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

Topic

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

Level

AnyCoreIntermediateHard

Type

AnyBrainteaserTechnicalCaseMarket viewFit

1. 001There are two bags of stones and you do not know how many black or white are in each. You draw two stones and both are black. What is the probability the next one is black, and will you bet on it?ProbabilityHardtechnicalCitadelQuantitative Trading · New York · 2025

   Say this

   Higher than a half, and yes I would bet on black. Because I do not know the composition, the two black draws are evidence about the composition itself, so I update towards bags that are black-heavy. The draws are not independent trials, they are a sample that teaches me about the urn.

   Then walk it

   1. Set it up properly: put a prior over the unknown mixture, say the proportion of black p is uniform on zero to one, and the draws are conditionally independent given p.
   2. Then this is Laplace's rule of succession. With k blacks out of n draws the posterior predictive probability of another black is (k+1)/(n+2). Two blacks out of two gives 3/4.
   3. The intuition without algebra: seeing black twice shifts the posterior mass towards high p, and the predictive probability is the posterior mean of p, which is now above a half.
   4. Compare it with the alternative model. If I were told the bag was exactly 50/50 and I was drawing with replacement, the answer would be exactly a half and the history would be irrelevant. The whole question is which model you are in.
   5. On the betting half: I would take anything better than even money on black, and I would size it small because 3/4 is a function of my prior, not of data. Two draws is almost no information. If the prior were concentrated near a half the answer moves back towards a half.

   Where candidates lose it

   Saying one half because the draws are independent. They are only independent conditional on the unknown composition, and the composition is exactly what you are learning. The second failure is giving 3/4 with no mention of the prior, as if it were a fact rather than the output of a uniform prior you chose.

   Expect next

   • What if the prior were Beta(2,2) instead of uniform?
   • Now make me a market on the probability and I will trade it.
   • Same question but sampling without replacement from a bag of 10 stones. Does the answer move?

   Reported by candidates at Citadel (Quantitative Trading, New York, 2025). Source: Wall Street Oasis.

2. 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.

3. 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.

4. 004Given a biased coin with probability p, how would you generate n independent fair coin tosses, and what is the lower bound on the number of biased tosses you need?Coins, cards and gamesHardsuperdayTower Research CapitalTrading · Princeton · 2018DED.E. ShawResearch · New York · 2026

   Say this

   The bound is information-theoretic. Each biased flip carries H(p) bits of entropy, where H(p) is minus p log2 p minus (1-p) log2 (1-p), and each fair flip you output consumes exactly one bit. So you need at least n divided by H(p) biased flips on average, and no scheme can beat that.

   Then walk it

   1. The argument is conservation of randomness. You cannot manufacture entropy, only repackage it, so expected input entropy must be at least expected output entropy.
   2. At p equal to a half, H is 1 and the bound is n flips, which is obviously right. At p equal to 0.1, H is about 0.47, so you need roughly 2.1 flips per fair bit at best.
   3. Von Neumann pairing achieves 1/(2p(1-p)) flips per bit, which at p equal to 0.1 is about 5.6. Miles off the 2.1 floor.
   4. To get close, you recycle the discarded information. Elias's and Peres's extractors take the sequence of discarded HH/TT outcomes and the positions of the successes, both of which still carry entropy, and feed them back in recursively. Peres's construction converges to the entropy bound as the block length grows.
   5. What I would actually say on a desk: for n fair bits I would use pairing because it is three lines of code and provably correct, and I would only reach for a Peres extractor if the biased source were expensive, which in practice it never is.

   Where candidates lose it

   Answering only with the construction and not the bound, or quoting the bound as n/H(p) without being able to say why entropy is the right currency. Also do not claim von Neumann is optimal. It is unbiased and simple, and it is provably wasteful, and saying so is the difference between having read the trick and understanding it.

   Expect next

   • Where exactly does the discarded entropy live in the von Neumann scheme?
   • Now the reverse problem: simulate a p-coin from fair coins.
   • What if p is unknown but you need to hit the entropy bound?

   Reported by candidates at Tower Research Capital (Trading, Princeton, 2018); D.E. Shaw (Research, New York, 2026). Source: Wall Street Oasis.

5. 005I shuffle a deck and turn cards face up one at a time. At any point you may say stop, and you win if the next card is red. What is your optimal strategy and what is your probability of winning?Coins, cards and gamesHardtechnicalJump TradingResearch · Chicago · 2018

   Say this

   Every strategy wins with probability exactly one half, so there is no optimal strategy. Stopping before the first card is as good as any clever rule based on the count.

   Then walk it

   1. The quick proof is a symmetry argument. Fix any stopping rule and imagine swapping the colour of every card in the deck. The rule's decisions are determined by cards already seen, and the swap turns every win into a loss and every loss into a win, so wins and losses are equally likely.
   2. The cleaner proof is a martingale. Let X be the fraction of red cards remaining. Before you see a card, the expected fraction of reds remaining after you see it is exactly the current fraction, because the card you turn is a uniform draw from what is left. So X is a martingale.
   3. Your win probability when you stop is X at the stopping time. Optional stopping says the expected value of a bounded martingale at any stopping time equals its starting value, which is 26/52, or a half.
   4. This is the whole lesson of the problem. Your information at the moment you stop is already priced into the state. There is no edge in a fair game no matter how you time it.
   5. One caveat that makes it a real problem rather than a trick: if you are forced to keep going to the last card, you still win a half, because the last card is red with probability a half. But the variance of the outcomes differs across strategies even though the mean does not, and if you had a utility function that is not linear you would care.

   Where candidates lose it

   Inventing a rule like wait until more blacks than reds have come out, and claiming it beats a half. That intuition feels right and it is wrong, because the situations in which the rule fires are exactly the situations where the deck was red-heavy from the start. Name the martingale and use optional stopping, or at minimum give the colour-swap symmetry argument.

   Expect next

   • Prove it with optional stopping, precisely.
   • Does the answer change if you can also bet on black?
   • Which strategy has the lowest variance of outcome?

   Reported by candidates at Jump Trading (Research, Chicago, 2018). Source: Wall Street Oasis.

6. 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.

7. 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.

8. 008What is the expected value of a roll of a fair six-sided die?Expected valueCorephone / first roundOld Mission CapitalTrading · Chicago · 2020

   Say this

   Three and a half. Each face has probability one sixth, and one through six sums to 21, so 21 over 6 is 3.5. Or faster: for a uniform run of integers the mean is the midpoint, which is (1 plus 6) over 2.

   Then walk it

   1. The sum one to n is n(n+1)/2, so the mean of a uniform die with n faces is (n+1)/2. For six faces that is 3.5.
   2. Variance is worth having ready too, because it is the immediate follow-up. E of X squared is (1+4+9+16+25+36)/6, which is 91/6, about 15.17. Subtract 3.5 squared, 12.25, and you get 35/12, about 2.917. Standard deviation is about 1.71.
   3. Also useful to know the general formula: the variance of a uniform die with n faces is (n squared minus 1) over 12. For n equal to 6 that is 35/12, matching.
   4. Two dice: expected sum is 7 by linearity, and variance is 35/6 because they are independent, so standard deviation is about 2.42.
   5. The reason a trading desk opens with this is speed and composure, not difficulty. Answer in under two seconds and have the variance ready before they ask, because the real question is the next one.

   Where candidates lose it

   Hesitating. This is a warm-up and any pause reads badly. The second trap is being caught flat-footed on variance, which follows more than half the time. Know 35/12 and know that the standard deviation of a die is about 1.71.

   Expect next

   • What is the variance?
   • What about the sum of two dice?
   • What is the expected value of the maximum of two dice?

   Reported by candidates at Old Mission Capital (Trading, Chicago, 2020). Source: Wall Street Oasis.

9. 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.

10. 010Same die game, but now you earn one dollar per dot and each re-roll costs you one dollar, with unlimited re-rolls. What is the value of the game and when do you stop?Expected valueHardsuperdayOld Mission CapitalFinance · New York · 2018

    Say this

    The game is worth 4 dollars if the first roll is free, and you stop on a 3 or better. The continuation value of choosing to roll again is exactly 3, so the acceptance threshold is 3, and the value of a free first roll is the average of 3, 3, 3, 4, 5 and 6, which is 4.

    Then walk it

    1. Set up the recursion. If you decide to roll, you pay 1, then with the threshold t you accept faces above or equal to t and otherwise roll again. So V equals minus 1 plus the average over faces of the max of the face value and V.
    2. Guess the threshold is 4, meaning V sits in the interval 3 to 4. Then faces 4, 5, 6 are accepted for 15 total, and faces 1, 2, 3 all continue at V each.
    3. V equals minus 1 plus (15 plus 3V)/6. Multiply through: 6V equals minus 6 plus 15 plus 3V, so 3V equals 9, V equals 3.
    4. Check consistency: V equal to 3 means you should accept anything at or above 3, not 4. Re-solve with threshold 3: accepted faces 3,4,5,6 sum to 18, continuing faces 1,2 give 2V. V equals minus 1 plus (18 plus 2V)/6 gives 6V equals 12 plus 2V, so V equals 3. Consistent, since 3 is in the interval 2 to 3 boundary case. So the value of choosing to roll is 3 and you stop on 3 or better.
    5. Check consistency, which is the step that matters: V equal to 3 means you accept anything at or above 3, so re-solve with threshold 3. Accepted faces 3, 4, 5, 6 sum to 18 and continuing faces 1 and 2 give 2V, so V equals minus 1 plus (18 plus 2V)/6, which gives 4V equals 12 and V equals 3. Now the assumed threshold and the solved value agree, so 3 is the answer. With a free first roll the game is worth the average of max(face, 3), which is 24 over 6, equals 4.

    Where candidates lose it

    Solving the fixed point once and not checking that the threshold you assumed is consistent with the value you found. That verification step is the whole exercise in an optimal-stopping problem, and skipping it is how candidates report a threshold of 4 with a value of 3 and never notice the contradiction.

    Expect next

    • What if the re-roll cost were 2 dollars instead?
    • At what cost per re-roll does the game become worthless?
    • How does this map to pricing an American option?

    Reported by candidates at Old Mission Capital (Finance, New York, 2018). 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.