When switching really does pay

Question

A host secretly picks an amount X from the list \{\10, \20, \40, \80, \160, \320, \640, \1280\}, all eight equally likely. She puts X in one envelope and 2X in the other, then hands you one envelope at random. You open yours and see \40. What is the expected value of the other envelope, and should you switch?

Accepted Answer

Unlike the paradox version, here there is a real, known distribution over amounts, so the calculation is legitimate. Seeing \40 narrows things to two possible pairs: - The pair is (\40, \80), meaning X = 40 and your envelope is the smaller one. Then the other holds \80. - The pair is (\20, \40), meaning X = 20 and your envelope is the larger one. Then the other holds \20. Are these equally likely given you see \40? The values X = 40 and X = 20 each have prior probability 	frac18, and in each case you were equally likely to be handed either envelope. So both scenarios are equally likely, 	fr…

When switching really does pay

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