# Die experiment

• Oct 16th 2011, 10:04 PM
agent2421
Die experiment

Consider an experiment where we roll a single die and observe the # of dots. Thus,m our sample space is S = 1,2,3,4,5,6 . Suppose that the die is not fair and that P (i) = k/i for i = 1,2,3,4,5,6. What is the value of k?

I don't understand what it means by the die not being fair. What would be the steps to solve such a problem? Would it be something along hte lines of:

P (i) = k/6

since it says we observe the # of dots do we just add the dots together?

P (i) = 21/6?

I'm just making a guess here but I'm not sure how to even start with this.

Thanks.
• Oct 16th 2011, 11:38 PM
CaptainBlack
Re: Die experiment
Quote:

Originally Posted by agent2421

Consider an experiment where we roll a single die and observe the # of dots. Thus,m our sample space is S = 1,2,3,4,5,6 . Suppose that the die is not fair and that P (i) = k/i for i = 1,2,3,4,5,6. What is the value of k?

I don't understand what it means by the die not being fair. What would be the steps to solve such a problem? Would it be something along hte lines of:

P (i) = k/6

since it says we observe the # of dots do we just add the dots together?

P (i) = 21/6?

I'm just making a guess here but I'm not sure how to even start with this.

Thanks.

For a fair die the probability of the outcomes are equaly likely. A die is not fair if the probability of each outcome is not the same. Since one of 1, .., 6 must be the result of rolling the die for a fair die p(i)=1/6, i=1..6.

This die has P(1)=k/1, P(2)=k/2, p(3)=k/3), p(4)=k/4, p(5)=k/5, p(6)=k/6

but when you roll the die the result must be one 1, 2, 3, 4, 5, 6 so:

p(1)+p(2)+..+p(6)=1

and you may use this to determine what value k must have.

CB
• Oct 17th 2011, 11:48 AM
agent2421
Re: Die experiment
I'm not sure if this is correct or not but I'm assuming you mean for a fair die each outcome will have a 1/6 probability. Since this is not fair it will be distributed through all the numbers meaning:

P (1)= k/1 + P (2) = k/2 + P (3) = k/3 + P(4) = k/4 + P(5) = k/5 + P (6) = k/6

Therefore

P (1) = 1/1 + P (2) = 2/2 + P (3) = 3/3 + P (4) = 4/4 + P (5) = 5/5 + P (6) = 6/6

would that be right or no? The problem is that k is different in each one so I think i'm still misunderstanding what you are trying to say.

Thanks.
• Oct 17th 2011, 11:57 AM
ebaines
Re: Die experiment
You've been told that P(i) = k/i. Therefore:

P(1) = k
P(2) = k/2
P(3) = k/3
P(4) = k/4
P(5) = k/5
P(6) = k/6

Since these 6 possibilities are the only ones that exist, you know that
P(1) + P(2)+ P(3) + P(4) + P(5) + P(6) = 1.

Now you can solve for k.
• Oct 17th 2011, 11:58 AM
agent2421
Re: Die experiment
Actually I didn't read that it all has to equal to 1. So just by plugging in numbers i came to the conclusion that k has to be somewhere around 0.4.

So i did:

0.4/1 + 0.4/2 + 0.4/3 + 0.4/4 + 0.4/5 + 0.4/6 = 0.98 so it's fairly close to 1... Is there an actual way of doing the math though to make it 1? 0.4082 reaches a bit over 1 but is there a way or formula to get a concrete answer... I'm pretty sure this is on the right track but once again it's just by plugging in numbers to try and make it equal to 1.
• Oct 17th 2011, 12:06 PM
Plato
Re: Die experiment
Quote:

Originally Posted by agent2421
Actually I didn't read that it all has to equal to 1. So just by plugging in numbers i came to the conclusion that k has to be somewhere around 0.4.

Really $\displaystyle \sum\limits_{n = 1}^6 {\frac{1}{n}} = \frac{{49}}{{20}}$ thus $\displaystyle k=\frac{20}{49}$.
• Oct 17th 2011, 01:26 PM
agent2421
Re: Die experiment
I'm sorry but where do you get the 49 from? I feel really stupid for not understanding this right away lol.
• Oct 17th 2011, 01:30 PM
Plato
Re: Die experiment
Quote:

Originally Posted by agent2421
I'm sorry but where do you get the 49 from? I feel really stupid for not understanding this right away lol.

$\displaystyle \left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1} {4}+\frac{1}{5}+\frac{1}{6}\right)~?$
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