The short version for those who don’t want to click the link:

I have 2 sealed envelopes, one of which has exactly twice as much money as the other.

I tell you if you pick one of the envelopes you can keep the cash.

After you pick an evelope I offer you the opportunity to take the other one instead.

You reason thusly:

Let X be the value of the envelope you are holding.

The other envelope must have either 2X or X/2 value with 50-50 probability.

The expected value of the other envelope is therefore 50%(2X) + 50%(X/2) = (5/4)X

The expected value of the other envelope is more than the one in your hand, therefore you should switch.

Doing so leaves you in the same position as before so according to the above analysis, you should continue to swap forever.

Im sure there have been many threads asking to explain how to resolve this problem, this is not one of those.

This thread is about what is wrong with what I am about to show.

Clearly the expected value (in reality) of the other envelope is the same as the one in your hand.

Let p be the probability that the other envelope contains twice as much money (2X) as the one on your hand. Therefore the probability of the other envelope containing X/2 is (1-p).

The expected value of the other envelope is p(2X) + (1-p)(X/2) which as stated above is equal to the value of the envelope in your hand.

So solve for p:

p(2X) + (1-p)(X/2) = X

2p + (1-p)/2 = 1

p = 1/3

So it turns out that the probability of the other envelope containing twice as much cash as the one in your hand is always 1/3 :dubious:

What is wrong with this reasoning?