You have two indistinguishable envelopes, each of which contains a positive sum of money, but one envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are given the chance to take the other envelope instead
Now suppose you reason as follows:
I denote by A the amount in my selected envelope.
The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
The other envelope may contain either 2A or A/2.
If A is the smaller amount, then the other envelope contains 2A.
If A is the larger amount, then the other envelope contains A/2.
Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
So the expected value of the money in the other envelope is (½)(2A)+(½)(A/2)=(5A/4)
This is greater than A, so I gain on average by swapping.
After the switch, I can denote that content by B and reason in exactly the same manner as above.
I will conclude that the most rational thing to do is to swap back again.
To be rational, I will thus end up swapping envelopes indefinitely.
As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.
Fun problem, try it out! Good luck!