Quant interview preparation

Questions

100

Traced to a firm

53

Firms

15

Updated

September 2026

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All firmsOld Mission Capital12Tower Research Capital10Jump Trading7Akuna Capital5Citadel4DED.E. Shaw3Jane Street3ACAQR Capital Management2DRW2Millennium Management2Schonfeld2SCSquarepoint Capital2Susquehanna International Group2Belvedere Trading1Optiver1

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All topicsProbability10Coins, cards and games6Expected value8Statistics11Market making15Estimation and mental maths4Stochastic processes4Regression5Machine learning6Time series6Programming10Options and derivatives8Fit and motivation7

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1. 002I hand you a coin that comes up heads one third of the time. How do you generate a fair coin flip from it?Coins, cards and gamesIntermediatephone / first roundDED.E. ShawResearch · New York · 2026Tower Research CapitalProp Trading · New York · 2019

   Say this

   Flip it twice. Call heads-then-tails a fair heads, tails-then-heads a fair tails, and if you get HH or TT throw the pair away and start again. HT and TH both have probability p times one minus p, so they are equally likely whatever p is.

   Then walk it

   1. With p equal to a third: HT is 1/3 times 2/3 which is 2/9, and TH is 2/3 times 1/3 which is also 2/9. Identical, so conditioning on one of the two having happened gives you exactly a half.
   2. That is von Neumann's trick. The point is that it needs no knowledge of p at all, which is what makes it useful. You never have to estimate the bias.
   3. It does need two things: the flips are independent, and p is strictly between zero and one. A coin that is genuinely two-headed breaks it, and so does a coin whose bias drifts flip to flip.
   4. Probability a given pair is useful is 2 times 2/9, which is 4/9. So you discard more than half your pairs at p equal to a third.
   5. The honest limitation: it is unbiased but wasteful. If the bias drifts slowly, you can protect yourself by pairing adjacent flips rather than flips far apart, so the drift cancels locally.

   Where candidates lose it

   Trying to estimate p first and then correct for it. That introduces estimation error and gives you an approximately fair coin, not a fair one. The elegant answer is exactly fair with zero knowledge of p, and the interviewer is looking for that symmetry argument.

   Expect next

   • What is the expected number of flips of the biased coin per fair flip?
   • How would you get a uniform random number on one to three from the same coin?
   • Can you do better than throwing HH and TT away entirely?

   Reported by candidates at D.E. Shaw (Research, New York, 2026); Tower Research Capital (Prop Trading, New York, 2019). Source: Wall Street Oasis.

2. 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.

3. 004Given a biased coin with probability p, how would you generate n independent fair coin tosses, and what is the lower bound on the number of biased tosses you need?Coins, cards and gamesHardsuperdayTower Research CapitalTrading · Princeton · 2018DED.E. ShawResearch · New York · 2026

   Say this

   The bound is information-theoretic. Each biased flip carries H(p) bits of entropy, where H(p) is minus p log2 p minus (1-p) log2 (1-p), and each fair flip you output consumes exactly one bit. So you need at least n divided by H(p) biased flips on average, and no scheme can beat that.

   Then walk it

   1. The argument is conservation of randomness. You cannot manufacture entropy, only repackage it, so expected input entropy must be at least expected output entropy.
   2. At p equal to a half, H is 1 and the bound is n flips, which is obviously right. At p equal to 0.1, H is about 0.47, so you need roughly 2.1 flips per fair bit at best.
   3. Von Neumann pairing achieves 1/(2p(1-p)) flips per bit, which at p equal to 0.1 is about 5.6. Miles off the 2.1 floor.
   4. To get close, you recycle the discarded information. Elias's and Peres's extractors take the sequence of discarded HH/TT outcomes and the positions of the successes, both of which still carry entropy, and feed them back in recursively. Peres's construction converges to the entropy bound as the block length grows.
   5. What I would actually say on a desk: for n fair bits I would use pairing because it is three lines of code and provably correct, and I would only reach for a Peres extractor if the biased source were expensive, which in practice it never is.

   Where candidates lose it

   Answering only with the construction and not the bound, or quoting the bound as n/H(p) without being able to say why entropy is the right currency. Also do not claim von Neumann is optimal. It is unbiased and simple, and it is provably wasteful, and saying so is the difference between having read the trick and understanding it.

   Expect next

   • Where exactly does the discarded entropy live in the von Neumann scheme?
   • Now the reverse problem: simulate a p-coin from fair coins.
   • What if p is unknown but you need to hit the entropy bound?

   Reported by candidates at Tower Research Capital (Trading, Princeton, 2018); D.E. Shaw (Research, New York, 2026). 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.