The St. Petersburg two-envelope paradox

Analysis 62 (274):155–157 (2002)
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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DOI 10.1111/1467-8284.00348
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Opening Two Envelopes.Paul Syverson - 2010 - Acta Analytica 25 (4):479-498.

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