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?

Open one envelope. If the amount inside is enough to make you feel like a winner, keep it; you don't need to risk it for more. If not, switch the envelope, since the amount wasn't enough anyway.

This riddle is inherantly unsolvable, mainly because most people overlook the fact that it doesn't matter how much is in the first envelope, or whether you switch or not. Switching or not switching still gives you exactly a 50% chance of getting the envelope with more money, since there's no way to possibly know without opening both envelopes.

Personally, I'd always switch, beacuse I could never live "not knowing" what was in the other envelope.

Posted by ccvannorman on 2004-11-11 03:07:22