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. 021If n(n+1)/2 is the sum of the integers from one to n, what is the formula for the sum of the squares, and can you derive it?Estimation and mental mathsCoretechnicalSCSquarepoint CapitalTrading · London · 2025

   Say this

   n(n+1)(2n+1)/6. The fastest derivation is telescoping: expand (k+1) cubed minus k cubed as 3k squared plus 3k plus 1, sum both sides from 1 to n, and solve for the sum of squares.

   Then walk it

   1. Left side telescopes to (n+1) cubed minus 1.
   2. Right side is 3S2 plus 3S1 plus n, where S2 is what you want and S1 is the known n(n+1)/2.
   3. So 3S2 equals (n+1) cubed minus 1 minus 3n(n+1)/2 minus n. Grind it out and you get S2 equals n(n+1)(2n+1)/6.
   4. Check at n equal to 3: 1 plus 4 plus 9 equals 14, and 3 times 4 times 7 over 6 equals 14. Always test a small case out loud, it costs three seconds and catches sign errors.
   5. The reason a desk asks this: it is the derivation that matters, not the formula. The same telescoping trick gives the sum of cubes, which is n squared (n+1) squared over 4, and it is the discrete analogue of integration by parts. And the practical use is immediate, because the variance of a uniform die and the variance of a linear time trend in a regression both fall straight out of this sum.

   Where candidates lose it

   Reciting the formula with no derivation. The interviewer already knows the formula, so the answer is worth nothing on its own. Show the telescoping and verify on n equals 3. Fumbling the algebra after setting it up correctly is forgivable; having no method is not.

   Expect next

   • Now the sum of cubes.
   • Use it to get the variance of a fair n-sided die.
   • What is the sum of 1 over k squared as n goes to infinity?

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

2. 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.

3. 023Monty Hall. Three doors, one car, you pick one, I open a door with a goat, do you switch?ProbabilityCorephone / first roundProp trading firmsQuant trading

   Say this

   Switch. Your original door wins one third of the time, so the other door wins two thirds. The host's choice is not random, and that is where the information comes from.

   Then walk it

   1. Condition on your first pick. One third of the time you picked the car, and switching loses. Two thirds of the time you picked a goat, the host is forced to reveal the only other goat, and switching wins.
   2. So switching wins two thirds. The Bayes calculation agrees: the likelihood of the host opening door 3 is 1/2 if the car is behind your door 1, and 1 if the car is behind door 2, which is what tilts the posterior two to one.
   3. The intuition people find convincing: extend it to a hundred doors. You pick one, the host opens 98 goats, and switching wins 99 times out of 100. The host did all the work of avoiding the car.
   4. The critical assumption, and this is what a quant interview is really checking: the host knows where the car is and always opens a goat. If the host opens a door at random and happens to show a goat, the posterior is fifty-fifty and switching gains nothing.
   5. So the honest answer is: switch, and the reason it works is that the host's constraint leaks information. Change the host's rule and the answer changes.

   Where candidates lose it

   Getting the right answer for the wrong reason, or failing to state the host's rule. Everyone knows the answer is switch, so the only thing being graded is whether you can name the assumption that makes it true. Say explicitly that the host knows and is forced to reveal a goat.

   Expect next

   • What if the host does not know where the car is?
   • What if the host only offers the switch when you picked the car?
   • Do it with a hundred doors.
4. 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?
5. 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.
6. 026You start with fifty dollars and bet a dollar on a fair coin each time. What is the probability you reach a hundred before going broke, and how does it change if the coin is slightly against you?Stochastic processesIntermediatetechnicalQuant tradingQuant research

   Say this

   In a fair game it is exactly one half, because your wealth is a martingale and the stopping value must average back to fifty. Tilt the odds slightly against you and the probability collapses, not linearly but exponentially in the number of steps.

   Then walk it

   1. Fair case: wealth is a martingale, so by optional stopping, 50 equals 100 times p plus 0 times (1 minus p), giving p equal to 0.5. In general starting at a with an upper barrier b, the probability is a over b.
   2. Biased case: with win probability q the hitting probability is (1 minus r to the a) over (1 minus r to the b) where r is (1-q)/q.
   3. Put a number on it. At q equal to 0.49, r is about 1.0408. With a equal to 50 and b equal to 100, the probability of reaching 100 drops to roughly 12 percent. A one percent edge against you turns a coin flip into a 1-in-8 shot.
   4. That sensitivity is the entire lesson. Expected value per bet is minus two cents, which sounds trivial, but over the hundreds of bets you need to walk the barrier it compounds into near certainty of ruin.
   5. And the practical version on a desk: expected time to absorption in the fair case is a times (b minus a), so 50 times 50 equals 2,500 bets. Casinos and market makers both live on this asymmetry. Small edge, high repetition, deep pockets.

   Where candidates lose it

   Giving a over b and stopping. The interesting content is how brutally the biased case differs, and candidates who cannot state the r to the power formula usually also guess that a one percent edge changes the answer by about one percent. It changes it from 50 percent to 12 percent. Put a number on it.

   Expect next

   • What is the expected number of bets until you stop?
   • What happens if you bet your whole stack each time instead?
   • How does this relate to a trader's drawdown limit?
7. 027What is a martingale, and how would you use optional stopping to solve a problem?Stochastic processesHardtechnicalQuant researchQuant trading

   Say this

   A martingale is a process whose expected next value, given everything you know now, equals its current value. Optional stopping says that for a suitably bounded stopping time, the expected value at the stopping time equals the starting value, which is what turns a hard path-dependent question into one line of algebra.

   Then walk it

   1. Formally: E of X_{n+1} given the filtration F_n equals X_n. No drift, conditional on history. It is not the same as independence, and increments need not be identically distributed.
   2. Optional stopping needs a condition, and you should name one: bounded stopping time, or bounded increments plus finite expected stopping time, or uniform integrability. Without it the theorem fails, and the classic failure is the doubling strategy, where a stopping time that is finite with probability one still produces E of X_tau equal to 1 rather than 0.
   3. How I use it: find a quantity that is conserved in expectation, then evaluate it at the stopping time. Gambler's ruin falls out immediately from wealth being a martingale.
   4. A second example, expected time in a symmetric random walk: W_n squared minus n is a martingale, so E of tau equals E of W_tau squared. With barriers at 0 and b starting from a, that gives E of tau equal to a(b minus a) in a line.
   5. And the reason it matters beyond puzzles: risk-neutral pricing is exactly the statement that the discounted price is a martingale under the pricing measure. Delta hedging is the construction of that martingale. If you can say that connection, the puzzle answer becomes a conversation about derivatives.

   Where candidates lose it

   Defining a martingale as a fair game and stopping there, or applying optional stopping without checking the integrability condition. Interviewers at the good shops will hand you the doubling strategy specifically to see whether you know why the theorem does not apply. Name the condition before you use the theorem.

   Expect next

   • Why does optional stopping fail for the doubling strategy?
   • Is the square of a martingale a martingale?
   • Connect this to risk-neutral pricing.
8. 028An ant walks randomly along the edges of a cube starting at one corner. What is the expected number of steps to reach the opposite corner?Stochastic processesHardsuperdayQuant tradingQuant research

   Say this

   Ten steps. Collapse the eight vertices into four states by distance from the start, then solve three linear equations. The symmetry reduction is the whole trick.

   Then walk it

   1. By symmetry, all that matters is your graph distance from the target: state 3 is the start, then 2, then 1, then 0 which is the target. Each vertex has three neighbours.
   2. From state 3 all three neighbours are at distance 2, so E3 equals 1 plus E2.
   3. From state 2, one neighbour is at distance 3 and two are at distance 1. So E2 equals 1 plus (1/3)E3 plus (2/3)E1.
   4. From state 1, one neighbour is the target and two are at distance 2. So E1 equals 1 plus (2/3)E2.
   5. Solve: substitute E3 equals 1 plus E2 into the second equation to get E2 equals 1 plus (1 plus E2)/3 plus (2/3)(1 plus (2/3)E2). That yields E2 equal to 9, so E1 equals 7 and E3 equals 10. Sanity check with the general theorem: for a random walk on a regular graph the expected return time to a vertex is the number of vertices, 8, which is the right order of magnitude for a 10-step commute across the diagonal.

   Where candidates lose it

   Trying to track all eight vertices individually and drowning in eight equations. Say the word symmetry, lump the states by distance, and you have three unknowns. The other error is miscounting neighbours in state 2, where it is one back and two forward, not two back and one forward.

   Expect next

   • What is the expected time to return to the starting corner?
   • Do it for a tetrahedron.
   • What if the walk is on a hypercube in n dimensions?
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. 030You draw n independent uniforms on zero to one. What are the expected values of the maximum and the minimum, and of the kth smallest?StatisticsIntermediatetechnicalQuant researchQuant trading

    Say this

    The maximum has mean n/(n+1), the minimum 1/(n+1), and the kth smallest k/(n+1). The n points cut the interval into n plus 1 gaps that are exchangeable, so each gap averages 1/(n+1).

    Then walk it

    1. Derive the max directly: P(max at most x) is x to the n, so the density is n x to the n minus 1, and the integral of x times that from 0 to 1 is n/(n+1).
    2. The gap argument is faster and generalises. The n order statistics plus the two endpoints create n plus 1 spacings, which are exchangeable with total length 1, so each has mean 1/(n+1). The kth order statistic is the sum of the first k spacings, hence k/(n+1).
    3. The kth order statistic is Beta(k, n minus k plus 1), which gives you the variance too: k(n-k+1) over ((n+1) squared (n+2)).
    4. Numbers: with 10 draws the max averages 0.909 and the min 0.091. With 100 draws the max averages 0.990. The max creeps to the boundary at rate 1/n, which is why extreme-value estimates converge slowly.
    5. Why a quant desk cares: the max of n draws is your model for the best of n signals, the worst drawdown of n periods, and the winning quote in an auction with n bidders. And it explains selection bias, because the best of a hundred backtests looks good even when none of them has any edge.

    Where candidates lose it

    Answering only for the max with a calculus derivation and then being stuck on the general kth. Learn the spacings argument, it gives all of them at once. And be ready to connect it to selection bias, because the practical follow-up is almost always about why the best of many strategies overstates its own quality.

    Expect next

    • What is the variance of the maximum?
    • What is the expected range, max minus min?
    • How does this explain the selection bias in picking the best of a hundred backtests?

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.