Quant trading interviews at Jane Street, Optiver, Citadel Securities, IMC, SIG, and the rest of the prop desks test three things: probability fluency, mental math under pressure, and disciplined market-making logic. These firms aren't trying to find the candidate who memorized the most puzzles — they're looking for the one who reasons cleanly when the numbers are unfamiliar.
Below are 18 brainteasers and probability questions firms actually ask. Each section gives the prompt, the clean derivation, the answer, and the follow-up the interviewer almost always asks.
If you'd rather drill these against an AI interviewer that pushes back on shaky reasoning and times your responses, run a quant trading mock.
1. You flip a fair coin until you see two heads in a row. What's the expected number of flips?
Let E be the expected total flips. Let E_1 be the expected flips remaining after just having seen one head.
From scratch:
- With prob 1/2 (tails), one flip wasted, restart: contributes (1/2)(1 + E).
- With prob 1/2 (heads), now in state E_1: contributes (1/2)(1 + E_1).
So E = 1 + (1/2) E + (1/2) E_1.
From state E_1:
- Heads (prob 1/2): done in 1 more flip.
- Tails (prob 1/2): wasted, back to start: contributes (1/2)(1 + E).
So E_1 = 1 + (1/2)(0) + (1/2) E = 1 + E/2.
Substitute: E = 1 + E/2 + (1/2)(1 + E/2) = 1 + E/2 + 1/2 + E/4. Solve: E/4 = 3/2, E = 6.
Follow-up: Three heads in a row? Same recursion structure: E_3 = 14. General formula for k heads in a row on a fair coin: 2^(k+1) − 2.
2. A drawer has 10 black socks and 10 white socks. You pull socks at random until you have a matching pair. What's the expected number of pulls?
Pull 1: any color, say black. Pull 2:
- Prob 9/19 it's black → done in 2 pulls.
- Prob 10/19 it's white → continue.
If you got mixed in the first 2, pull 3:
- Now 9 black + 9 white left. Either color matches one of the two you've seen. Prob match = 18/18 = 1. Done in 3 pulls.
So expected pulls = 2 × (9/19) + 3 × (10/19) = (18 + 30)/19 = 48/19 ≈ 2.53.
Follow-up: 3 colors with 10 each? Pull 1: any. Pull 2: prob 9/29 same color (done in 2), 20/29 different (state: 1 of two colors). From state, pull 3: 9/28 same as pull 1, 9/28 same as pull 2, 10/28 third color. So either prob 18/28 done in 3, or 10/28 in state 3-distinct. Pull 4: any color matches. EV = 2(9/29) + 3(20/29)(18/28) + 4(20/29)(10/28). Compute: ≈ 2.62 + 0.78 ≈ 3.4.
3. You have 25 horses and a racetrack that holds 5 at a time. No timer — you only learn the relative finish order of each race. What's the minimum number of races to find the top 3 fastest horses?
Answer: 7.
- Races 1–5: Race them in 5 groups of 5. Now you know the rank within each group.
- Race 6: Race the 5 group winners against each other. The winner of this race is the overall fastest horse.
- After race 6, eliminate horses that can't possibly be in the top 3: the entire group whose winner finished 4th and 5th (10 horses out), the bottom 3 of the group whose winner came 3rd (3 out), the bottom 2 of the group whose winner came 2nd (2 out), the bottom 2 of the group of the overall winner (2 out — overall winner is in, top 3 of that group are still candidates for top 3 overall).
- Remaining candidates for places 2 and 3: 5 horses (top 3 of group 1 minus the winner already known, top 2 of group 2, top 1 of group 3 = 2 + 2 + 1 = 5).
- Race 7: Race those 5. Top 2 are the overall 2nd and 3rd.
Follow-up: top 4? Need an 8th race in some configurations. The answer 7 holds for top 3 specifically.
4. You're flipping a fair coin. If it comes up heads, you win $1. If tails, you flip again — but if tails again, you lose $2. Otherwise (heads on second flip) you win nothing. What's the EV of one round?
Outcomes:
- Heads on first flip (prob 1/2): +$1.
- Tails then heads (prob 1/4): $0.
- Tails then tails (prob 1/4): −$2.
EV = (1/2)(1) + (1/4)(0) + (1/4)(−2) = 1/2 + 0 − 1/2 = $0. Fair game.
Follow-up: What if the second flip is biased 60/40 toward tails? Probabilities become (1/2)(1) + (1/2)(0.4)(0) + (1/2)(0.6)(−2) = 0.5 + 0 − 0.6 = −$0.10. You'd pay to play it positive, not negative.
5. A factory makes 100 products. 1 is defective. You sample 10 at random without replacement. What's the probability you find the defective one?
Symmetry argument: each product is equally likely to be in your sample. The defective one is in your sample with probability 10/100 = 0.10.
Follow-up: What if 5 are defective and you sample 10? P(no defectives) = C(95,10)/C(100,10) = (95!/85!) / (100!/90!) = (95×94×93×92×91)/(100×99×98×97×96). Compute: 95/100 × 94/99 × 93/98 × 92/97 × 91/96 ≈ 0.95 × 0.9495 × 0.9490 × 0.9485 × 0.9479 ≈ 0.584. So P(at least 1 defective) ≈ 0.416.
6. You have a 3-gallon jug and a 5-gallon jug. How do you measure exactly 4 gallons?
Fill the 5-gallon. Pour into the 3-gallon until the 3 is full — leaves 2 gallons in the 5. Empty the 3-gallon. Pour the 2 from the 5 into the 3 (the 3 now has 2). Fill the 5-gallon again. Pour from the 5 into the 3 until the 3 is full — that takes 1 more gallon. The 5-gallon now has 4 gallons.
Follow-up: 7-gallon and 11-gallon jugs, measure 6? Same idea, more steps. Fill 11, pour into 7 (7 full, 4 in 11). Empty 7. Pour 4 from 11 into 7. Fill 11, pour into 7 until 7 is full — uses 3, leaves 8 in 11. Empty 7. Pour from 11 into 7 — 8 won't fit, 7 fills, 1 left in 11. Empty 7. Move 1 from 11 to 7. Fill 11. Pour into 7 until full (uses 6, leaves 5 in 11). Hmm — better answer involves GCD and Bezout's identity.
7. Three light switches outside a room control three light bulbs inside, but you can't see the bulbs from outside. You can only enter the room once. How do you figure out which switch controls which bulb?
Turn switch 1 on for 10 minutes. Turn it off. Turn switch 2 on. Enter the room.
- The bulb that's on is switch 2.
- The bulb that's off and warm is switch 1.
- The bulb that's off and cold is switch 3.
Follow-up: 4 switches, 4 bulbs? Two bits of state per bulb (on/off + hot/cold) gives 4 possible signatures. Need to encode 4 distinct signatures across 4 switches. Doable — turn 1 on for 20 min then off, turn 2 on for 10 min then off, turn 3 on (no toggle), leave 4 off. The two "off" bulbs differ by temperature; the two "on/off recently" bulbs are 1 and 3 (3 is on, 1 is hottest off). Works.
8. You're playing a game where you draw 2 cards from a standard 52-card deck. Both must be red to win. What's the probability of winning?
P(first red) = 26/52 = 1/2. P(second red | first red) = 25/51.
P(win) = (26/52) × (25/51) = (1/2) × (25/51) = 25/102 ≈ 24.5%.
Follow-up: 3 red? (26/52)(25/51)(24/50) = (1/2)(25/51)(12/25) = 12/102 = 2/17 ≈ 11.8%.
9. A bag contains an unknown mix of red and blue marbles. You draw 5 and they're all red. What's your best estimate of the proportion of red marbles?
Frequentist answer: 5/5 = 100% — useless.
Bayesian answer: assume a uniform prior on the proportion p of red marbles. Posterior given 5 reds out of 5 is Beta(6, 1) (using Beta(α+k, β+n−k) with α = β = 1 prior and k=5, n=5). Posterior mean = 6/(6+1) = 6/7 ≈ 0.857.
Why does this matter for trading? The posterior mean reflects the right level of uncertainty. With 5 reds the true proportion could easily be 0.7 or 0.5; reporting 100% would lead to overconfident position sizing. The Beta-Binomial framework is how prop traders update on small samples.
Follow-up: 50 reds out of 50? Posterior Beta(51, 1), mean ≈ 0.981. Still not 1.0 — Bayesian humility.
10. You're a market maker quoting bids and asks on a coin flip game. The coin is fair. The game pays $1 for heads, $0 for tails. What's your bid-ask?
True EV = $0.50.
A risk-neutral market maker should quote symmetrically around the EV with a spread that covers their costs and adverse selection risk. Reasonable spread: 5–10 cents. So bid 0.45 / ask 0.55.
Follow-up: Now the customer can buy or sell unlimited size at your quote. Do you widen? Yes — adverse selection means a counterparty trading huge size against you probably has information. Spread widens with size. A more sophisticated answer: I'd quote bid 0.40 / ask 0.60 for sizes above some threshold, or refuse to quote unlimited size.
Follow-up 2: The customer trades at your ask. Do you adjust your quote? Yes — they bought from me, so they might know something. I lean my quote up — maybe 0.48 / 0.58 — until I see more flow or learn whether their trade is informed.
11. There are 3 boxes: one has only red marbles, one has only blue, one has a 50/50 mix. The labels have all been swapped (every label is wrong). You can pull one marble from one box without looking inside. How do you figure out which box is which?
Pull one marble from the box labeled "mixed."
The labels are all wrong, so this box is not mixed — it's either all red or all blue. The marble you pull tells you which one.
If you pulled a red marble, that box is "all red." Now: the box labeled "blue" can't be blue (label wrong) and can't be red (we just identified that one), so it must be mixed. The remaining box (labeled "red") must be blue.
Why it works. Pulling from "mixed" gives one bit of information that resolves all three because the constraint "every label wrong" is so tight.
12. You have a 100-sided die. You roll it. If you don't like the roll, you can re-roll once. The number on your final roll is your payout in dollars. What strategy maximizes EV?
Re-roll if and only if the first roll is below your re-roll EV. Let r be the threshold.
If first roll ≥ r: keep it. EV contribution per first-roll value v (for v ≥ r): v × (1/100).
If first roll < r: re-roll, EV of re-roll is the unconditional mean of the die = 50.5. Contribution: 50.5 × (probability first roll < r) = 50.5 × (r−1)/100.
Total EV: Σ(v=r to 100) v/100 + 50.5 × (r−1)/100.
Optimize over r. The optimum is r = 51: re-roll if you roll ≤ 50, keep if you roll ≥ 51.
EV at r=51: Σ(51 to 100) v/100 + 50.5 × 50/100 = (50 × 75.5)/100 + 25.25 = 37.75 + 25.25 = 63.
Follow-up: Two re-rolls allowed? Same recursion: re-roll if first roll < EV-with-1-re-roll-remaining (=63). Keep if ≥ 63. EV-with-2-re-rolls = Σ(63 to 100) v/100 + 63 × 62/100 ≈ 30.97 + 39.06 = 70.0.
13. You have an urn with N coins, of which one is biased to land heads with probability 0.75 and the rest are fair. You draw a coin at random and flip it twice. Both come up heads. What's the probability you drew the biased coin?
Bayes: P(biased | HH) = P(HH | biased) × P(biased) / P(HH).
P(HH | biased) = 0.75² = 0.5625P(biased) = 1/NP(HH | fair) = 0.5² = 0.25P(HH) = (1/N)(0.5625) + ((N−1)/N)(0.25) = (0.5625 + 0.25(N−1))/N
So P(biased | HH) = 0.5625 / (0.5625 + 0.25(N−1)).
For N = 10: 0.5625 / (0.5625 + 2.25) = 0.20.
For N = 100: 0.5625 / (0.5625 + 24.75) = 0.022.
Follow-up: How many heads in a row to be 95% sure it's biased with N=100? (0.75/0.5)^k × (1/99) ≥ 0.95/0.05 = 19. 1.5^k ≥ 19 × 99 = 1881. k ≥ log(1881)/log(1.5) ≈ 18.6, so 19 heads in a row.
14. You're trading a stock currently at $100. You're told the stock will, at the end of one year, equal one of {80, 90, 100, 110, 120} with equal probability. What's the EV and what's the volatility?
EV = (80+90+100+110+120)/5 = 100.
Variance = E[(X − μ)²] = (400 + 100 + 0 + 100 + 400)/5 = 200. Standard deviation = √200 ≈ 14.14.
Follow-up: What's a reasonable price for an at-the-money call expiring in 1 year? Call value = E[max(X − 100, 0)] = (max(0, 0) + max(10, 0) + max(20, 0))/5 = (0+0+0+10+20)/5 = 6.
So you'd pay roughly $6 for the at-the-money call, and the implied volatility from a Black-Scholes back-solve would be around 14% (matches the standard deviation).
15. There are 100 prisoners and 100 boxes, each containing one of the prisoners' numbers in random order. Each prisoner enters the room alone, opens up to 50 boxes, and must find his number. They can't communicate after the game starts. They survive only if every prisoner finds his number. What strategy maximizes survival probability?
Strategy. Each prisoner opens the box with his own number first. If it doesn't contain his number, he opens the box matching the number he just found. He continues this chain until he finds his own number or runs out of opens.
This works because the boxes form a permutation, and the "follow the chain" rule traces the cycle containing the prisoner's number. The prisoner finds his number iff the cycle has length ≤ 50.
P(every prisoner survives) = P(every cycle in random permutation of 100 has length ≤ 50). The probability that a random permutation has some cycle of length > 50 = sum over k=51..100 of (1/k). That's H_100 − H_50 ≈ 5.187 − 4.499 = 0.688. So P(all cycles ≤ 50) ≈ 31.2%.
Random strategy gives (1/2)^100 ≈ 10^−30. The chain strategy is exponentially better.
Why this matters. It's the canonical "structure beats brute force" interview question — interviewers want to see you spot that the random permutation has cycle structure they can exploit.
16. A and B take turns flipping a fair coin. The first to flip heads wins. A goes first. What's P(A wins)?
A wins on turn 1 with prob 1/2. If both A and B miss (prob 1/4), the situation is identical to the start with A about to flip again.
P(A) = 1/2 + (1/4) × P(A). Solve: P(A) × (3/4) = 1/2, P(A) = 2/3.
Follow-up: A, B, C take turns? P(A) + P(B) + P(C) = 1. By similar recursion, P(A):P(B):P(C) = 1 : 1/2 : 1/4 = 4:2:1. So P(A) = 4/7, P(B) = 2/7, P(C) = 1/7. Going first is worth ~57% in a 3-player game.
17. You have a deck of 52 cards. You draw cards one at a time without replacement. After each card, you can choose to "stop" — and your payoff is the fraction of black cards remaining in the deck. What's the optimal strategy and the EV?
Surprising answer: the optimal strategy guarantees exactly 0.5.
Proof sketch: if you never stop, you end with 0 cards left and 0 fraction. If you stop after card 1, fraction = 26/51 if first card was red, 25/51 if first was black. EV across both = (1/2)(26/51) + (1/2)(25/51) = 51/102 = 0.5.
Inductively, the EV of optimal play from any deck state equals the current fraction of black cards remaining. So whatever state you're in, stopping gives you exactly the EV of continuing optimally — there's no edge.
This is a martingale problem. Tracking it through is the kind of question that separates candidates with real probability intuition from those who memorized formulas.
Follow-up: What if the payoff is the fraction of cards drawn so far that were black? Now stopping earlier doesn't matter because each draw has expected fraction 0.5 anyway. EV is still 0.5.
18. Market making — you're quoting on the over/under for the sum of two fair dice. Customer can buy "over 7" or sell "under 7" at your quotes. What do you quote?
True P(sum > 7) = 15/36, P(sum = 7) = 6/36, P(sum < 7) = 15/36. By symmetry, no edge in either direction.
If you're trading on a binary "is sum > 7?" with payout $1 for yes, EV = 15/36 ≈ $0.417.
So bid/ask 0.40 / 0.45 or so. Tight spread because there's no information asymmetry on a dice game; you only need to cover transaction costs and inventory risk.
Follow-up: Customer buys "over 7" at 0.45. Then someone else asks for the same. Do you adjust? Slightly — you've taken on inventory in one direction. Lean your quote down (0.43/0.48) to encourage offsetting flow. This is exactly how delta hedging works in options market making, just on a discrete distribution.
Follow-up 2: Customer buys "over 7" at 0.45 and you find out he had inside information that one of the dice is loaded toward 6. The probability assumptions break — you're now short a position with bad EV. The lesson: market making isn't risk-free; adverse selection is the dominant cost. Real desks size positions based on suspected information asymmetry, not just the textbook EV.
FAQ
How fast should I be answering these?
Aim for 60–90 seconds per question, with another 30–60 for the follow-up. Optiver and SIG specifically time these and grade speed; Jane Street is more lenient on time but harder on the reasoning quality.
Do interviewers care if I get the answer wrong?
They care more about how you got there. A wrong answer with clean reasoning ("I think it's X because of Y, though I'm not certain Y is the right framing") gets respect. A right answer recited from memory with no reasoning gets you the next question fast.
How much mental math do I need?
At Optiver and SIG, fluent up to 3-digit × 2-digit and percentage-of-percentage in your head. At Jane Street and Citadel, less raw math but more probability reasoning. Drill mental math separately — it's a muscle that builds in days, not weeks.
What's the worst thing to say in a quant interview?
"I don't know" without reasoning. Quant trading is fundamentally about reasoning under uncertainty — interviewers want to see you make a guess, name the assumption, and update on feedback. Refusing to commit looks like you don't have a model of the world.
Do these firms care about my technical / coding background?
Yes, but the brainteasers are the differentiator. Plenty of strong CS candidates wash out in trader interviews because they can't reason cleanly under time pressure, and plenty of quant candidates with weaker CS get hired because they nail the probability and the market making. The brainteasers are the bar.