The Exchange Paradox

Two sealed envelopes are handed out, one to you and one to a competitor. One contains $m$ dollars and the other $2m$ dollars, with $m$ unknown. You reason as follows. Let $A$ be the amount in my envelope. The other holds either $2A$ or $\tfrac{A}{2}$ , each with probability $\tfrac12$ , so its expected value is $\tfrac12(2A) + \tfrac12\big(\tfrac{A}{2}\big) = \tfrac54 A$ , which beats $A$ , so I should switch. Your competitor reasons identically about their envelope. Both cannot be right. What is wrong, and should you switch?

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#The symmetric truth

Fix the pair, a smaller amount $m$ and a larger amount $2m$ . You are equally likely to be holding either one, so the expected contents of your envelope are

\E[\text{yours}] = \tfrac12 m + \tfrac12 (2m) = \tfrac32 m, \tag{1}

and the other envelope has the very same expectation $\tfrac32 m$ . Switching trades a $\tfrac32 m$ for a $\tfrac32 m$ , an expected gain of $0$ . By symmetry neither player can hold an edge.

Either you hold the smaller envelope and switching wins you m, or you hold the larger and switching costs you m. Both worlds are equally likely, so the expected gain from switching is zero and the two envelopes are worth the same.

#Where the quarter-gain argument breaks

The tempting step writes the other envelope as $2A$ or $\tfrac{A}{2}$ and treats $A$ as one fixed number across both branches. It is not. The two branches are two different worlds.

If you hold the smaller envelope then $A = m$ and the other holds $2m$ . If you hold the larger then $A = 2m$ and the other holds $m$ . So the single symbol $A$ secretly means $m$ in the first branch and $2m$ in the second. Averaging $2A$ against $\tfrac{A}{2}$ as though $A$ were constant adds amounts pinned to different values of $m$ . Carried out honestly the other envelope holds

\E[\text{other}] = \tfrac12 (2m) + \tfrac12 (m) = \tfrac32 m = \E[\text{yours}]. \tag{2}

The phantom factor $\tfrac54$ is pure equivocation on $A$ .

#Result

There is no advantage to switching. Both envelopes have expected contents $\tfrac32 m$ , and the $\tfrac54 A$ calculation only looks compelling because one symbol $A$ is quietly standing for two different amounts at once.