The St. Petersburg game, what would you pay?

Question

A casino offers this game: it flips a fair coin until the first heads appears. If the first heads is on flip $n$ , you are paid $\$ 2^n $. So heads on flip$ 1 $pays$ $2 $, on flip$ 2 $pays$ $4 $, on flip$ 3 $pays$ $8$, and so on.

What is the expected payout, and what would you actually pay to play?

Accepted Answer

Expected value is infinite. The first heads lands on flip n with probability (1/2)^n, paying 2^n: E[	ext{payout}] = \sum{n=1}^{\infty} \left(\frac{1}{2}ight)^n 2^n = \sum{n=1}^{\infty} 1 = \infty. Every term contributes exactly \1 of expected value, and there are infinitely many. By the expected-value criterion alone, you should pay any finite price. This is the St. Petersburg paradox, because obviously you would not pay \1{,}000, let alone everything you own. Why the sensible answer is small, two independent resolutions. 1. Diminishing marginal utility (Bernoulli's resolution). You don't …

The St. Petersburg game, what would you pay?

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