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. 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.

3. 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.

4. 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.