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

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Type

AnyBrainteaserTechnicalCaseMarket viewFit

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. 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.

7. 024I have two children and at least one is a boy. What is the probability both are boys?ProbabilityIntermediatetechnicalProp trading firmsQuant trading

   Say this

   One third, if the information came from a statement about the pair. The sample space is BB, BG, GB, GG, the condition kills GG, and one of the three survivors is BB. But the answer becomes a half if you learned it by meeting one specific child.

   Then walk it

   1. Equally likely and independent births give four ordered outcomes, each one quarter. Conditioning on at least one boy leaves three, of which one is BB. So one third.
   2. Now the version that makes it a real question. Suppose instead I introduce you to my elder child and he is a boy. Now you have conditioned on the elder being a boy, which leaves BB and BG, so the answer is one half.
   3. Same words in English, different conditioning event, different answer. The phrase at least one is a boy is a statement about the pair; this is my son is a statement about a position.
   4. The famous extension is the Tuesday boy: at least one is a boy born on a Tuesday. Now the answer is 13/27, because the extra detail changes how many pairs satisfy the condition and it breaks the symmetry between the two children.
   5. What I would actually say in an interview: the answer is one third under the standard reading, and then immediately name the ambiguity, because the entire point of the question is whether you notice that the conditioning event is underspecified.

   Where candidates lose it

   Answering one half on instinct, or answering one third and stopping. Both are half answers. Give one third with the sample space, then say precisely which conditioning event gives a half, because a quant interviewer is testing whether you can spot an ill-posed conditioning statement, which is a daily hazard in real data work.

   Expect next

   • Now: at least one is a boy born on a Tuesday.
   • What if I tell you my eldest is a boy?
   • How does this relate to survivorship bias in a dataset?
8. 025What is the expected number of fair coin flips to see two heads in a row, and how does it compare to heads followed by tails?Stochastic processesIntermediatetechnicalQuant tradingQuant research

   Say this

   Six flips for HH and four for HT. They differ because HH can destroy its own progress: a tail after a single head sends you back to nothing, while for HT a head after a head keeps you one step from done.

   Then walk it

   1. Set up states for HH. Let A be the expected flips from scratch and B from having one head. A equals 1 plus half A plus half B. B equals 1 plus half times 0 plus half A.
   2. Substitute: B equals 1 plus A/2, so A equals 1 plus A/2 plus (1 plus A/2)/2, which gives A equals 1.5 plus 0.75A, so 0.25A equals 1.5 and A equals 6.
   3. Now HT. Let A be from scratch, B from having a head. A equals 1 plus half A plus half B. But B equals 1 plus half times 0 plus half B, because another head leaves you still in state B rather than resetting. So B equals 2.
   4. Then A equals 1 plus A/2 plus 1, so A/2 equals 2 and A equals 4.
   5. The lesson worth saying out loud: patterns with self-overlap take longer. Both patterns have probability 1/4 per pair of positions, yet the waiting times differ, and that is purely about overlap structure. It generalises: the expected wait for a pattern equals the sum of 2 to the power of the length of each of its self-overlapping prefixes. HH gives 4 plus 2 equals 6, HT gives 4 plus 0 equals 4.

   Where candidates lose it

   Assuming both answers are 4 because each two-flip pattern has probability a quarter. That is the intuition the question is designed to break. Set up the state equations explicitly and pay attention to where a failed attempt lands you, because that is the only difference between the two problems.

   Expect next

   • Now do HHH.
   • In a race between HH and HT, which appears first and with what probability?
   • Derive it with the martingale approach instead.
9. 029There are n distinct types of card in cereal boxes, uniformly at random. How many boxes do you expect to buy to collect all n?Expected valueIntermediatetechnicalQuant tradingQuant research

   Say this

   n times the harmonic number H_n, which is roughly n times (ln n plus 0.577). For 50 cards that is about 225 boxes, so four and a half times the number of cards.

   Then walk it

   1. Decompose by waiting times. Once you hold k distinct cards, the chance the next box is new is (n minus k)/n, so the wait for the next new card is geometric with mean n/(n minus k).
   2. Sum over k from 0 to n minus 1: n times (1/n plus 1/(n-1) up to 1/1), which is n H_n.
   3. Numbers: n equal to 6 gives 14.7 boxes, n equal to 50 gives 224.9, n equal to 365 gives about 2,364. The last one is the expected days to see every birthday.
   4. The tail is where the cost is. Getting the first half of the set takes about 0.69n boxes; the last single card alone takes n boxes in expectation. Most of the pain is the final few.
   5. Variance is worth flagging: it is about n squared times pi squared over 6, so the standard deviation is roughly 1.28n. For n equal to 50 that is 64 boxes, which is enormous relative to the mean of 225. Quoting the mean without the spread would be misleading if you were budgeting for it.

   Where candidates lose it

   Trying to compute it by inclusion-exclusion over the whole collection. The decomposition into independent geometric waits plus linearity of expectation is the intended route and it takes twenty seconds. Also note the harmonic sum by name, because the log growth is the insight the interviewer wants.

   Expect next

   • What is the variance?
   • What if the cards are not equally likely?
   • How many boxes for a 90 percent chance of completing the set?
10. 031Break a stick at two uniformly random points. What is the probability the three pieces form a triangle?ProbabilityIntermediatetechnicalProp trading firmsQuant trading

    Say this

    One quarter. Let the cuts be x and y on a stick of length one. The triangle condition is that no piece exceeds one half, and that region is a quarter of the unit square.

    Then walk it

    1. The triangle inequality for three pieces summing to 1 reduces to a single condition: every piece must be strictly less than 1/2. If any piece is at least a half it is at least as long as the other two together.
    2. Draw the unit square in x and y. Take x less than y without loss of generality, which is the lower triangle of area 1/2. The three pieces are x, y minus x, and 1 minus y.
    3. The three conditions x less than 1/2, y minus x less than 1/2, and 1 minus y less than 1/2 carve out the middle triangle with vertices at (0, 1/2), (1/2, 1/2) and (1/2, 1). That has area 1/8.
    4. Double it for the other ordering and divide by the total area 1, giving 1/4.
    5. Different setup, different answer, and this is the part worth saying: if instead you break the stick once and then break the longer piece, the probability drops to 2 ln 2 minus 1, about 0.386. The phrase break at two random points must mean both cuts on the original stick, and you should confirm that reading before you compute.

    Where candidates lose it

    Not reducing the three triangle inequalities to the single condition no piece over a half. Candidates who try to handle three inequalities geometrically in one pass usually get 1/2 or 1/8. Also state the sampling scheme, because the sequential-break version has a completely different answer and interviewers use the ambiguity deliberately.

    Expect next

    • Now break the stick once and then break the longer piece.
    • What is the expected length of the longest piece?
    • What is the probability the triangle is obtuse?

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.