Probability Puzzles for Quant Interviews: 5 Must-Know Problems
These five puzzles appear repeatedly at Jane Street, SIG, Citadel, and Optiver. They are not just math problems — they test your ability to reason carefully about information, priors, and the limits of expected value as a decision framework.
You are on a game show. There are 3 doors: behind one is a car, behind the other two are goats. You pick Door 1. The host (who knows what is behind every door) opens Door 3, revealing a goat. He then asks: do you want to switch to Door 2, or stay with Door 1?
Switch. Your probability of winning jumps from 1/3 to 2/3 by switching.
- 1.Initially, P(car behind Door 1) = 1/3 and P(car behind Door 2 or 3) = 2/3.
- 2.The host's action transfers information. By opening a goat door from {2, 3}, he concentrates the 2/3 probability onto the one door he did not open.
- 3.After the host opens Door 3: P(car behind Door 1) = 1/3 (unchanged), P(car behind Door 2) = 2/3.
- 4.Switching wins in exactly the cases where you initially picked a goat — which happens 2/3 of the time.
The trap is intuiting that the host's reveal is uninformative because you already knew at least one of the other doors was a goat. The key is that the host's choice is constrained: he cannot open the car door. That constraint carries information.
In a room of n people, what is the probability that at least two share a birthday? How large does n need to be for this probability to exceed 50%?
With n = 23 people, P(shared birthday) > 50%. With n = 70, it exceeds 99.9%.
- 1.Compute the complement: P(no shared birthday among n people).
- 2.Person 1 can have any birthday: 365/365. Person 2 must differ: 364/365. Person 3: 363/365. And so on.
- 3.P(all different) = (365/365) × (364/365) × (363/365) × ... × ((365 − n + 1)/365)
- 4.For n = 23: P(all different) ≈ 0.493, so P(at least one match) ≈ 0.507.
The result shocks most people because they implicitly compare against a specific target birthday (P ≈ 6% for a specific person matching yours with 23 others). The birthday problem asks about any pair, and there are C(23,2) = 253 pairs — far more than people intuit. In quant interviews, this problem tests whether you think about complements and avoid the common base-rate mistake.
Two envelopes each contain money. You are told one has twice as much as the other. You pick one and see it contains $100. Should you switch? It seems like the other envelope contains either $50 or $200, giving an expected value of ½($50) + ½($200) = $125 > $100. So always switch — but then by the same logic, switch again forever. What's going on?
The switching strategy is not actually dominated. The paradox dissolves when you apply a proper prior on the amounts.
- 1.The flaw is in assuming P(other envelope = 2×) = P(other envelope = ½×) = ½ independently of the observed amount.
- 2.Call the smaller amount X (unknown). You observe either X or 2X with equal probability. If you see X, switching gives 2X (gain X). If you see 2X, switching gives X (lose X). The expected gain from switching is 0 — not positive.
- 3.The seeming paradox arises because you cannot simultaneously have P(lower amount = 50 | saw 100) = ½ and P(lower amount = 100 | saw 100) = ½ for every possible observation. A valid probability distribution over the amounts would break this symmetry.
- 4.Without a proper prior (e.g., uniform on some bounded set), the calculation is ill-defined.
Quant interviewers use this to test whether you can identify when expected value calculations are ill-posed. The correct answer is not a number — it is a diagnosis of why the calculation fails, and a statement of what additional information would make it well-defined.
Sleeping Beauty is put to sleep on Sunday. A fair coin is flipped. If Heads: she is woken once on Monday. If Tails: she is woken on Monday and Tuesday (with her memory wiped between awakenings). Upon waking, with no knowledge of the day or the outcome, she is asked: what is your credence that the coin landed Heads?
This is a genuine open problem in probability philosophy. The two defensible answers are 1/2 (the "halfer" position) and 1/3 (the "thirder" position).
- 1.Halfer argument: The coin is fair, so P(Heads) = 1/2. Waking up provides no information about the flip outcome — it happens regardless. Credence = 1/2.
- 2.Thirder argument: Consider all possible awakenings: (Heads, Monday), (Tails, Monday), (Tails, Tuesday). Beauty is equally likely to be in any of these three situations, so P(Heads) = 1/3.
- 3.The disagreement is about what "credence" means: is it the probability of the event, or a self-locating belief about which possible awakening you are currently in?
- 4.Most working probabilists lean toward 1/3 because it leads to correct betting behavior (if Beauty is paid $1 for correctly guessing the outcome, she should bet Tails at any credence < 1/2).
You will not be expected to resolve this definitively. The point is to articulate both views clearly, explain the underlying disagreement (objective probability vs. self-locating belief), and not confuse yourself. Showing you can hold two coherent positions simultaneously is the signal the interviewer is looking for.
A casino offers the following game: a fair coin is flipped until Tails appears. If the first Tails is on flip n, you win $2ⁿ. What is a fair price for this game?
The expected value is infinite, yet no rational person would pay more than $20–$30 to play. The paradox illustrates the failure of expected value as the sole guide to rational decisions.
- 1.E[payout] = Σ P(Tails on flip n) × 2ⁿ = Σ (½)ⁿ × 2ⁿ = Σ 1 = ∞.
- 2.Despite infinite EV, the practical value is low: 50% of the time you win $2, 75% of the time you win at most $4, 87.5% of the time at most $8.
- 3.Resolution via expected utility: if utility is concave in wealth — as with logarithmic utility u(w) = ln(w) — the expected utility is finite. Bernoulli showed that with u(w) = ln(w), a player should pay about ln(2) × $wealth ≈ roughly $25 for a starting wealth of $1,000.
- 4.Resolution via variance: the game has extreme variance. Risk-averse agents discount high-variance bets — the option price of a bet is not just its EV.
This problem comes up in trading interviews as an introduction to risk-aversion and utility theory. The key points: (1) expected value alone is insufficient for decision-making under risk, (2) in practice, tail events are only worth their discounted expected value to a risk-averse agent, (3) this is the origin of Bernoulli's logarithmic utility and the expected utility framework still used in portfolio theory today.
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