The two envelopes paradox

Question

Two sealed envelopes contain money. One contains twice as much as the other. You choose one at random and are offered the chance to switch to the other before opening either.

Here is the argument for switching. "Call the amount in my envelope A. The other envelope contains either 2A or A/2, each with probability 1/2. So its expected value is (2A + A/2)/2 = 1.25A, more than A. I should switch." But the same argument applies after switching, so you would switch back, and so on forever.

Where exactly is the flaw? Then: if you are allowed to open your envelope and see the amount before deciding, is there a situation in which switching genuinely helps?

Accepted Answer

The flaw is the assumption that, whatever amount A you hold, the other envelope is equally likely to hold 2A or A/2. No way of filling the envelopes can make that true for every A. Why the symmetric argument is wrong. Suppose the envelopes were filled by choosing a smaller amount S and putting S in one and 2S in the other. Before you look, the pair is (S, 2S) and your envelope holds S or 2S with probability 1/2 each. The expected gain from switching is then (1/2)(2S minus S) + (1/2)(S minus 2S) = 0. Nothing to gain, as symmetry demands. The seductive calculation uses the letter A to stand for…

The two envelopes paradox

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