You are told there are two envelopes. One contains twice as much money as the other one. You pick one but are allowed to change your mind after picking it. (You are equally likely to pick the one with less money as the one with more money.)

To figure out how much on average the other envelope should contain, one might average x/2 and 2x because one is equally likely to pick one as picking the other. Since this comes out to 5x/4, one might always change his or her mind. But wouldn't this end up with one never making up his or her mind?

(In reply to uniform distribution....? by ronen)

>the paradox would have been true, had there been a uniform distribution of x from 0 to infinity.

The solution to this puzzle is often stated in terms of probability distributions, but I'd still be wary of your idea that the puzzle would work if x can stretch between 0 and infinity.

Suppose instead of real money I just had a piece of paper in each envelope with an arbitrarily large random number on it. Would the paradox work then? If so you would always switch, but this is clearly absurd (not least because you could apparently "pump up" the value of the envelopes by switching back and forth multiple times.

Or, if you still think that numbers on a piece of paper fall under such a probaility distribution, what if I had money in just one envelope, and the other had a piece of paper that either said "x/2" or "2x", with a 50% probability for each. Now we really can't bring in probability distributions, but we still can't suppose that the paradox works.

Posted by Sam on 2004-05-10 20:34:00