Technically there are two mistakes though. First is letting x be two things at once. However, even if you want to play the game of letting x be two things at once, you reach an infinite series and cannot perform any calculations. I.e. if the second is 2x or 0.5x , then by definition, we were wrong that the first was x. It must be 4x,x,x or 0.25x. Which then means we were wrong about the second, which is either 8x,2x,2x,0.5x,2x,0.5x,0.5x or 0.125x. Should be clear at this point that something has gone horribly wrong. The second mistake was to presume it was ok to calculate the expected value before updating the possible values in the envelopes infinitely many times.

You have a 50% chance of getting 2x and a 50% chance of getting x when you first pick swapping does nothing to change that, the other envelope also has 50% chance of each, the reasoning is wrong because x is defined as 2 numbers at once (try inserting 1 into the equation, 2 times 1 is 2 while 0.5 times 1 is 0.5 which is four times less than 2 instead of the stated 2 times less)

I also don’t like the explanation given above, “x stands for two different things in the equation … the smaller amount [or] the larger amount.” The point of a random variable is that it “stands for” the set of all possibilities. So while this is attempting to explanation the error, it is technically incorrect.

I prefer an example of what that explanation was trying to say, which is that treating the envelope as having a specific value, even the unknown X, requires knowing how X was picked.

1) Say I prepare two envelopes, with $5 and $10, and give them to you for the above experiment. I tell you these values. But since the envelopes are sealed, all it means to you is that one has twice as much as the other. The point is that you can’t consider the any one value as having the potential to both win, and lose. That is, your envelope has $5, switching can only increase the value.

2) Say I prepare two pairs, one with ($5,$10) and one ($10,$20). You pick one at random for the above experiment. Again, all you know for certain is that is that one of the two has twice as much as the other. If your envelope has $10, then the “you should switch” argument above is actually correct! You have a 50% chance to lose $5, and a 50% chance to gain $10, so the expected gain is $2.50. But if you have $20, you lose $10.

3) Say I prepare ten pairs, nine with ($5,$10) and one ($10,$20). Now, if you have $10, there is a 90% chance you will lose $5, and only a 10% chance you will gain $10. So the expectation is a loss of $3.50.

The point is that the 50%:50% split applies only to whether your envelope is the higher or lower envelope, not to whether the specific value in the envelope is the higher or lower value. That is determined by both the 50%:50% split, and the relative probabilities that the envelopes were prepared with ($X/2,$X) or ($X,$2X).