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The two-envelope paradox

Asked at Jane Street

Two sealed envelopes are placed in front of you; one contains exactly twice as much money as the other. You pick one, open it, and find XX dollars. You reason:

The other envelope holds either 2X2X or X/2X/2, each equally likely. Its expected value is 12(2X)+12 ⁣(X2)=54X>X\tfrac12(2X) + \tfrac12\!\left(\tfrac{X}{2}\right) = \tfrac{5}{4}X > X, so I should switch.

But by symmetry the same argument applies before you look at all, and after switching you'd want to switch back forever.

Where exactly is the reasoning wrong?

Show a hint

Write down what the symbol XX actually denotes in each of the two branches "the other is 2X2X" and "the other is X/2X/2." Is it the same number in both?

Your answer

This one is open-ended. Work it through, then check your reasoning against the full solution.

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