text: German Tanks, Traffic Wardens, and the End of the World
  The
  November 18, 2014 at 10:45 pm
  “For me, a valid argument about Doomsday would have to involve a particular physical cause for the extinction of humanity”
  Well if the LHC black hole (or worse) doesn’t get us the triffids surely will:
  But a lifeform with a soul (unlike a plain lifeform) can be surprised to find itself incarnated in a naturalistic but fine-tuned and life-supporting cosmos and can infer that Doomsday is probably nothing to worry about anyway:
  November 19, 2014 at 8:28 am
  “Not knowing what this was”
  You hereby own up to not being a regular reader of Cusp’s blog:
  November 19, 2014 at 8:32 am
  ?”
  I think this objection is bogus. If all Brits and all Chinese apply this argument, then only a small fraction conclude that they are atypical, which is the whole point if being atypical is defined as belonging to the minority. It is extremely unlikely to win the lottery, yet someone wins every week. If I say in advance “person X will win the lottery” and indeed person X does, then this is indeed surprising. However, no news reporter would visit someone who has won the lottery to report on a very unlikely event.
  November 19, 2014 at 8:35.”
  This is a bit misleading. When Gott visited the Berlin wall, or a Broadway play, then his assumption that he was there at a random time is valid. However, human population has been growing exponentially for a wrong time, so you should think of yourself being picked from this distribution, and not a uniform one. This crucial difference is the essence of the Doomsday argument. You can’t apply Gott’s argument to argue that the Doomsday argument gives too short a timescale since the underlying assumption of Gott’s argument is a uniform distribution in time. (If you modify this for an exponential distribution then your are back at the Doomsday argument.)
  November 19, 2014 at 9:42 am
  What do you mean by “at a random time”? The phrase seems entirely meaningless to me in this context.
  November 19, 2014 at 9:59 am
  I mean the probability of Gott visiting Berlin in 1969 was the same as in 1974 or 1982.
  November 19, 2014 at 8:44 am
  “If I happen to walk past a building just as it is being finished, presumably I should hang around and watch its imminent collapse….”
  Not really. This would be expected to happen once in a while. You have to integrate over all buildings you have walked past in your life. In a few cases, you would expect this to happen.
  This is related to the discussion of whether a certain statistical anomaly is interesting. An isolated event with a probability of being due to chance of only 1/1000 might be worth looking into. If I investigate 1000 such possible events, then it is not. Same thing.
  November 19, 2014 at 8:52 am
  “The allied command structure had thought the heavy Panzer V (Panther) tanks”
  I’m not an expert on military history, and the tanks might well be called “Panther” in English. However, the similarity to “Panzer” is coincidental. “Panzer” means “body armour”, be that of a knight or a tortoise, and by extension an armoured vehicle. The German word “Panther” is the same as the English word “panther”*. To confuse things, there are various German tanks known as “leopards” (which is spelled the same way in German as in English).
  *For some reason, this cat has many names in English: panther, cougar, mountain lion, catamount, puma, just to name a few. (German has just Panther and Puma.)
  November 19, 2014 at 9:44 am
  The nickname “Panther” was applied exclusively to the Panzer V and is not connected to the name Panzer. There were also “Tiger” tanks.
  November 19, 2014 at 9:58 am
  OK. As I said, I’m not an expert on military history.
  November 19, 2014 at 10:01 am
  I’m looking forward to your blog post on the Sleeping-Beauty problem” (yes, the dash essentially changes the meaning).
  November 19, 2014 at 10:52 am
  “the thing that makes me most uncomfortable about these Doomsday arguments is that they attempt to determine a probability of an event without any reference to underlying mechanism.”
  Yes, that is the point. As you then say:
  .”
  And also the prior. But yes, this is the basic error in the reasoning that you are (rightly) criticising.
  November 19, 2014 at 1:01 pm
  Yes, I should have mentioned the prior in that paragraph – the nature of the prior is however implicitly related to the form of the model.
  November 19, 2014 at 11:52 am
  Gott’s argument is essentially that a typical observer, one observing something at a random time (assuming a flat prior), will not observe this at a special time.* One can turn this around and say that the typical state of some object is that which a typical observer (one observing at a typical time) will see. In the case of the Berlin wall or a Broadway play, this doesn’t provide one with any extra information, because these objects didn’t change much during their existence. However, if the state of an object changes with time, in particular if there is a special, relatively short time when it is in an atypical state, then a typical observer would not expect to observe this state. For this reason, one would not expect a typical observer to observe a large value of Omega in a universe which will collapse in the future, even though Omega becomes infinite (but, of course, for a very short time). I claim that neglecting this effect leads almost everyone to misunderstand one aspect of the flatness problem. I’m still waiting to be proved wrong. 😐 (In a nutshell, the time evolution of a pencil falling after having been almost balanced on its tip, or similar behaviour on the part of a tightrope walker, is quantitatively different than the expansion of a Friedmann universe; this is one example where arguing too far from analogy leads to a wrong impression.)
  *Gott’s argument has provoked a huge amount of discussion, but in its basic form is trivially true: Go to Egypt and you will see the pyramids and you will see an apartment house built in the 1970s. Which will be there 1000 years hence? If I want to leave the room only after I can be reasonably sure that the baby will sleep for an hour, I can be more confident if I leave after the baby has been asleep for 5 minutes than if I leave after it has been asleep for 5 seconds. People do this intuitively even if they have never heard of Gott.
  November 19, 2014 at 12:39 pm
  That’s because they know that the baby’s internal sleep mechanisms are such that if it has been asleep only 5 seconds it is not necessarily fully quiescent whereas if it has been asleep 5 mins then it is. In your other example, stone lasts longer than brick or mud buildings because the mechanisms by which buildings decay are slower for stone. It’s always about prior information about mechanisms and if you are deprived of that then you are in trouble.
  An observation is an observation. Nobody has any right to assume that they are a “typical observer”.
  November 19, 2014 at 1:01 pm
  In both cases, the logic is that a state which has existed for a longer period of time will probably continue to exist for a longer period of time. There are of course explanations for each individual case, but one can come to this (statistically correct) conclusion without knowing the mechanism in each individual case.
  I think if I offered you 1000:1 odds that something which has been in existence for 1000 times longer than something else would last longer into the future than this something else, then you should take the be (if you are interested in maximizing your income) even if you know nothing about the mechanisms determining the longevity of the objects.
  November 19, 2014 at 1:07 pm
  …even if you understand what “typical observer” means!
  November 19, 2014 at 2:34 pm
  A typical observer is one who considers himself to be a typical observer. 🙂
  November 19, 2014 at 3:27 pm
  I think that’s a deluded observer…
  November 19, 2014 at 3:29 pm
  Not necessarily a contradiction.
  November 19, 2014 at 7:25 pm
  “I think if I offered you 1000:1 odds that something which has been in existence for 1000 times longer than something else would last longer into the future than this something else, then you should take the bet (if you are interested in maximizing your income) even if you know nothing about the mechanisms determining the longevity of the objects.”
  I disagree. I think that you are being influenced by your experience of situations in which you know something about the mechanism, such as the two examples you gave. As the point is to strip out all prior information, the problem you invent regarding time is logically equivalent to another regarding space, namely how far you are from the endpoint of a lengthy ruler. In the absence of any observation you are equally likely to be anywhere.
  November 20, 2014 at 10:46 am
  Let me phrase that spatial analogue more clearly. You arrive at a railway station in a country about which you know nothing at all. A local tells you that there are no branches on this line. You ask him how many stops there are to the end of the line in one direction. He tells you – say, 17. Can you infer anything from this about the number of stops to the end of the line in the other direction? In other words, can you infer anything from his reply about the number of stops on the entire railway line (since you can just subtract 17 from that)?
  November 20, 2014 at 3:59 pm
  With 95% confidence, more than 17 but less than 663 total stops.
  November 20, 2014 at 5:58 pm
  Please show me the calculation.
  November 20, 2014 at 8:01 pm
  Following Gott: I don’t know how many stops there are, I know only that I at at the 17th stop. Assuming that all are likely, then with 95 per cent confidence I am in the middle 95 per cent of the stops. Thus, there must be less than 39*17=663 otherwise I would be in the first two-and-one-half per cent. There must also be more than 17 (MAX(17,17/39)), trivially. (If I were at, say, stop 780, then the minimum would be 780/39=20, otherwise I would be in the last two-and-one-half per cent.)
  November 21, 2014 at 10:06 pm
  Suppose the railway (no branches) comprises N stations including the termini; N >2 so that there is at least one intermediate stop. You assign a prior probability Prior(N|I)=f(N) where I is your prior information about railway lines. Suppose the names of all stations including the termini are put in an urn and one is withdrawn, after which you are set down at that station (a “randomly selected” station, meaning that any station is equally likely); this is information J. You ask the stationmaster, “How many stations are there down the line to the left when looking at the track from your office on the platform?” He answers, “n”. (NB n may be zero; the track is such that you cannot tell if you are at a terminus because the buffers are just out of sight beyond the platform). The selection process means that p(n|N,J) is independent of n; as there are N possible answers (0,1,…(N-1)) it follows that p(n|N,J)=1/N for n=0,1…(N-1) and =0 for n>(N-1); in other words p(n|N,J)=(1/N)H(N-n-1) where H(*) is zero if its (integer-valued) argument is negative and is unity if its argument is zero or positive. (It is of course assumed that all variables are non-negative.) Bayes’ theorem tells us that p(N|n,I,J) is proportional to p(n|N,I,J)p(N|I,J) where the constant of proportionality comes from normalisation over N. Now, I is redundant in the conditioning information in p(n|N,I,J), and J is redundant in p(N|I,J), so that p(N|n,I,J) is proportional to p(n|N,J)p(N|I), which upon substitution is proportional to (1/N)f(N)H(N-n-1). Suppose now that there are m stations along the track to the right. We can change variable from N to m, since N=m+n+1; the result is that the probability distribution for m is proportional to 1/(m+n+1) * f(m+n+1)H(m) where the constant of proportionality comes from normalising over m from 0 to infinity. As all variables are understood to be non-negative the H-function need not be written and the posterior for m is equal to (1/K)1/(m+n+1) * f(m+n+1) where K is given by normalising over all non-negative values of m. This is the Bayesian solution and is the basis for analysing all other intuitive analyses, which hopefully I’ll do within 24 hours.
  November 23, 2014 at 7:12 pm
  Now the continuous case. You are put down on a non-branching railway track at a “random position” defined as being equally likely anywhere. You assign a prior density f(L) for the total length L of the railway based on your knowledge of railway geography; \int dL f(L) =1 integrated from 0 to infinity. You have a railway surveyor handy and you ask him how far to the end of the line looking to the left from the side of the tracks you are standing on. He answers “x” (in the same units of distance as L). What is your posterior density for y, the distance to the right to the buffers, ie y=(L-x) ?
  An analogous analysis gives the answer
  (1/K) 1/(x+y) * f(x+y)
  where K is a function of x, given by
  \int dy 1/(x+y) * f(x+y) from 0 to infinity.
  Now we can start comparing with intuitive analyses. The prior f(*) is bound to contain some characteristic distances (at the least, its mean) and the mode for y can be compared to these characteristic distances for various forms of f.
  November 24, 2014 at 12:23 pm
  So, you agree? 🙂
  November 24, 2014 at 12:28 pm
  Busy weekend Philip, as you can judge from my warning of delayed responses on Friday. Then and in the continuum case I set the scene by giving the Bayesian (ie, correct) answer. Before I analyse your argument, here’s a question: What if, before the ‘experiment’, you were 99.999% certain that there were 997 stations?
  November 24, 2014 at 12:31 pm
  I don’t really see a conflict. The Gott case is that you are there at a random time and don’t have any further information. Of course the conclusion changes if your prior information changes, which is why it doesn’t work in the case of a human, for example (finite maximum age and a pretty sharp cutoff at that). Obviously, if I am 99.999% certain that there were 997 stations then if I find myself at number 996 I will reach a different conclusion if I didn’t have this prior information.
  November 24, 2014 at 1:36 pm
  OK Phillip, let me sharpen the question. Suppose your prior info is such that you are 100% certain that the railway has 997 stops. This is still a special case of the probabilistic situation, and can therefore be used to test assertions based on intuition rather than Bayes’ theorem. In that case, are you still 95% confident that you are in the middle 95% of all the stops?
  November 24, 2014 at 2:40 pm
  No, but I never said I was. 100 per cent sure is strong prior information. The whole point is whether one can say anything at all if one has essentially no information.
  November 24, 2014 at 2:54 pm
  Phillip, Your statement “Assuming that all are likely, then with 95 per cent confidence I am in the middle 95 per cent of the stops” is not contingent on the prior information you have, and I am showing you by means of extreme prior information that it can therefore fail. It is therefore untrustworthy – which, in problems involving inference, is another way of saying wrong.
  November 24, 2014 at 3:15 pm
  The point is that this statement applies in the lack of any prior information. Gott visiting the Berlin Wall in 1970 or whatever with no further information, and Gott visiting the Berlin wall knowing when it was built, or with intelligence indicating the rise of Gorbachev, is a different situation. If a conclusion holds in the absence of prior information, then one can’t say that providing prior information changes the inference. This is true, but not relevant to the original problem.
  I get off of a bus and look at the house number on a street (where houses are consecutively numbered). What is the highest number on the street? I can only guess, of course, and for a 95 per cent confidence answer I will be wrong 5 per cent of the time. However, if I see the number 2 I will reach a different conclusion than if I see the number 200. All Gott’s argument does is quantify this.
  November 24, 2014 at 3:53 pm
  Phillip,
  My station problem is already a simple model problem and I don’t think your new model problem is any simpler. I’m reluctant to analyse it because there can be subtle differences between a thing and an analogy to it. We are in disagreement about the railway problem; let’s try to settle it there.
  Let proposition A = “the middle 95% of stations are in the urn” and B = “my station is in the middle 95%”.
  Then you are reasoning (I think) that
  p(B|A) = 0.95 = p(A|B).
  This is the usual error appearing in intuitive non-Bayesian reversion; not all people who are female are pregnant, but all people who are pregnant are female.
  November 25, 2014 at 10:56 am
  Phillip: to make my critique more specific, with 95% probability you are in the first 95% of the track, or the middle 95%, or the end 95%, or – crucially – the subset of track comprising the first 47.5% plus the last 47.5%.
  November 25, 2014 at 4:44 pm
  Right, but all are equally correct. When making a contour plot, there are in general an infinite number of 95 per cent contours. It is convention to draw the smallest such contour, but in the case of a constant pdf there is not a unique such contour.
  November 25, 2014 at 8:43 pm
  “100 per cent sure is strong prior information. The whole point is whether one can say anything at all if one has essentially no information.”
  The whole point is to make a general statement that doesn’t fail in special cases.
  November 25, 2014 at 9:09 pm
  Phillip, I confess that I don’t fully understand your brief argument at 8.01pm on 20th, and as a Bayesian who finds that it fails in special cases I take that to mean it contains a subtle error. Unless you fill it out, the only thing I can do is to give the Bayesian version, as follows:
  Define the proposition
  A = “my stop is in the middle 95% of all stops”
  Then prior(A|J) = 0.95. Define
  B = “my stop is the 18th”
  To calculate posterior(A|B,J) we must use Bayes’ theorem:
  posterior(A|B,J) = (1/K) prior(A|J)likelihood(B|J,A)
  where
  K = prior(A|J)likelihood(B|J,A) + prior(~A|J)likelihood(B|J,~A)
  The priors are known, the likelihoods can be found, but even then you have to play intuitive tricks to get the probability for the number of stations – when Bayes’ theorem allows you to consider this proposition directly.
  November 25, 2014 at 9:52 pm
  Maybe the difference is that Gott doesn’t think of being in the middle 95 per cent as a prior. Maybe that is wrong, I don’t know.
  Say there are 100 children in a school, and the are numbered by consecutive integers. Picking one at random, there is a 90 per cent probability that the number is between 6 an 95 (inclusive). I hope we can agree on that.
  Suppose I am give the number n of a child at random and want to estimate the size of the class. AT 90 per cent confidence, it is between 19/18*n and 19*n.
  That’s it. Is this a different answer than the Bayesian answer?
  November 25, 2014 at 11:47 pm
  Never mind Gott, the argument at 8.01pm on 20th was yours; can you fill it out please?
  November 27, 2014 at 1:27 pm
  This is just a direct translation of Gott’s time argument to railway stops; if I’m at stop 17 then I am pretty sure that there are less than 800; this just quantifies it.
  November 28, 2014 at 7:34 pm
  I’m not asking you to translate it across model problems; I’m asking you to explain it in more detail than you have yet done above, if you are willing.
  November 19, 2014 at 6:48 pm
  “Indeed it’s a reasonable “best guess” that T=2t. This makes intuitive sense because the observed value of t then lies right in the middle of its range of possibilities.”
  A quick Monte Carlo program suggests to me your best guess was wrong:
  Mode result: 1.0 appeared 33300.0 times.
  2.0 appeared 26797 .0 times.
  (Although feel free to point out mistakes in the program, or if I misunderstood your point!)
  from numpy.random import randint, seed
  from numpy import zeros, size
  from scipy.stats import mode
  seed()
  #Number repeats
  N = 100000
  results_array = zeros(N)
  Number_2 = 0
  for i in range (0,N):
  #pick the size of the range
  range_size = randint(2,1e9)
  #pick a number from the range
  random_number = randint(1, range_size -1)
  result = float(range_size)/random_number
  results_array[i] = int(round(result,0))
  if (results_array[i] == 2) : Number_2 +=1
  print “Number: “, random_number
  print “Coles Estimate of range: “, 2*random_number
  print “Actual range: “, range_size
  print ” “, result, int(round(result,0))
  print ” ”
  print ” ”
  print “Mode result: “, mode(results_array)[0][0],” appeared “, mode(results_array)[1][0],” times.”
  print “2.0 appeared “, Number_2,”.0 times.”
  November 19, 2014 at 6:50 pm
  It depends by what you mean by “best” – it’s not necessarily the most probable
  November 19, 2014 at 7:11 pm
  So what do you mean by “best”?
  November 21, 2014 at 11:50 pm
  It surprises me that “doomsday” still gets so much discussion. Ken Olum pointed out the logical flaw quite a few years ago:
  November 24, 2014 at 8:55 am
  I’m not disagreeing, John, but the problem with pointing to such a reference is that someone has surely criticized Ken Olum since then.
  In general, when there is a huge literature on a topic, it is difficult to address it in blog comments. 😐
  Someone recently pointed to an article where Mike Turner was the second of two co-authors, which claimed that CDM naturally produces MOND phenomenology. Of course, this article has been criticized in turn. Even if one does not agree with the MOND folks, I think it is fair to say that the Turner paper at best gets an order-of-magnitude estimate right, but is far from explaining all MOND phenomenology.
  November 25, 2014 at 12:12 pm
  […] week I posted an item that included a discussion of the Doomsday Argument. A subsequent comment on that post mentioned a […]