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?
First of all, isn't the average of x/2 and 2x 5x/4? I suppose that could be written as the fraction 5/4 times x, but it looks wrong in the question.
Secondly, I think attempting to solve this problem by averaging is the wrong approach. You definitely WILL NOT receive 5x/4 amount of money, as the only options that can be in the second envelope are x/2 and 2x. One possibility is as likely as the other (50% chance of getting 2x as opposed to x/2) so I guess you just have to choose whether or not you're satisfied with the amount of money in the first envelope. If you are, keep it; if not, switch.
Penny, also, this question did remind me of the Monty Hall question, although the two are actually very different. It took me a while to understand why you should switch your choice when they show you what's behind a door you didn't pick, but I enjoyed wrapping my brain around it!
Lose the math
