Two envelopes hold 2^k and 2^(k+1) dollars, where k is drawn with P(k) = (1/3)(2/3)^k for k = 0, 1, 2, and so on. You open one envelope at random and see 2^n with n >= 1. What is the expected amount in the other envelope, as a multiple of what you see?

Two envelopes hold 2^k and 2^(k+1) dollars, where k is drawn with P(k) = (1/3)(2/3)^k for k = 0, 1, 2, and so on. You open one envelope at random and see 2^n with n >= 1. What is the expected amount in the other envelope, as a multiple of what you see?

Approach: Condition on the value you saw. Two values of k are consistent with it, so weight them by the prior and by the 1/2 chance you opened either side of the pair, then average the double and the half.

11/10. Seeing 2^n with n >= 1 is consistent with k = n-1 or k = n. The prior ratio is P(n)/P(n-1) = 2/3 and each pair is opened either way with probability 1/2, so the posterior probability that you hold the smaller amount is (2/3)/(1 + 2/3) = 2/5. Conditional on that, switching pays 2x with probability 2/5 and x/2 with probability 3/5, giving (2/5)(2) + (3/5)(1/2) = 11/10. Switching at every observed value is consistent here because the unconditional mean is an infinite expectation: the average amount grows like sum (4/3)^k, so there is no finite quantity for the two envelopes to trade against each other.

Follow-up: For which priors on k does the switching argument stop favouring a switch at every observed value?

Key concepts: posterior probability, prior distribution, conditional expectation, infinite expectation.