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. 011A random variable is uniform on the interval zero to ten. What are its expected value and variance?StatisticsCorephone / first roundOld Mission CapitalFinance · New York · 2018

   Say this

   Mean 5, variance 100 over 12, which is 8.33, so standard deviation about 2.89. For a uniform on a to b the mean is the midpoint and the variance is (b minus a) squared over 12.

   Then walk it

   1. Mean by symmetry: the midpoint of 0 and 10 is 5. No integration needed.
   2. Variance from the formula (b-a) squared over 12: 100 over 12 equals 8.33, standard deviation 2.887.
   3. If you want to derive it, E of X squared is the integral of x squared over 10 from 0 to 10, which is 1000/30 equals 33.33. Subtract 25 and you get 8.33. Good to be able to do it either way.
   4. The 1/12 is worth carrying in your head because it recurs: a fair n-sided die has variance (n squared minus 1)/12, and the rounding error of a value rounded to the nearest tick has variance tick squared over 12. That last one comes up in real microstructure work.
   5. Practical note: the uniform has thin support and no tails, so it is a bad default for anything financial. The moment somebody hands you a uniform in a trading context, ask what it is meant to represent.

   Where candidates lose it

   Reaching for integration under time pressure and fumbling the arithmetic. Know the (b-a) squared over 12 form cold. Also do not quote variance when they asked for standard deviation or the other way round, and say which one you are giving.

   Expect next

   • What is the expected value of the maximum of two independent draws?
   • What is the distribution of the sum of two independent uniforms?
   • What is the variance of the rounding error when you round to the nearest penny?

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

2. 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.

3. 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.

4. 014Here is a game. What is the expected value of winning under three different strategies, and which one would you choose?Expected valueHardsuperdayJane StreetTrading · London · 2025OptiverGeneralist · Chicago · 2025

   Say this

   Set up the state and the decision rule before you compute anything, price each strategy with a clean conditional expectation, then choose on expected value first and on variance and ruin risk second. Say the comparison out loud as you go so the interviewer can follow your bookkeeping.

   Then walk it

   1. Step one, define the state precisely: what you know when you decide, and what the payoff function is. Most errors in these problems are specification errors, not arithmetic.
   2. Step two, price each strategy by conditioning on the first move. E of payoff equals the sum over first outcomes of probability times conditional value. If the game is repeated or recursive, write V in terms of V and solve the fixed point.
   3. Step three, do the arithmetic in fractions, not decimals. Fractions let the interviewer audit you and they do not accumulate error.
   4. Step four, choose. If one strategy dominates on expected value, say so and stop. If they are close, break the tie on the second moment: I would take the lower-variance strategy at the same expected value, and I would pay a small amount of expected value to avoid a path that can lose more than my stake.
   5. Then state the assumption you are relying on, unprompted: whether you may stop adaptively, whether the game is repeated, and whether the payoff is linear in money. Those three change the answer more than the arithmetic does.

   Where candidates lose it

   Diving into arithmetic before defining the state, and then losing track of which branch you are on. The other failure is picking the highest expected value without a word about variance. A trading floor cares about the distribution of outcomes, so say which strategy you would actually run with real money and why.

   Expect next

   • Now suppose you can play the game a hundred times. Does your choice change?
   • What if the payoff were doubled but the probability halved?
   • What is the variance of your preferred strategy?

   Reported by candidates at Jane Street (Trading, London, 2025); Optiver (Generalist, Chicago, 2025). Source: Wall Street Oasis.

5. 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.

6. 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.

7. 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.

8. 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.

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. 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.

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.