Two envelopes I have 2 envelopes; one has twice as much money in it as the other. I choose one at random (both equally likely), and give it to you. Do you want to switch envelopes?
schmoe
The flaw is you're using the same letter to refer to two different amounts. In one case the envelope has $2N, in the other case the envelope has M/2, but M=2N.
Peter Meilstrup
You are right in that a nicer way to answer the problem is to say that the envelopes have amounts n and 2n, and by switching, I'd either gain n or lose n, with equal likelihood. This description is very clear.
schmoe
I think the error is in that your two possibilities suppose a different amount of total money in the game. You are considering n to be a constant, and comparing 3n/2 with 3n. Obviously in this case switching would seem to be advantageous. The total amount of money in the game is fixed - say x. If the other envelope contains 2n dollars, then n was only x/3. If the other envelope contains n/2 dollars, then n was 2x/3.
Paul Viney
If you are allowed to open the envelope, and you see $50 in the envelope, AND (your unstated assumption) you know that envelopes containing $50 and $100 are just as likely as envelopes containing $25 and $50, then you should switch.
Peter Meilstrup
I have to say I find this one hard to get to grips with. I think several of the posters have good explanations. I'm just posting my version because I like to talk.
David Clayworth
"If you were to tell the guesser that N was evenly distributed between 1 and 1 billion (say) then it would be correct to switch for any number less than half a billion."
schmoe
The problem with the expectation calculation is to do with dividing by zero. If you expand the full Bayes theorem calculation (probability of the other envelope having $200 given that I have an envelope with $100) you will find that you are dividing by the a priori probability that the first envelope has $100. For an unbounded even distribution, this probability is zero. If you can divide by zero you can make anything equal anything (e.g. 2x=x => 2=1). So the expectation calculation is meaningless.
David Clayworth
It ought to be possible to come to the "correct" answer without opening the envelopes, and without assuming anything about the prior distribution of money in envelopes, and by using values of "money in this envelope" and "money in the other envelope" since it's just as valid a distinction as "smaller envelope" versus "bigger envelope."
Peter Meilstrup
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