When the paradox quietly disappears

Question

A fair coin is tossed until the first head. In the St. Petersburg game the payout on toss n is 2^n, and the expected value is infinite. Now change only the payout: if the first head comes on toss n, you win 
 instead (so head on toss 1 pays \1, toss 2 pays \2, toss 3 pays \3). What is the expected payout now, and why has the paradox vanished?

Accepted Answer

The first head lands on toss n with probability \left(	frac12ight)^n, and now the payout is n, so the expected value is E[	ext{payout}] = \sum{n=1}^{\infty} \frac{n}{2^n}. This is a standard convergent series. A quick way to evaluate it: let S = \sum{n\ge1} n x^n = \frac{x}{(1-x)^2} for x < 1. At x = 	frac12, S = \frac{1/2}{(1 - 1/2)^2} = \frac{1/2}{1/4} = 2. So E[	ext{payout}] = \2. A perfectly sensible, finite price, and no paradox at all. Why the difference? In the original game each term contributed \frac{1}{2^n}\cdot 2^n = 1, and infinitely many 1's diverge. Here each term is \frac…

When the paradox quietly disappears

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