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?

Ha ha - sneaky... This problem proposes to use x/2 and 2x to represent the 2 possible amounts of money in the other pouch... but basically x is not clearly defined to represent the same thing in both situations.

Suppose you have an envelope with $10, and one with $20:

The problem suggests that if you pick up the $10 envelope first, then the amount in the other envelope can be represented as 2x

- "x" then being $10 (the smaller unknown)

Alternately, if you happen to pick up the $20 envelope first, then the amount in the other envelope can supposedly be represented as x/2

- suddenly "x" here is meant to represent $20 (the larger unknown)

And then the problem goes on to use these 2 non-corresponding definitions of x in a single equation to find an average. 

INSTEAD:

To solve the problem of finding the average amount in the second envelope, you need to fix x as representing EITHER the smaller unknown OR the larger unknown.  

So, for example, let x = the smaller unknown amount;

If you choose the smaller amount first, the amount in the other envelope can be represented as 2x;

Alternately, if you choose the larger amount first, the amount in the other envelope can be represented simply as x (NOT x/2, as suggested in the problem).

Finally, by averaging the equal chances of having x and 2x, the average amount in the other pouch is 1.5x (Again, remembering that x represents the smaller unknown amount, NOT the necessarily amount you're holding in your hand when you choose the first envelope.)

And the logic follows when you plug numbers in that if you have a $10 envelope and a $20 envelope, the average amount in the envelope you don't choose will be $15 (same as the average amount for the envelope you do choose... equal odds, of course.)

Posted by ladyfly on 2005-04-28 08:01:53