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 Mean-ing by Tristan)

>Well, now that we know that the average presented in the puzzle is wrong, can we find why it is wrong... besides the fact that it gets the wrong answer?

 

I believe, though I can't say for sure, that the problem is that which I wrote in the first reply - you are using x as if it were a constant, grounded variable, whereas it actually depends on which envelope you pick.

The puzzle states that "...one might average x/2 and 2x ... this comes out to 5/4x."

I don't think that this is exactly true. If the other envelope contains x/2, then you picked the larger x. If the other envelope contains 2x then you picked the smaller x. You can't treat the x's as if they were the same. This is why I said that you have to pick a single x, say the smaller value. Then in envelope B you have either x or 2x. Using these values, the probability works out correctly.

Posted by Sam on 2004-05-10 20:40:12