The St. Petersburg Paradox, Explained

After reading this you will know why a coin game with infinite expected value almost always pays 2 or 4 coins, why its running average grows like the logarithm of the number of plays, and how to price it sensibly anyway.

What the game is

Toss a fair coin until it comes up heads. Count the tosses. If the first head appears on toss k, you win 2^k coins. So heads on the first toss pays 2. Tails then heads pays 4. Tails, tails, heads pays 8, and so on.

Here is the hook. The expected prize is infinite, yet you would not pay 100 coins to play a single round. Most rounds end after one or two tosses and pay 2 or 4. The paradox is the gap between the average, which is infinite, and the typical outcome, which is tiny.

Daniel Bernoulli wrote about this in St. Petersburg in 1738, and the name stuck. It is one of the oldest known cases where expected value fails as a guide to what a gamble is worth.

Where the infinity comes from

The probability that the first head lands on toss k is the chance of k-1 tails followed by one head:

Here k is the toss number of the first head, and P(k) is its probability. So P(1) = 1/2, P(2) = 1/4, P(3) = 1/8. These add to 1, as they must.

The expected prize multiplies each payout by its probability and sums:

Every term is exactly 1. The prize doubles at each step, and the probability halves, so the two cancel. An infinite sum of ones diverges. The mean prize is \infty.

Why the typical payout is tiny

Infinite mean does not mean infinite winnings. It means a small number of enormous jackpots drag the average up without limit. Look at how the probability mass sits.

Add the first four rows: 50 + 25 + 12.5 + 6.25 = 93.75% of games pay 16 or less. The median payout is 2, since half of all games end on the first toss. A payout of 1024 needs the first head on toss 10, which happens once in 1024 games on average. The infinity lives entirely in the far tail.

What the running average actually does

Play N games and average the payouts. The average does not settle. It creeps upward roughly in step with \log_2 N. The reason is that in N games you expect to see a first head as late as toss \log_2 N, and that rare game contributes a payout near N, which alone adds about 1 to the average.

A useful rule of thumb: the running average after N games sits near

where \bar{X}_N is the mean payout over N games. This is an approximation with large swings, not a tight law. Each time a fresh record jackpot arrives, the average jumps, then drifts down slowly until the next record.

A worked run reproducing the demo

Run the simulator with the default settings. It plays the game many times and plots the running average. Here is a concrete synthetic run so you can check the arithmetic against what you see on screen.

Now watch what a long run does. If you extend to 100 games, that lone jackpot's influence shrinks toward its expected share, and the average tends back toward roughly \log_2 100 \approx 6.6 until the next big hit arrives.

How to price the game with utility

Bernoulli's own fix was to say people value coins by usefulness, not face amount, and that usefulness grows slower than money. Use the logarithm of wealth as utility. The expected utility of the prize then converges even though the expected prize does not.

Here X = 2^k is the prize and \log_2 X = k. The sum \sum k/2^k equals exactly 2. So the expected log-payout is 2 log-coins, which corresponds to a prize of 2^2 = 4 coins. A person valuing money on a log scale would pay about 4 coins to play, not infinity. That matches intuition far better.

Utility is one repair. Another is a bankroll cap: no casino can pay 2^k without limit. Cap the payout at 2^{40} (about a trillion), and the fair price drops to 40 coins, since the sum now has 40 terms of 1. Finite banks force finite prices.

Common mistakes

Trusting the mean. Expected value assumes you can average over many independent plays and that no single play can bankrupt you. Neither holds here. The mean is the wrong summary.

Expecting the average to converge. The Law of Large Numbers needs a finite mean. This distribution has none, so the ordinary law does not apply and the running average keeps climbing.

Confusing infinite with large-but-typical. Infinite mean coexists with a median of 2. The game is cheap almost always and astronomically rich almost never.

Related tools

To see averages that actually settle, watch the Law of Large Numbers pull a die's running mean onto 3.5. To understand why sums of many small independent draws smooth into a bell shape (and why this game breaks that), try the Central Limit Theorem Demo. For a hands-on tour of estimating quantities by random sampling, the Monte Carlo Playground shows the method working when the mean is finite. And for another probability puzzle where intuition and arithmetic disagree, play the Monty Hall Simulator.

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