The Two-Envelope Paradox: A Complete Analysis?

March 22, 1994 A wealthy eccentric places two envelopes in front of you. She tells you that both envelopes contain money, and that one contains twice as much as the other, but she does not tell you which is which. You are allowed to choose one envelope, and to keep all the money you find inside. This may seem innocuous, but it generates an apparent paradox. Say that you choose envelope 1, and it contains $100. In evaluating your decision, you reason that there is a 50% chance that envelope 2 contains $200, and a 50% chance that it contains $50. In retrospect, you reason, you should have taken envelope 2, as its expected value is $125. If your sponsor offered you the chance to change your decision now, it seems that you should do so. Now, this reasoning is independent of the actual amount in envelope 1, and in fact can be carried out in advance of opening the envelope; it follows that whatever envelope 1 contains, it would be better to choose envelope 2. But the situation with respect to the two envelopes is symmetrical, so the same reasoning tells you that whatever envelope 2 contains, you would do better to choose envelope 1. This seems contradictory. What has gone wrong? The paradox can be expressed numerically. Let A and B be the amounts in envelope 1 and 2 respectively; their expected values are E(A) and E(B). For all n, it seems that p(B > AjA = n) = 0:5, so that E(BjA = n) = 1:25n. It follows that E(B) = 1:25E(A), and therefore that E(B) > E(A) if either expected value is greater than zero. The same reasoning shows that E(A) > E(B), but the conjunction is impossible, and in any case E(A) = E(B) by