This is one of my favorite problems. Here is one variation can help clarify thinking about it: "There's two envelopes. One contains twice as much money as the

This is one of my favorite problems. Here is one variation can help clarify thinking about it: "There's two envelopes. One contains twice as much money as the other. No envelope contains more than $N." This changes the problem dramatically. If your envelope contains more than $N/2, of course you should not switch: the other envelope necessarily contains less. If it contains $N/2 or less, perhaps you should switch: the other envelope may or may not contain more. Say your envelope contains $x, and you don't know what $N is. There's two possibilities: 1. $x <= $N/2. If you switch, your expectation value is $3x/2, by the original argument. 2. $x > $N/2. If you switch, you will get $x/2, since no envelope contains more than $N. If we assume the distribution is uniform on the range [0, N], then these possibilities are equally likely. Therefore the total expectation value is the average of the EVs of the two possibilities (3x/2 and x/2), which is $x. So we recovered the naive expectation of "it doesn't matter" from this variation. Now we can take the limit as $N goes to infinity, and while the EV of $x approaches infinity, the fact that switching does not matter does not change. As others have said, the underlying problem is the assumption that a uniform probability on an infinite set makes sense, which it does not. However, we can instead take the limit for finite sets, in which case we recover the intuitive result that switching does not matter.