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?
I currently think that the paradox arises from the probability distribution of the two envelopes. The two envelopes cannot have the same probability distribution, so given any x, the probability of choosing either envelope is
not always equal.
Yes, we have equal chance of picking either envelope, but generally, lower values of x are more likely to come from the smaller amount of money, while higher values are more likely to come from the larger amount of money.
If we did in fact know the probability distribution, and the value of x, it would often be best to switch envelopes. But since we don't know x, the average gain should be zero.
Here is a bit of mathematical justification that the two distributions cannot be equal (ignoring the trivial case where both envelopes have zero):
Let f(x) equal the probability distribution for both envelopes. The integral from zero to infinity should equal 1. Now, I realize that the function may contain infinite values (e.g. perhaps the probability of being exactly one dollar is .5), and thus not be integrable. But if f(x) is infinite, then so is f(2x), f(4x), etc., leading to a total probability greater than 1contradiction.
We know that when the envelope with more money has x, the one with less has x/2. Therefore, f(x) = 2f(2x).
Let n = the integral of f(x) dx from 1 to 2. n must be positive, or f(x) would equal zero for all positive real numbers. Because f(x) = 2f(2x), we can use a little calculus to show that the integral of f(x) dx from 2 to 4 is equal to n, and likewise from 4 to 8, 8 to 16, etc. Therefore, the integral from 1 to infinity of f(x) is infinitecontradiction.
Therefore, the probability distributions of the two envelopes cannot be the same.
Edit: I think there may be a problem with the statement "The average gain should be zero," at the end of the third paragraph, since this does not seem to be true for the case given "The grass is always greener II." I may look more into this later. Or not.
Edited on October 15, 2006, 2:01 am

Posted by Tristan
on 20061013 14:35:30 