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. 012Four points are chosen at random on the surface of a sphere. What is the probability that the tetrahedron they form contains the centre?ProbabilityHardsuperdayOld Mission CapitalProp Trading · Chicago · 2018

   Say this

   One eighth. The clean argument: take three random points and their three antipodes, giving eight candidate tetrahedra from the eight sign choices, and exactly one of the eight contains the centre.

   Then walk it

   1. Build the construction. Draw three random points P1, P2, P3 and three random diameters through them. The fourth point is then the head or tail of an independent diameter, and by symmetry each of the eight sign combinations of the three diameters is equally likely as the configuration.
   2. For almost every set of three diameters, exactly one of the eight tetrahedra formed by choosing one endpoint from each diameter, plus the fourth point, contains the centre. So the probability is 1/8.
   3. Warm up with the two-dimensional version first if you are stuck. Three points on a circle contain the centre with probability 1/4, by the same argument with two diameters and four sign choices.
   4. The pattern generalises: n plus 1 points on the surface of an n-sphere contain the centre with probability 1 over 2 to the n. Two to the power n sign choices, one winner.
   5. Say the 2D case out loud before the 3D case. It is the same proof at half the cognitive load, and it shows the interviewer your method rather than a memorised number. The number alone is worthless here because the answer is famous.

   Where candidates lose it

   Attempting to integrate over solid angles. It is a five-line symmetry argument and any attempt at brute-force geometry will run out of time. The other trap is stating one eighth flatly, which reads as recall. Construct the antipodal argument, because with a famous answer the reasoning is all they can grade.

   Expect next

   • Do the circle case in two dimensions.
   • What is the expected volume of that tetrahedron?
   • Three random points on a circle: what is the probability the triangle is acute?

   Reported by candidates at Old Mission Capital (Prop Trading, Chicago, 2018). Source: Wall Street Oasis.

2. 017Five pirates must split a hundred gold coins. The most senior proposes a split, everyone votes, and if at least half agree it passes, otherwise he is thrown overboard and the next most senior proposes. How should the senior pirate split the coins to survive and maximise his take?Coins, cards and gamesHardsuperdayOld Mission CapitalProp Trading · New York · 2014

   Say this

   98 for himself, 0 to the second, 1 to the third, 0 to the fourth, 1 to the fifth. Solve it by backward induction from two pirates, because each pirate's vote depends only on what they would get if the current proposer dies.

   Then walk it

   1. Two pirates left: the senior of the two takes 100, votes for himself, and half of two is one vote, so it passes. Pirate 4 gets 100 and pirate 5 gets 0.
   2. Three left: pirate 3 needs one more vote. Pirate 5 gets nothing in the two-pirate world, so 1 coin buys him. Split is 99, 0, 1.
   3. Four left: pirate 2 needs one more vote out of four. He buys pirate 4, who gets 0 in the three-pirate world, for 1 coin. Split is 99, 0, 1, 0.
   4. Five left: pirate 1 needs two more votes. The pirates who get 0 under pirate 2's plan are 3 and 5, so he buys both for 1 coin each. That gives 98, 0, 1, 0, 1.
   5. The whole method is: work out what each pirate gets if the proposal fails, then pay each cheap vote exactly one coin more than that. The assumptions matter and you should state them: pirates are perfectly rational, prefer gold, prefer to live, and prefer fewer rivals if otherwise indifferent.

   Where candidates lose it

   Trying to reason forwards from five pirates, which is impossible. State that you are doing backward induction and start from the base case of two. The second trap is the tie rule. Half of an even number counts as passing here, and if you assume a strict majority the whole answer shifts, so say your reading of the rule out loud before you solve.

   Expect next

   • What happens with two hundred pirates and a hundred coins?
   • How does the answer change if a tie means the proposer dies?
   • What if pirates value killing above one extra coin?

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

3. 018You have n cars, each with fuel for a thousand miles, and you can transfer petrol between them mid-journey. What is the maximum distance one car can travel, and what happens as n goes to infinity?Expected valueHardsuperdayMillennium ManagementInvestments · London · 2024

   Say this

   A thousand times the harmonic sum: 1000 times (1 plus 1/2 plus 1/3 up to 1/n). It diverges, so as n goes to infinity the distance is unbounded, but only logarithmically, which is the interesting part.

   Then walk it

   1. Think in stages, working from the start. With all n cars moving together, you burn n tanks per 1000 miles of travel, so you can go 1000/n miles before you can consolidate one car's worth of fuel out of the collective and abandon it.
   2. After that leg, n minus 1 cars carry on, each full, and you get 1000/(n-1) more miles before dropping the next. Continue until one car is left, which contributes 1000/1.
   3. Sum the legs: 1000 times the sum of 1/k for k from 1 to n. That is 1000 times H_n.
   4. H_n grows like the natural log of n plus gamma, about 0.577. So with 10 cars you get roughly 2,929 miles, with 100 cars about 5,187, and with a million cars only about 14,392.
   5. That is the point worth making: the distance is unbounded but painfully inefficient. To double your range from 100 cars you need about 100 squared cars. This is the same log scaling as the coupon collector problem, and it is a good example of a divergent series that is useless in practice.

   Where candidates lose it

   Getting the legs backwards, i.e. putting the long leg first. The many-car legs are short because you are burning fuel n times as fast. Also do not answer infinite and stop. The number they want is 1000 H_n with the log growth spelled out, because the divergence-but-barely is the whole insight.

   Expect next

   • How many cars to reach ten thousand miles?
   • What if the cars must all return to the start?
   • Where else does the harmonic series show up in probability?

   Reported by candidates at Millennium Management (Investments, London, 2024). Source: Wall Street Oasis.

4. 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.

5. 020How many zeros are at the end of a thousand factorial?Estimation and mental mathsCorephone / first roundJump TradingTrading · Chicago · 2013

   Say this

   249. A trailing zero needs a factor of ten, which needs a two and a five, and fives are scarcer than twos, so just count the fives: 200 plus 40 plus 8 plus 1 equals 249.

   Then walk it

   1. Trailing zeros equal the number of times 10 divides the number, which is the minimum of the exponent of 2 and the exponent of 5 in the prime factorisation. In a factorial, 5 always binds.
   2. Legendre's formula: sum of floor(1000 divided by 5 to the k). That is floor(1000/5) equals 200, floor(1000/25) equals 40, floor(1000/125) equals 8, floor(1000/625) equals 1, and floor(1000/3125) equals 0.
   3. 200 plus 40 plus 8 plus 1 gives 249.
   4. Why the higher powers: 25 contributes two fives, not one, so it must be counted again. Missing that is the single most common error and it costs you 49.
   5. Quick sanity check on the order of magnitude: roughly 1000/4 is 250, because each multiple of five contributes one and a bit. 249 sits right where it should.

   Where candidates lose it

   Answering 200 by counting only the multiples of five. Multiples of 25, 125 and 625 carry extra factors of five and each must be counted again. Say out loud why five binds rather than two, because that is the part of the reasoning being graded.

   Expect next

   • How many zeros in 100 factorial?
   • How many digits does 1000 factorial have?
   • What is the last non-zero digit of 100 factorial?

   Reported by candidates at Jump Trading (Trading, Chicago, 2013). Source: Wall Street Oasis.

6. 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.

7. 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.
8. 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?
9. 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.
10. 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?

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.