The Two-Envelope Paradox

Question

Two envelopes: one holds twice as much money as the other. You pick one at random and find 100. You may switch. The tempting argument: "the other holds 50 or 200 with equal probability, EV = 125 > 100, so switch, but that logic applies before opening too, for any amount, so you'd switch forever." Where exactly does the argument break, and should you switch?

Accepted Answer

The flaw: "50 or 200 with equal probability" assumes a prior under which every amount is equally likely at every scale: a "uniform distribution over all positive numbers," which doesn't exist (it can't integrate to 1). The conditional probabilities after observing 100 depend on the prior over the envelope pair (x, 2x), and for any proper prior they are generally not 50/50. Concretely: seeing 100, the relevant comparison is between priors on pairs (50,100) vs (100,200): P(textother = 200 mid textsaw 100) = fracpi(100,200)pi(50,100) + pi(100,200), where pi is the prior on pairs. Under any prior that decays (as real-world money must, bounded bankrolls), large pairs are less likely, this probability drops below frac12 at high observed amounts, and switching at large values is bad. For bounded priors, the optimal rule is a threshold: switch when you see a small amount, keep when you see a large one. Unconditionally (never opening), symmetry makes switching exactly neutral. Interview-grade summary: the paradox is a disguised improper-prior error. The reflexive lesson for trading: "the move is up or down 50/50" style reasoning smuggles in priors, conditional EV claims are only as good as

The Two-Envelope Paradox

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