red75 comments on Probability puzzle: Coins in envelopes - Less Wrong

8 Post author: HonoreDB 02 December 2011 05:58AM

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Comment author: red75 03 December 2011 07:06:57AM * 2 points [-]

Problem 2 by Bayes rule.

N is a random variable (RV) of number of filled envelopes.

C is a RV of selected envelope contains coin. P(C) means P(C=true) when appropriate.

Prior distribution

P(N=n) = 1/(m+1)

by the problem setup

P(C|N=n) = n/m

by the rule of total probability

P(C)=sum_n P(C|N=n)P(N=n) = sum_n (n/m/(m+1))=m(m+1)/2/m/(m+1)=1/2

by Bayes rule

P(N=n|C) = P(C|N=n)P(N=n)/P(C) = 2n/m/(m+1)

Let C' is a RV of picking filled envelope second time.

by the problem statement

P(C'|N=n,C) = (n-1)/m

by the rule of total probability

P(C'|C)=sum_n P(C'|N=n,C)P(N=n|C) = ... substitutions and simplifications ... = 2(m-1)/(3m)

solving P(C'|C)=P(C) obtains

m=4