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. 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.

2. 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.

3. 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.

4. 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.

5. 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.

6. 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.

7. 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.

8. 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?

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.