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The Two Envelopes Argument

Asked atCitadelHudson River Trading

Two envelopes contain money; one holds twice as much as the other. You pick one and are offered the chance to switch.

The usual argument: if your envelope holds $x$ , the other holds $2x$ or $x/2$ with equal probability, so switching has expected value $\tfrac{1}{2}(2x) + \tfrac{1}{2}(x/2) = 1.25x > x$ . So you should switch — and by symmetry, switch back forever.

Where exactly does the argument break?

Solution

Try it yourself first. Open a hint if you get stuck.

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The mistake: $x$ is treated as a fixed number, but it is random. Call the two amounts $A$ and $2A$ . What $x$ equals depends on which envelope you picked:

you hold the smaller envelope: $x = A$ , and switching gains $A$ ;
you hold the larger envelope: $x = 2A$ , and switching loses $A$ .

Each case has probability $\tfrac12$ , so the expected gain from switching is

\tfrac12(+A) + \tfrac12(-A) = 0.

The argument writes the gain as $x$ and the loss as $x/2$ , as if both used the same $x$ . They don't: the gain uses $x = A$ and the loss uses $x = 2A$ , so both are worth $A$ .

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