The St. Petersburg two-envelope paradox

Analysis 62 (2):155–157 (2002)
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Abstract

I reason: (1) For any x, if I knew that A contained x, then the odds are even that B contains either 2x or x/2, so the expected amount in B would be 5x/4. So (2) for all x, if I knew that A contained x, I would have an expected gain in switching to B. So (3) I should switch to B. But this seems clearly wrong, as my information about A and B is symmetrical.

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David Chalmers
New York University

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