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?

To find out how much, on average, the other envelope should contain, one might average x and 2x, NOT x/2 and 2x. The average is therefore 3/2x. Obviously, because one envelope contains 'x' amount of money and the other '2x' amount of money, the averge will be greater than x. But in this problem, averaging both envelopes is not a solution. There is a 50% chance of picking up the envelope with the most amount of money; whatever you do after that is, probabilistically speaking, useless. You are always 50% as likely to pick up the one with x money as you are of picking up the one with 2x money. Switching envelopes still gives you a 50% probability of picking either envelopes.

Posted by Henry on 2004-07-23 11:23:49