# Posterior probability or Prior probability

I have an arguement with my friends on a probability question.

Question: There are lots of stone balls in a big barrel A, where 60% are black and 40% are white, black and white ones are identical, except the color.

First, John, blindfolded, takes 110 balls into a bowl B; afterwards, Jane, blindfolded also, from bowl B takes 10 balls into cup C -- and find all 10 balls in C are white.

Now, what's the expectation of black balls in bowl B?

It seems there are 3 answers

Answer 1 My friend thinks the probability of black stones in bowl B is still 60%, or 60% * 100 = 60 black balls expected in bowl B.

Answer 2 However, I think 60% is just prior probability; with 10 balls all white from bowl B, we shall calculate the posterior probability. Denote $B_k$ as the black ball numbers in bowl B, and $C_{w}=10$ as the event that 10 balls in cup C are all white.

$$E(B_B | C_w=10) = \sum_{x=10}^{110}\left[x P(B_k = x | C_w=10)\right] = \sum_{x=10}^{110} \left[x \frac{P(B_k = x \text{ and } C_w=10)}{P(C_w=10)}\right] = \sum_{x=10}^{110} \left[x \frac{P(B_k = x) P( C_w=10 | B_k=x)}{P(C_w=10)}\right]$$

, where $$P(C_w=10) = \sum_{x=10}^{110} \left[ P(B_k = x) P(C_w = 10 | B_k=x) ) \right]$$

, and according to binomial distribution $$P(B_k=x) = {110 \choose x} 0.6^x 0.4^{110-x}$$ , and $$P(C_w = 10 | B_k=x) = \frac{1}{(110-x)(110-x-1)\cdots (110-x-9)}$$

Answer 3 This is from Stefanos below: You can ignore the step that John takes 110 balls into a bowl B, this does not affect the answer. The expected percentages in bowl $B$ are again $60\%$ and $40\%$ percent, i.e. 66 black balls and 44 white balls. Now, after Jane has drawn 10 white balls obviously the posterior probability changes, since the expected number of balls is now 66 and 34. So, you are correct.

Which answer is right?

I sort of don't agree with Stefanos that, the black ball distribution from bowl B could vary a lot from barrel A, as the sampling distribution could be different from the universe distribution.

in other words, if Janes draws a lot balls and all are white, e.g. 50 balls are all white, I fancy it's reasonable to suspect that bowl B does not have a 60%-40% black-white distribution.

## 2 Answers

You can ignore the step that John takes 110 balls into a bowl B, this does not affect the answer. The expected percentages in bowl $B$ are again $60\%$ and $40\%$ percent, i.e. 66 black balls and 44 white balls. Now, after Jane has drawn 10 white balls obviously the posterior probability changes, since the expected number of balls is now 66 and 34. So, you are correct.

• pls see my update of the question. hopefully I explained myself better this round. – athos Oct 25 '14 at 14:27
• Wow, nice work!! Yeah, I agree with you, but the expected(!) number is 66 and 44 as the expected number of a binomial distribution is np. Perhaps if you finich the calculations you will find the same, I am not sure. But, that the expected number of balls in bowl should be 66 and 44 that is intuitively as well as thoretical sound! – Jimmy R. Oct 25 '14 at 14:44
• basically, our difference is, what Jane met (10 or 50 white balls in a row), should it affect the estimation of the bowl B black-white distribution? your answer is no, but I doubt -- if John is a dealer, Jane has reason to believe he's cheating! – athos Oct 25 '14 at 15:00
• It is not in his hand to cheat, remember he is blindfolded. I pretty sure that I am correct. Fair disagreement though, everyone has his points ;) – Jimmy R. Oct 25 '14 at 15:24

You are (almost) right. However, your phrase "60% is just prior probability" is wrong/makes no sense. You need to calculate the prior probability distribution for the bowl B, it will have expectation 0.6 * 100 = 60.

• where is the 10 from? – athos Oct 25 '14 at 14:26
• P.S. Stefanos is incorrect that you can ignore the step that John takes 110 balls into bowl B. If you want a precise (mathematical) answer, then the prior distribution of the number of black balls in the bowl B is hypergeometric with parameters N, K and 100, where N = number of balls in the barrel A and K = number of black balls in the barrel A. – Valentas Oct 25 '14 at 14:48