Quant interview preparation

Questions

100

Traced to a firm

53

Firms

15

Updated

September 2026

Asked at

All firmsOld Mission Capital12Tower Research Capital10Jump Trading7Akuna Capital5Citadel4DED.E. Shaw3Jane Street3ACAQR Capital Management2DRW2Millennium Management2Schonfeld2SCSquarepoint Capital2Susquehanna International Group2Belvedere Trading1Optiver1

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All topicsProbability10Coins, cards and games6Expected value8Statistics11Market making15Estimation and mental maths4Stochastic processes4Regression5Machine learning6Time series6Programming10Options and derivatives8Fit and motivation7

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1. 002I hand you a coin that comes up heads one third of the time. How do you generate a fair coin flip from it?Coins, cards and gamesIntermediatephone / first roundDED.E. ShawResearch · New York · 2026Tower Research CapitalProp Trading · New York · 2019

   Say this

   Flip it twice. Call heads-then-tails a fair heads, tails-then-heads a fair tails, and if you get HH or TT throw the pair away and start again. HT and TH both have probability p times one minus p, so they are equally likely whatever p is.

   Then walk it

   1. With p equal to a third: HT is 1/3 times 2/3 which is 2/9, and TH is 2/3 times 1/3 which is also 2/9. Identical, so conditioning on one of the two having happened gives you exactly a half.
   2. That is von Neumann's trick. The point is that it needs no knowledge of p at all, which is what makes it useful. You never have to estimate the bias.
   3. It does need two things: the flips are independent, and p is strictly between zero and one. A coin that is genuinely two-headed breaks it, and so does a coin whose bias drifts flip to flip.
   4. Probability a given pair is useful is 2 times 2/9, which is 4/9. So you discard more than half your pairs at p equal to a third.
   5. The honest limitation: it is unbiased but wasteful. If the bias drifts slowly, you can protect yourself by pairing adjacent flips rather than flips far apart, so the drift cancels locally.

   Where candidates lose it

   Trying to estimate p first and then correct for it. That introduces estimation error and gives you an approximately fair coin, not a fair one. The elegant answer is exactly fair with zero knowledge of p, and the interviewer is looking for that symmetry argument.

   Expect next

   • What is the expected number of flips of the biased coin per fair flip?
   • How would you get a uniform random number on one to three from the same coin?
   • Can you do better than throwing HH and TT away entirely?

   Reported by candidates at D.E. Shaw (Research, New York, 2026); Tower Research Capital (Prop Trading, New York, 2019). Source: Wall Street Oasis.

2. 003Using that procedure with p equal to one third, what is the expected number of biased flips you need to produce one fair flip?Expected valueIntermediatetechnicalDED.E. ShawResearch · New York · 2026Tower Research CapitalProp Trading · New York · 2019

   Say this

   Four and a half. Each pair succeeds with probability 2p(1-p), which is 4/9 here, so the number of pairs is geometric with mean 9/4, and each pair costs two flips. Two times 9/4 is 4.5 flips.

   Then walk it

   1. A geometric with success probability q has mean 1/q. Here q is 4/9, so you expect 2.25 pairs before one is usable.
   2. Two flips per pair gives 4.5 flips per fair bit. Say the arithmetic out loud so the interviewer sees the two-step structure: geometric on pairs, then a constant multiplier.
   3. Sanity check the extremes. At p equal to a half, q is 1/2 and the cost is 4 flips per fair bit, which is the cheapest this method ever gets. As p goes to zero the cost blows up like 1/p, which matches the intuition that a near-deterministic coin carries almost no information.
   4. Compare that to the theoretical floor. A p equal to 1/3 coin carries about 0.918 bits of entropy per flip, so in principle you need only about 1.09 flips per fair bit. Von Neumann at 4.5 is four times worse than optimal.
   5. The gap is the interesting part, and it is where the follow-up goes: you are throwing away the information in the discarded HH and TT pairs, and better extractors recycle it.

   Where candidates lose it

   Forgetting to double. Candidates compute 9/4 as the number of trials and stop, when a trial is a pair of flips. Also worth stating the entropy bound unprompted, because the interviewer is almost certainly going to ask whether you can do better, and knowing the floor is how you answer that credibly.

   Expect next

   • What is the information-theoretic minimum number of flips?
   • Describe a scheme that gets closer to that bound.
   • What is the variance of the number of flips, not just the mean?

   Reported by candidates at D.E. Shaw (Research, New York, 2026); Tower Research Capital (Prop Trading, New York, 2019). Source: Wall Street Oasis.

3. 006You want to draw a black card followed by a red card. One deck is a full 52-card deck, another has had some cards removed. Which deck do you choose and why?Coins, cards and gamesIntermediatetechnicalOld Mission CapitalQuantitative Research · New York · 2014

   Say this

   Write the probability down before you pick. For a deck with b blacks and r reds, drawing black then red is b/(b+r) times r/(b+r-1). Then just compare the candidate decks on that expression, and you will find you want the deck that is as balanced as possible and as small as possible.

   Then walk it

   1. Full deck: 26/52 times 25/51, which is 0.5 times 0.490, about 24.5 percent.
   2. Now try a tiny balanced deck, one black and one red. That is 1/2 times 1/1, which is 50 percent. Far better.
   3. So the direction is clear. Removing cards helps if it keeps the deck balanced, because the second draw's conditional probability improves once the black card you removed is a bigger fraction of a smaller deck.
   4. Unbalancing hurts. A deck of 26 blacks and 1 red gives 26/27 times 1/26, which is 1/27, about 3.7 percent. Almost all your probability mass dies on the second draw.
   5. So: balanced beats unbalanced, small beats large, and the extreme is one black plus one red at fifty percent. Say the formula first, then test the corners. That is faster and less error-prone than trying to reason about it verbally.

   Where candidates lose it

   Reasoning in words about whether removing cards helps or hurts, and getting tangled. Write b/(b+r) times r/(b+r-1) immediately, then plug in three corner cases. Also do not forget the minus one in the denominator, because sampling without replacement is the entire content of the question.

   Expect next

   • What deck maximises the probability of black then red then black?
   • What if you wanted two cards of the same colour instead?
   • Now make me a market on the probability for the standard deck.

   Reported by candidates at Old Mission Capital (Quantitative Research, New York, 2014). Source: Wall Street Oasis.

4. 007What is the probability of being dealt four of a kind in a five-card poker hand?Coins, cards and gamesIntermediatetechnicalOld Mission CapitalQuantitative Research · New York · 2014

   Say this

   624 hands out of 2,598,960, which is about 0.024 percent, or one in roughly 4,165. Thirteen choices of rank for the quad, times 48 remaining cards for the fifth card.

   Then walk it

   1. Denominator: 52 choose 5 is 2,598,960. Worth memorising, it comes up constantly.
   2. Numerator: pick the rank of the four of a kind, 13 ways. All four suits are forced. Then the fifth card is any of the 48 cards left, so 13 times 48 is 624.
   3. 624 over 2,598,960 simplifies to 1 over 4,165. Call it one in four thousand.
   4. The counting discipline that matters: the kicker is 48, not 12. If you write 13 times 12 you are counting ranks not cards, and you would be off by a factor of four.
   5. Quick cross-check against a fact you might already know: a full house is 3,744 hands and a straight flush is 40. Four of a kind sitting between them at 624 is consistent with the standard hand ranking, which is ordered by exactly this rarity.

   Where candidates lose it

   Double counting, or using 12 instead of 48 for the fifth card. The other classic error is dividing by 5 factorial somewhere by accident. Use combinations consistently in both numerator and denominator, and state the denominator before you start so the interviewer can follow.

   Expect next

   • Now do a full house.
   • What is the probability of a flush, excluding straight flushes?
   • How would that change in a seven-card game like Texas hold'em?

   Reported by candidates at Old Mission Capital (Quantitative Research, New York, 2014). Source: Wall Street Oasis.

5. 009You roll a fair die and may choose to re-roll once, taking the second value if you do. What is the expected value of the game, and what is your strategy?Expected valueIntermediatetechnicalOld Mission CapitalFinance · New York · 2018

   Say this

   4.25. Re-roll on a 1, 2 or 3, keep a 4, 5 or 6. The continuation value is 3.5, so you keep anything strictly above 3.5 and re-roll anything below.

   Then walk it

   1. Work backwards. If you re-roll you face a plain die, worth 3.5. So the rule is: keep the first roll if it beats 3.5.
   2. With probability one half you roll 4, 5 or 6 and keep it. The conditional mean of those three is 5.
   3. With probability one half you roll 1, 2 or 3 and re-roll, collecting 3.5.
   4. So the value is 0.5 times 5 plus 0.5 times 3.5, which is 2.5 plus 1.75, equals 4.25.
   5. Sanity check the bounds before you commit: the answer must sit between 3.5, which is the value of no re-roll option, and 6, which is the value of a free choice of face. 4.25 sits sensibly in between, and the option to re-roll is therefore worth 0.75 to you.
   6. The general principle, and the thing they are actually testing: the threshold is the continuation value, always. This is the same logic as an American option's exercise boundary. Exercise when the intrinsic value exceeds the value of holding on.

   Where candidates lose it

   Taking the average of the two rolls, or re-rolling a 4 because it is below the maximum. The rule is compare against the continuation value, not against the best possible outcome. Also say the threshold out loud before computing, because the interviewer wants to hear the backward-induction step, not just the number.

   Expect next

   • Now allow two re-rolls. What is the value and the thresholds?
   • What if you get n re-rolls, as n goes to infinity?
   • What if the re-roll costs you a dollar?

   Reported by candidates at Old Mission Capital (Finance, New York, 2018). Source: Wall Street Oasis.

6. 013You have a feed of a hundred thousand data points and you know fifteen of them are missing, recorded as zeros at the end. If you pull a window, what is the probability of at least one missing value?ProbabilityIntermediatetechnicalJump TradingProp Trading · Remote · 2022

   Say this

   Use the complement. For a sample of n points drawn without replacement from 100,000 of which 15 are bad, the probability of at least one bad is one minus the hypergeometric probability of none, which is one minus the product over i of (99,985 minus i)/(100,000 minus i). For small n that is well approximated by one minus (1 minus 0.00015) to the n.

   Then walk it

   1. Always compute at least one as one minus none. Summing the cases is the slow road and it invites double counting.
   2. The exact object is hypergeometric: choose n from 99,985 good over choose n from 100,000. For n much smaller than 100,000 the with and without replacement answers agree to several decimals.
   3. Numbers give it life. p is 15 over 100,000, which is 0.00015. For a window of 1,000 points, one minus 0.99985 to the 1000 is about 13.9 percent. For a window of 100 it is about 1.5 percent. So this is a real problem, not a rounding issue.
   4. Useful shortcut: for small p and moderate n the answer is roughly n times p, capped by 1. A thousand times 0.00015 is 0.15, close to the exact 0.139, and the Poisson approximation 1 minus e to the minus 0.15 gives 0.1393, which is very close.
   5. The thing I would say next on a desk, because it is the real question: they are at the end of the series, which is not random at all. If they are the most recent 15 points, then any window containing the tail hits all 15 with certainty and every other window hits none. Position matters more than the count.

   Where candidates lose it

   Treating the missing points as randomly scattered when the question says they sit at the end. That is the detail being tested. Give the hypergeometric answer for the random case, then flag the structural point: trailing zeros are usually a feed-truncation artefact, so the right fix is to detect and drop the tail, not to price the probability.

   Expect next

   • How would you detect that the zeros are missing values rather than genuine zeros?
   • What is the Poisson approximation and when does it break?
   • How do you handle those points in a model without leaking future information?

   Reported by candidates at Jump Trading (Prop Trading, Remote, 2022). Source: Wall Street Oasis.

7. 015Two games have exactly the same expected value. Which one would you choose to play?Expected valueIntermediatetechnicalAkuna CapitalSales and Trading · Chicago · 2025Belvedere TradingProp Trading · Chicago · 2022

   Say this

   If the expected values tie, I choose on variance, on how many times I get to play, and on whether any outcome can wipe me out. As a one-off with a fixed stake I take the lower-variance game. Repeated many times with the ability to size, I might prefer the higher-variance one.

   Then walk it

   1. First, ask the question the interviewer wants you to ask: how many times do I get to play, and can I choose my size? Those two facts change the answer completely.
   2. One shot, fixed size: take low variance. Same mean, less dispersion, strictly better under any concave utility, and a trader's utility is concave because a bad first day costs them their limits.
   3. Repeated, and I can size: variance becomes something I can dial. Kelly says bet a fraction proportional to edge over variance, so the high-variance game just gets a smaller position. Per unit of risk they may be identical.
   4. Then the killer criterion, which is ruin. If one game has any probability of a loss larger than my capital, its long-run growth rate is minus infinity regardless of its expected value. Expected value is a bad objective when the bet is not repeatable.
   5. One more real consideration: correlation with everything else I have on. A game with the same mean and variance but zero correlation to my book is worth more than one that doubles my existing exposure. On a desk that is usually the deciding factor.

   Where candidates lose it

   Saying I am indifferent because the expected values are equal. That answers the arithmetic and fails the question, which is about risk preference. Also do not just say I prefer lower variance and stop, because the interesting answer depends on repetition, sizing and ruin. Ask the clarifying question first.

   Expect next

   • What if you could play one of them a thousand times?
   • How would you size each one?
   • Explain the Kelly criterion and why traders bet less than Kelly.

   Reported by candidates at Akuna Capital (Sales and Trading, Chicago, 2025); Belvedere Trading (Prop Trading, Chicago, 2022). Source: Wall Street Oasis.

8. 016You win a hundred dollars if you roll a ten with two dice. How much would you risk to play?Market makingIntermediatetechnicalAkuna CapitalTrading · Chicago · 2025

   Say this

   Fair value is eight dollars and a third. Three of the 36 outcomes make ten, so probability is 1/12 and the expected payoff is 100 over 12. I would pay up to about seven to leave edge, and if I am being asked to make a two-way price I would quote around 7 at 9.

   Then walk it

   1. Count the outcomes: 6-4, 4-6, 5-5. Three ways out of 36, so 1/12, about 8.33 percent.
   2. Expected payoff 100 times 1/12 equals 8.33. That is fair value, and fair value is where you break even, not where you trade.
   3. So I need edge. I would bid 7 and offer 9 if I have to two-way it, which is about a dollar and a half of edge either side, roughly fifteen percent of fair value. That width reflects the fact that I cannot hedge a one-off die roll.
   4. Size matters as much as price. This bet has a standard deviation of about 28 dollars against a mean of 8.33, which is a terrible ratio. I would do it small even at a good price, and I would want to repeat it many times rather than do it once large.
   5. If the game is repeatable and I can do it a thousand times, I pay closer to 8. The edge I demand is compensation for variance I cannot diversify, and repetition diversifies it.

   Where candidates lose it

   Answering with the fair value of 8.33 as if that were your bid. A trader never pays fair value, and saying eight and a third is what I would risk tells the interviewer you do not understand where the money comes from. Quote a price below fair value, name your width, and say your size.

   Expect next

   • Now make me a two-way market on it and I will trade you.
   • What if I could roll a hundred times?
   • What is the standard deviation of your P&L on one play?

   Reported by candidates at Akuna Capital (Trading, Chicago, 2025). Source: Wall Street Oasis.

9. 019I owe you pi dollars, and we can only exchange whole cents. How do you settle the debt fairly?ProbabilityIntermediatetechnicalMillennium ManagementQuantitative Research · Hong Kong · 2025

   Say this

   Randomise the last cent. Pi is 3.14159 and change, so pay 3.14 with probability 1 minus 0.159 and 3.15 with probability 0.159. The expected payment is exactly pi, so the settlement is unbiased even though every individual payment is wrong.

   Then walk it

   1. The general rule: to pay an amount x on a grid, pay floor(x) with probability 1 minus the fractional part and floor(x) plus one tick with probability equal to the fractional part.
   2. Check it: 3.14 times 0.841 plus 3.15 times 0.159 equals 3.141590 to six places. Unbiased by construction.
   3. How do you generate the 0.159? Flip a fair coin repeatedly and read off binary digits until the number you have built is decisively above or below 0.159. That terminates with probability one and needs about two flips on average.
   4. This is called randomised rounding and it is not a party trick. It is exactly how you settle fractional share allocations, how you break ties in sub-penny pricing, and how stochastic rounding keeps low-precision numerics from accumulating drift.
   5. The limitation, said before they ask: unbiasedness buys you a variance of about a quarter of a cent squared per payment. If we settle once, I have injected noise to remove bias. If we settle a thousand times, the bias from always rounding down would be 1.59 dollars while the randomised total is within a few cents of correct. So the method pays off under repetition, not on a single trade.

   Where candidates lose it

   Answering just round to 3.14, which is the boring answer and is systematically biased against one party. Also do not overreach into give me an IOU, which dodges the question. The interviewer wants unbiased-in-expectation, and wants to hear you construct the random bit from fair coins.

   Expect next

   • How many fair coin flips do you need to generate that probability?
   • What is the variance of your payment?
   • Where does randomised rounding matter in a real trading system?

   Reported by candidates at Millennium Management (Quantitative Research, Hong Kong, 2025). Source: Wall Street Oasis.

10. 022If X and Y are dependent, does that tell you anything about the relationship between X and Z?ProbabilityIntermediatetechnicalTower Research CapitalProp Trading · New York · 2019

    Say this

    Nothing at all. Dependence is not transitive and it says nothing about a third variable you have not mentioned. X can be dependent on Y and completely independent of Z.

    Then walk it

    1. Trivial counterexample: let X and Y be the same fair coin and let Z be a separate independent coin. X and Y are maximally dependent, X and Z are independent.
    2. The deeper point is that even if X depends on Y and Y depends on Z, X need not depend on Z. Let Y be X plus Z with X and Z independent. Y is dependent on both, and X and Z remain independent of each other.
    3. Correlation is a bit more constrained than dependence because the correlation matrix must be positive semi-definite. If corr(X,Y) is 0.9 and corr(Y,Z) is 0.9, then corr(X,Z) is bounded below by about 0.62. So high correlations do restrict the third pair, but only through that PSD constraint, and dependence in general carries no such bound.
    4. The formula for the bound: rho_xz is at least rho_xy times rho_yz minus the square root of (1 minus rho_xy squared)(1 minus rho_yz squared). Plug in 0.9 and 0.9 and you get 0.81 minus 0.19, which is 0.62.
    5. Why this matters on a desk: people assume that if two assets both correlate with a factor they must correlate with each other. If the loadings are moderate, say 0.5 and 0.5, the bound is minus 0.5, so they can be strongly negatively correlated. That mistake shows up in risk models constantly.

    Where candidates lose it

    Answering yes because it feels like dependence should chain. Give the counterexample in one breath, then earn the extra credit with the correlation bound, because the interviewer's follow-up is almost always the correlation version. And be precise that zero correlation does not mean independence, only the converse holds.

    Expect next

    • Now with correlations. If corr(X,Y) is 0.9 and corr(Y,Z) is 0.9, what do you know about corr(X,Z)?
    • Give me an example of zero correlation with strong dependence.
    • What is conditional independence and why does it matter for factor models?

    Reported by candidates at Tower Research Capital (Prop Trading, New York, 2019). Source: Wall Street Oasis.

Firm tags come from public, anonymous candidate reports on Wall Street Oasis: strong signal, not sworn testimony. Firms are named as the places a question was reported, not as partners of Fin Maverick. Answers are written for this page to show how to think out loud; they are not scripts to recite.