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?
Hmmm. I'm not sure if I'm on the right track, but I'd start by saying that x/2 and 2x is needlessly confusing.
Call the smaller amount x. Whichever envelope you pick up, you either have x or 2x. The other envelope also has either x or 2x. The average amount you'd expect to have if you swapped envelopes would therefore be 1.5 x.
Ah ha, so you have above x, so swap, right? No, this is where I think the puzzle is confused. You aren't comparing the quantity in envelope 2 with x, you're comparing it to what you have in envelope 1. Envelope 1 also contains an average of 1.5 x (and together they make 3x, the total amount between the envelopes).
So envelopes 1 and 2 both contain the same average amount. No reason to swap.
So why the paradox? Possibly I'm missing somthing (quite likely), but I think that this is one of those math problems that sount confusing when you state them one way and are completely banal when you state them another. I think that stateing the amount you get in envelope 2 in terms of what you get in envelope 1, instead of grounded at a single x, sets you on a faulty path.
Posted by Sam
on 2004-05-06 14:34:39