .3) for the three races. Hence the probability of having the disease is Pff = 1 1 1 × = .30 pm Sum = 1 1 1 1 1 1 + + + + + = 1.Solutions to Problems 207 mutually exclusive so the probability of getting a sum of 6 is Psum=6 = 5 × 5 1 = . 2.
3!3! One of those non-ordered selections is red + green + blue so the 1 probability of picking that is 20 = 0.45 pm Sum = Sum = 1 1 1 1 1 1 + + + + + = 1. 4. 3 4 4 5 7 9 1 1 1 1 1 = 0.3. and 1 blue. The sum is the greatest for the second race and hence is the best from the bookmaker's point of view (and worst for the punters' chances of winning!). Number of ways of selecting three balls. taking the order into account.2873. The number of different orders is 5! = 5 × 4 × 3 × 2 × 1 = 120. so the probability is 120 . in that order. Problems 4 4.2. Ways of selecting three balls without regard to order is 6 C3 = 6! = 20. One of those ways is red. 4 is four.208 Everyday Probability and Statistics 3. + + + + 3 4 5 9 13 The third race does not satisfy the golden rule. 4×3×2×1 1 There is one way to get the first prize so the probability is 4. A third prize is won with three correct numbers and one of the 15 numbers that is neither a correct number nor the bonus ball . The number of ways of picking four numbers from 20 is 20 C4 = 20×19×18×17 = 4.05. The number of ways is 10 C4 = 6!4! = 210. 845. with three correct numbers plus the bonus ball. 4. The number of different selections of three correct numbers from four is 4 C3 = 4.845 . The probability of getting a second prize is thus 4.9714. so the number of different ways of getting a second prize.845 . green. is 6 × 5 × 4 = 18.104.22.168 pm 3. 10! 4.
(i) The number of ways they can have birthdays is 100 × 100.1. Problems 5 5.9411. Hence the probability they have different birthdays is 100×99 100×100 = 0.6282 0. The table similar to Table 5. (iii) We calculate the series of terms 100 99 98 97 × × × × ··· 100 100 100 100 until it falls below 0. The probability they have different birthdays is 100×99×98×97 100×100×100×100 = 0.4428 The number of Arretians for a greater than 50% chance that two have the same birthday is 13.5653 n 3 6 9 12 Probability 0. 60 4 so the probability of winning a third prize is 4.9702 0.6903 0. (ii) The number of ways they can have birthdays is 100 × 100 × 100 × 100.5. . At this number the probability that two Arretians have the same birthday is greater than 0.5.9900 0. The number of ways of obtaining this is 4 × 15 = 60.99. The number of ways they can have different birthdays is 100×99×98×97.8583 0. The number of ways they can have different birthdays is 100 × 99.1 is n 2 5 8 11 Probability 0.7503 0.9411 0.9035 0.Solutions to Problems 209 number.845 = 323 .8068 0.5032 n 4 7 10 13 Probability 0.
18.104.22.168 = 106. 14 86 14 86 .1. The number cured with drug a 130 × 0. The table similar to Table 5. Consider a notional 200 patients with the symptoms. drug b 130 × 0.2 is Outcome Probability Profit (loss) on first game (−25) 5 5 1 14 Profit (loss) on C&A (−25) 5 5 2 13 0 0 5 6 5 6 1 6 1 6 × × × × 5 6 1 6 5 6 1 6 = = = = 25 36 5 36 5 36 1 36 0 6 6 6 0 6 Net loss Problems 6 6. Of these 130 will have condition A and 70 condition B.2. Hence the doctor should prescribe drug a.210 Everyday Probability and Statistics 5.3 + 70 × 0.6 + 70 × 0. The O and E tables are Buy old No buy Buy No buy 8 92 O old 14 86 E new 20 80 new 14 86 These give χ2 = 36 36 36 36 + + + = 5.9 = 102.
There is one path to A so the probability of getting there (or to E) is 1 . costing 90p.2. Problems 7 7.10256. (i) The probability of setting a point of 4 and then losing by throwing a 7 is P4. Each path from the top point to a cup has a probability of 1 3 1 9 × 1 3 . a total of 80p in all.. × 2652 52 51 4 16 4 × = = 0. The probabilities of the various ways of getting 20 are 10 + 10 11 + 9 9 + 11 probability probability probability 16 15 240 = = 0.00603.7 = Probability of getting 4 × Probability of getting a 7 before a 4 = Probability of getting 4 × (1 − Probability of getting a 4 before a 7) 1 1 1 × 1− = . the 3 player receives back 40p from the ball reaching B and D and 40p from the ball reaching A or E. Hence the indication is that the new packaging is effective.09050. There are two paths to B so the probability 9 of getting there (or to D) is 2 . 7. This is much less than the decision limit of 0.05.02.1 indicates that the probability of getting this value or above is between 0.00603.e.3. 7. There are three paths to C so the 9 probability of getting there is 1 . In nine games.01 and 0.1. 52 51 2652 4 16 4 × = = 0. 52 51 2652 Hence the total probability is 0. i. = 12 3 18 .Solutions to Problems 211 Table 6.
is .7 + P9.7 + P8.7 . P9.7 . Problems 8 8.12.7 = Probability of getting 6 × Probability of getting a 7 before a 6 = Probability of getting 6 × (1 − Probability of getting a 6 before a 7) 5 5 5 × 1− = . 18 15 66 66 15 18 and P8. = 36 11 66 We also have P10.7 + P5. 100 100 100 100 100 100 There are five degrees of freedom and according to Table 8.3960.212 Everyday Probability and Statistics (ii) With the notation of (i) we need to find P4. or some greater value.7 .7 = P4.7 + P10.9 the probability of getting this value of χ2 .7 + P9.7 .7 + P5.7 = Probability of getting 5 × Probability of getting a 7 before a 5 = Probability of getting 5 × (1 − Probability of getting a 5 before a 7) 2 1 1 = .7 = P5.7 + P8.7 + P10.1. The expected number for each side is 100 so that χ2 = 144 169 361 81 1 256 + + + + + = 10.7 1 1 5 5 1 1 = + + + + + = 0. = × 1− 9 5 15 P6.7 + P6. Hence P4. P5.7 = P6.7 + P6.
05 and 0.2. . Hence.2222 for fair dice).3. is close to 0.28. we may deduce that the wheel is almost certainly biased. The observed and expected tables are 0 Observed 20 non-0 350 0 Expected 10 non-0 360 This gives χ2 = 100 + 100 = 10. The probability of getting either 2 + 5 or 5 + 2 is 1 2 × 1 × 1 = 18 . 24 18 18 36 8. 8. This is less that the 10% criterion so we conclude that the die is probably unfair.1.Solutions to Problems 213 between 0. The probability of getting either 1 + 6 or 6 + 1 is 1 1 1 2 × 4 × 12 = 24 . 6 Hence the probability of getting a natural is Pnatural = 1 1 1 1 + + + = 0.001. or some greater value. For one degree of freedom the 10 360 probability of getting this value of χ2 . 6 6 The probability of getting either 3 + 4 or 4 + 3 is 1 2 × 1 × 1 = 18 .1806 (0. 6 6 The probability of getting either 5 + 6 or 6 + 5 is 1 1 2 × 1 × 12 = 36 .
Distribution of dress sizes 0.1.996 so the total profit would fall. The average is 2+4+6+7+8+10+12 7 = 49 7 =7 V= σ= (2 − 7)2 + (4 − 7)2 + (6 − 7)2 + (7 − 7)2 + (8 − 7)2 + (10 − 7)2 + (12 − 7)2 √ 7 10 = 3.2. Sales of 47. the area under the normal curve from the mean to 21/2σ is 0. 10.49379 . The changed profit would be of 0. and (iii) 31.214 Everyday Probability and Statistics Problem 9 The histogram representing the distribution of dress sizes is shown below. (i) 311 .83 × 220.127.116.11 0.1.05 0 6 8 10 12 14 16 size 18 20 22 24 26 Problems 10 10. the profit per dress would change by a factor of 1.15 proportion 0.2 0. If only sizes inside the range 8 to 18 were stocked then the number of dresses sold would change by a factor of 0.000 are 21/2σ below the mean.2.3.2 = 0. However. (ii) 312 . = 10 10. From Table 10.3 .
0902 + 1. Hence the number of days per year with sales below 47.4112 = 1. Sales of 55.00621 × 365 = 2.0862 + 1. School 4 h2 = 0.8362.. 11. Problems 11 11.9988) + (50 × 1. from Table 10.000 are 11/2σ from the mean.2672 = 1.3522 = 1. School 2 h2 = 0. 0.0912 + 1. Hence the number of days per year with sales above 55. School 3 h2 = 0. = = 1. Numbers Squares 2 4 4 16 6 36 7 49 8 64 10 100 12 144 Mean = 7 Mean square = 59 Variance = mean square – square of mean = 59 − 72 = 10.4332 so the area beyond 11/2σ is 0.9988.000 will be 0.372) 62 + 47 + 54 + 50 ha.352) + (47 × 1.Solutions to Problems 215 so the area beyond 21/2σ is 0.8409 The mean height of all the children is (62 × 1. .1.8362) + (47 × 1.2.353 m.0668.8905.411) + (50 × 1. to the nearest whole number. The area from the mean out to 11/2σ is.0892 + 1.267) + (54 × 1.1.6127.0668 × 365 = 24. We first work out the mean square height for each school from ¯ V + h2 .3722 = 1. to the nearest whole number. The mean square height for all the children is (62 × 1.8905) 62 + 47 + 54 + 50 h2 = all = 1.6127) + (54 × 1.000 will be 0. School 1 h2 = 0.00621.
014. the probability of which is 0.3679. Hence the most probable support for party A is 0.52. For this Poisson distribution the average a = 1 −1 ×10 −1 ×11 −1 ×13 (i) e 0! = 0.216 Everyday Probability and Statistics The variance of the heights for all the children is Vall = h2 − hall = 1.2.986. The average number of asteroids per 1.1 ×0.0 − 0. The −0. (ii) e 1! = 0.0 − 0. 0.19σ may be found 9 from Table 10.1.9048. The proportion of those polled supporting party A is 0.486 = 0. or 13 days to the nearest whole number. 12. Hence the probability that at least one will fall is 1. For party B to win the actual result would have to be more than 2. 3000 The indicated support is 0. Hence the probability that party A will have more than 50% of the popular vote is 1.00912. Problems 13 13.486.10 probability that no large asteroids will fall is e 0! = 0.52 × 0.00912 The area under the normal curve out to 2.48 = 0. −1 5 ×1 and (iv) e 5! = 0.19σ from the 50% level.1σ toward 2.2σ.014 = 0.1 by going 10 of the way from 2.50 σ = 2.01029 m2 all Hence the standard deviation is σall = √ Vall = 0.3.9048 = 0.7.0952.5 − 0.000.19σ from the mean.000 years is 0. The number of days that 15 would fail is 365 × e 15! = 12.1014 m.0613.1. (iii) e 3! = 0. .52 with a standard deviation σ= 0. This area is 0. −10 ×1015 12.3679. 2 Problems 12 12.3532 = 0.0031.8409 − 1.1.52−0.
This is found by adding the squares of their heights and dividing by 20. or further.0352. From Table 10. Problems 14 14.85694 m2 .0 m is 2.2.097 m.14σ from the 50% level. 0. we first need to find the mean square height of the men.022 m.041.9615 σap = 1.1. = 1.2σ. the probability of this is 0.9615 m.128. This gives h2 = 3. from the sample mean is found from Table 10.0995 m. This gives the mean height as h = 1.54 200 = 0.14σ from the mean. from (14.46 0. (i) The mean height is found by adding the given heights ¯ and dividing by 20.0094577 m2 √ giving σh = V h = 0. 19 (v) 2.022 The probability of being this far. Hence the variance of the height is ¯ Vh = h2 − h2 = 3.75 σap from the sample mean.1.459 = 0. (iii) Applying the Bessel correction the estimated standard 20 deviation for the whole population is σp = 19 × 0. The standard deviation of this estimate is σ = 0.46. From the sample the proportion of male snakes is 0.097 = 0. interpolating between 1.96152 m2 = 0. (ii) To find the variance.0−1.Solutions to Problems 217 13.1 as 0.0094577 = 0.5 − 0372 = 0.85694 − 1.4) σap = Vh = 19 0. The sample proportion is For more than 50% of snakes to be male the sample average must be more than 1.0352 σ 0.1σ and 1. (iv) The estimated standard deviation of the sample mean is. .5 − 0.5−0.46×0.
22 g.5 The actual difference is 0. Problem 15 15.04899 g giving a standard deviation of 0.12 +2. The probability of being that far.218 Everyday Probability and Statistics 20 14. since 100 fish were tagged the total number of fish in the pond is 500.5 − 0. Ratio = 0. Estimating figures from a graph is not something that can be done precisely but the following estimates are reasonable. 0.0236. The proportion of the sample tagged is 100 = 0. The difference in mean weights from the two samples is 0. We take the null hypothesis that there is no difference in the feeding regimes. or further. 2 × (0.30 The number of cars on the road in the period in question has greatly increased. from Table 10. which makes the reduction in road deaths more remarkable.4882) = 0. The variance of the difference Vdiff = 2. (ii) 1964 1982 2005 Pedestrians Total Ratio 3000 7700 0.2. (i) Number killed in 1964 = 7700. Problem 16 16.22 = 2. The MOT test for cars was introduced in 1960 for cars over 10-year-old but the age at which the test was applied was reduced until by 1967 it was 3 years.1.5 g.1. Assuming that this is true for all the fish then.39 1800 5950 0. Number killed in 2005 = 3200.2.30 850 3200 0.1. Cars have .32 2 100+100−2 = 0.27. from the mean in both directions is.42.
0 without smoking and 4.56. This is now less than the probability of a bad outcome by invading. The probability of disease is increased from 0.1.6. (i) 0. (ii) 0. and hence p9 = 1. a factor of 1.8 i.095. power-assisted steering and improved suspensions give better vehicle control giving extra safety both to those in the car and to pedestrians.0 with smoking. Problem 17 17.24 The exposure level is 2. Problem 18 18. particularly on icy roads.e.8 × 0.28.2 × 0 = 0.Solutions to Problems 219 become much safer with the introduction of mandatory safety belts and of airbags that protect both the driver and passengers.175 to 0.75 = 1.. . 2. With p10 = 0. the probability of a bad outcome by not invading is reduced to 0.22.7 + 0. and (iii) 0. Front-wheel drive.1. Advanced braking systems (abs) have also contributed greatly to safety giving braking in shorter distances without skidding.
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146 difference of sample means. 30 Bortkiewicz. 204 choosing best medication. 55 ERNIE. 81 Dewey. 162 Fitzgerald. 35 choices. 153 with many parties.Index American craps natural. 198 fish in a pond. 145 combining polls. 183 British Crime Survey. 149 effect of poll size. 84 arranging n objects. 59. 172 burglary. 25. 18 drugs testing. 35 background radiation. 137 both–and probability. 192 Edison. 149 influences on. 58 early religion. 17 dominant gene. 17 combinations. 21 point. 63. 191 five-card trick. 81. 57 chromosomes. 142 Calphurnia. 139 221 . 49 de Moivre. 13 election polling. 175 call-center service. 194. Thomas E. 114 factorial. 21 probability of winning. 201 degrees of freedom. 45 blackjack. 21 with loaded dice. 15 breast cancer. 187 baseline measurement. 179 crown-and-anchor. 70 flying bombs. Ladislaus. 151 email usernames. 166 disease clustering. 5 chi-squared (χ2 ) test. 188 carcinogens. 69 crime and politics. 141 DNA. 118 birthday problem. 72 snake eyes. 184 carrier. 83 choices and selections. 108 decision making. 39 combining probabilities. 97 bookmaker's golden rule. 184 sensitivity-exposure relationship. UK. 120 either–or probability. 118 exponential e. Abraham. 37 Falklands war. 40 empirical probability. Edward. Thomas. 196 decision-tree for war. 70 block diagram (histogram). 193 cancer. 78 Everest. George. 18 certainty.
19 gene dependent. 156 Gauss. Arnold. 70 post-invasion conflict in Iraq. 192 hemophilia. 161 random errors.222 Everyday Probability and Statistics normal distribution. 67 probability range. 15 invasion of Iraq. 153 light bulb failures. 108. 171 roulette wheel. 100 Oracle of Apollo. 156 variance. 6 independent events. 106 bell-shaped. 131 pontoon. 191 Saddam Hussein. Sir Walter. 89 logical probability. 155 average. 107. Carl. 115 premium bonds. 132 formula. 160 Hawkins. 99 for continuously-variable quantity. 112 formula. 7 Old Faithful geyser. 194 pi (π). 195 Phuket. 16 Gossett. 199 powers of numbers. 192 Kinnock. 120 nature and nurture. 80 Rayleigh. 118 random numbers generation. 108 Gaussian distribution (normal distribution). WS. 156 standard deviation. 183 Heisenberg Uncertainty Principle. 97. Neil. 185 hay-fever. 89 numerical probability. 143 mutually exclusive events. 108 genes. 117 null hypothesis. 137 impossibility. 195 National Health Service. 168 . 91 Rubáiyát of Omar Khayyám. UK. 102 homicide. Sir John. 78 probability machine. 5 punter's best condition. 79. 8 probability. 31 quality control. 55 Monte-Carlo method. 98 bimodal. 147 area under curved. 19 histogram. 135 Poisson. Siméon-Denis. 192 Palmer. 185 reporting of crime. 177 horse-kick fatalities. 78 mutation frequency. 198 judgmental probability. 89 testing. 60. 97. 25 Julius Caesar. 77 mathematics of war. 198 sampling. 14 Napoleon. 158 population. 84 testing for. 200 medical diagnosis. 17 genetically inherited disease gender dependent. 156 numerical example. 25. 145. 142 loaded dice. 161 numerical example. 115 Poisson distribution.
Harry S. 173 Tote. S. 147 statistics. 153 systematic errors. UK. 37 Shipman. 95 of sample mean. 41. 82 UK National Statistics Office. UK. 157 vehicle crime. 153 UK national lottery. 196 . 127 Student t-distribution. 78 223 WMD (Weapons of Mass Destruction). 70 violent crime. 119 tail of distribution. J. 110. 185 restrictions. 199 variance. 198 York City v Arsenal. 131 tribal loyalties. UK. 52 total crime. 201 World War II. 62 significance of differences. 178 sickle-cell anemia. 186 Ulam. Harold. 197 Truman. 167 smoking in the UK. 160 Sun newspaper. 133 UK 1992 general election. 112 Tay-Sachs disease. 16 significance level. 186 standard deviation.Index selections non-ordered. 146 typing-error distribution. 186 variation with time. 78 UN Resolution 1441. 125 combining information. 110. 176 Von Neumann. 17 television box-opening games. 166 variation. 126. 38 ordered. 174 vingt-et-un. 165 significant difference. 25 traffic flow.
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