red75 comments on Probability puzzle: Coins in envelopes

red75 3 Dec 2011 7:06 UTC
3 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