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.
Well, not everything—it isn’t Laplace’s rule of succession, but if you correct the mistake, you’ve pretty much solved part 2. Instead of a fixed value you get an equation you can solve for m.
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.
(X-1)/m, because the emptied envelope is shuffled back into the set.
Oh, whoops, I didn’t read the question correctly. Drat, then it’s not Laplace’s rule of succession.
In fact, that messes up pretty much everything—I’ve finally found a use for the retract button.
Well, not everything—it isn’t Laplace’s rule of succession, but if you correct the mistake, you’ve pretty much solved part 2. Instead of a fixed value you get an equation you can solve for m.
That’s true. It also invalidates my answer for part 1, which is a bit trickier to correct, because you no longer have the nice symmetry.