The St. Petersburg Paradox, for quant interviews

Infinite expected value, and nobody will pay $20

A house offers you this game. It flips a fair coin until the first Heads appears. If that first Heads lands on flip $k$, it pays you $\$2^{k}$: two dollars for Heads on flip one, four for Heads on flip two, eight for flip three, and so on, doubling forever. What is the most you would pay for a single ticket?

Compute the expected payout and you get a number that ends the conversation: infinity. By the naive rule “pay anything below expected value,” you should hand over your house, your savings, and your firstborn for one ticket. You wouldn’t pay $20. Neither would the interviewer, and neither would any desk. That gap — between an infinite average and a two-figure willingness to pay — is the entire question. They are not testing whether you can evaluate the sum. They are testing whether you understand that expected value is a tool with a domain of validity, and whether you can name what breaks it here.

Why the expected value really is infinite

Do the arithmetic honestly first, because the answer is genuinely $\infty$ and pretending otherwise is a worse mistake than believing it. The payoff on flip $k$ is $\$2^{k}$ and it occurs with probability $2^{-k}$ (that is $k-1$ Tails then a Head: $(1/2)^{k-1}\cdot(1/2)=2^{-k}$), so each term of the expectation is a product of a shrinking probability and an exploding prize:

Look at what happened: the probability halves every round and the payoff doubles every round, so they cancel exactly. Every single round contributes the same $\$1$ to the expected value, and there is no final round to stop the sum. The partial sum after $n$ rounds is exactly $\$n$ — it grows without bound. The infinity is not a rounding artifact or a trick of notation; it is a real feature of a payoff that grows as fast as its probability decays. Any correct answer starts by conceding this, then explains why it doesn’t imply an infinite price.

What the interviewer is actually testing: the tail is the whole distribution

Expected value is a mean, and a mean is only a good summary when the distribution isn’t dominated by a remote tail. Here it is all tail. Ask what actually happens when you play. With probability $\tfrac12$ you win just $\$2$. With probability $\tfrac34$ you win $\$4$ or less. With probability $\tfrac78 \approx 87.5\%$ you walk away with $\$8$ or less. The chance of winning $\$64$ or more is $2^{-5} \approx 3.1\%$.

So the typical experience of this game is winning pocket change; the infinite mean is manufactured entirely by outcomes like “1,024 Tails in a row,” each astronomically unlikely and astronomically large. The interviewer wants you to say: expected value is being hijacked by a measure-zero-in-spirit tail, so a decision-maker should price the game by something that discounts that tail — bounded utility, a finite bankroll, or the long-run growth rate. That single sentence is what separates a candidate who memorized “St. Petersburg = infinity” from one who understands why the paradox is a paradox.

Bernoulli’s resolution: value the prize, not the dollars

Daniel Bernoulli’s 1738 fix, and still the cleanest interview answer: people value marginal dollars less as they get richer. Replace the linear “money” yardstick with a concave utility $u(x)$ and price the game by expected utility. Take the canonical choice $u(x)=\ln x$. Then the utility of a $\$2^{k}$ prize is $\ln(2^{k})=k\ln 2$ — it grows only linearly in $k$ while the prize grows geometrically. Weight those utilities by their probabilities:

The series $\sum_{k\ge1} k\,2^{-k}$ converges to $2$ (it is $\tfrac{x}{(1-x)^2}$ at $x=\tfrac12$), so the expected log-utility is finite: $2\ln 2$. Convert back to a dollar figure via the certainty equivalent, the guaranteed amount with the same utility: $\ln(\text{CE}) = 2\ln 2$, hence

A log-utility player should pay about $4 for a ticket — finite, small, and much closer to human behavior than $\infty$. Concavity tamed the tail: doubling a prize adds a fixed $\ln 2$ of value, not double the value, so the exploding payoffs stop mattering.

The bankroll bound: no real house can pay $2^{60}$

The utility argument is about your preferences. The blunter, more quant-flavored resolution is about the counterparty: no real house can honor a payoff of $\$2^{60}$. Suppose the house can pay at most $\$2^{L}$ and caps the prize there. Then for $k\le L$ you get $\$2^{k}$, and for every longer run you get the capped $\$2^{L}$. The expectation collapses to something tiny:

The whole infinite tail past round $L$ contributes exactly $\$1$. So the fair price is $L+1$ dollars, where the bankroll is $2^{L}$. Put real numbers on it: a house with a $1 million bankroll ($2^{20}\approx 1.05\text{M}$) makes the game worth about $21. Bump the house to $1 billion ($2^{30}$) and it’s worth $31. Give it a $1 trillion war chest ($2^{40}\approx 1.10\text{T}$) and the ticket is still only worth about $41. The value grows like $\log_2(\text{bankroll})$ — doubling the entire bankroll adds exactly one dollar of fair value. “Infinite” was always an idealization that assumed an infinitely deep counterparty.

Variants that change the answer

The result is fragile in instructive ways; expect the interviewer to poke it.

• Super-exponential payoffs defeat log utility. Change the prize to $\$2^{(2^{k})}$. Now $\ln(\text{payoff})=2^{k}\ln 2$, so each utility term is $2^{-k}\cdot 2^{k}\ln 2=\ln 2$ and the expected utility diverges again. Log utility is not a magic shield. This is Karl Menger’s 1934 point: only a bounded utility function resolves every version of the paradox, because any unbounded utility can be out-run by a fierce enough payoff schedule.
• A different growth base. If the prize grows slower than the odds shrink — say payoff $c^{k}$ with $c<2$ — each term is $2^{-k}c^{k}=(c/2)^{k}$, a convergent geometric series, and the EV is finite and ordinary. The paradox is specifically the knife-edge where payoff growth exactly matches probability decay.
• Repeated play / time average. Averaging over infinitely many parallel players (the ensemble average) gives $\infty$, but a single player compounding a fixed fee over time faces a finite — often negative — growth rate. The ergodicity lens is the modern “why EV alone can mislead”: the number you should optimize is the time-average growth rate, not the ensemble mean.

Common mistakes

• Denying the infinity. Insisting the EV is “really finite” because “you can’t flip forever” without doing the capped-bankroll computation. State $\E[X]=\infty$ for the idealized game, then introduce the cap.
• Stopping at “utility fixes it.” Naming diminishing marginal utility but not producing a number. The interviewer wants the certainty equivalent — here $e^{2\ln 2}=\$4$ — and wants to hear that unbounded utility can still be broken.
• Confusing “rare” with “ignorable.” The tail here is rare and enormous; that is exactly the regime where the mean is a bad summary. Reach for the median ($2), a quantile, or a risk-adjusted value instead.
• Forgetting the counterparty. Pricing purely off your own utility and never asking who is on the other side. The bankroll cap is often the cleanest, most “trader” answer: $L+1$ dollars for a $2^{L}$ house.