Resolving the two envelope paradox

arXiv2101.03484MaRDI QIDQ6357945

Author name not available (Why is that?)

Publication date: 10 January 2021

Abstract: Consider the following game: You are given two indistinguishable envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope, you are given the chance to switch envelopes. Should you switch? The intuitive answer is that it makes no difference, since you are equally likely to have picked the envelope with the higher or the lower amount. However, a naive expected value calculation implies you gain by switching, since you have $50\mathrm{\%}$ chance of doubling and $50\mathrm{\%}$ chance of halving your current winnings, and so if the first chosen envelope contains X, then switching gives an expected final value of $(X/2+2X)/2>X$. That seems like a paradox. We prove that the former is the correct answer, and show how the apparent "paradox" can be resolved.

Has companion code repository: https://github.com/nemozen/math/blob/master/envelope_paradox.ipynb

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