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 The faulty logic
by Carlos Da Peanut)
If one envelope contains twice the amount of the other, and the lesser contains 6$, then x/2 or x(2) would solve the amount in their respective situations:
If you had envelope with the greater amount, which would be $12, then you could use x/2 to find the lesser amount, which would be $6.
If you had the lesser amount, then you could use x(2) to find the greater amount.
Posted by Balance
on 2004-08-12 00:29:55