Probability puzzle: Coins in envelopes

Cool puzzle!

DanielLC has solved this already, but for 1: simplest solution is m=2 and random function is "with 50% probability fill either 0 or 2 envelopes".

I notice solving problem 2 also solves problem 1.

The solution to two:

sum(f(k)) = sum of f(k) for each value of k from 0 to m inclusive

1/2 = sum(k^2/m)/sum(k)-1/m

= (m(m+1)(2m+1)/6)/(m^2(m+1)/2)-1/m

= (2m+1)/3m-1/m

= (2m+1)/3m-3/3m

= (2m-2)/3m

= 1/2

4m-4 = 3m

m = 4

Four envelopes.

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

1) m=2, p(0)=11/18, p(1)=2/9, p(2)=1/18

2) m=4

I must be missing something: both 1 and 2 appear to be impossible. He picked a random envelope, it had a coin, he removed it, she put that envelope, now empty, back with the others. How can the expected value of picking an envelope not have decreased? There are the same number of envelopes, and one fewer coin.

Interesting, there was recently a somewhat related question posed here.

Using my experience from that question I can give a pretty large group of answers for problem 1. Pubbfr nal z, naq nffvta n ahzore bs pbvaf tvira ol n ovabzvny qvfgevohgvba jvgu a=z naq fbzr c xabja gb Obo, fb gung vg'f nf vs lbh syvccrq na vaqrcraqrag pbva sbe rnpu rairybcr.

2 is a bit tricky, since perfect mathematicians don't just eliminate the obviously wrong, they also update against the unlikely but right. Bayes' rule says that if you get a coin, you multiply your probabilities by P(got coin | X envelopes filled)/P(got coin). P(got coin) is 1/2, your chance of getting a coin if there are X coins is X/m, and your prior that there's X coins is 1/(m+1) (since 0 is a valid number of coins too). So after getting a coin, the hypothesis that there are X envelopes with coins in them gets probability 2X/m 1/(m+1).

Gut check stop. This means that for m=2, Bob would say, P(0) = 0, P(1) = 1/3, and P(2) = 2/3 after getting one coin. Looks right.

Each hypothesis leads to an expected value of (X-1)/(m-1). So we take the sum of 2X(X-1) / (m-1)m(m+1) (thanks wolfram) to get 2/3. No matter the m, the expected value for the second draw is 2/3! It's Laplace's rule of succession! Cool, huh? I'm going and giving damang an upvote just for how helpful his post was for this one. Shame about it making the problem unanswerable :P

EDIT: part two was answering the wrong question, see comment.