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

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Type

AnyBrainteaserTechnicalCaseMarket viewFit

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. 022If X and Y are dependent, does that tell you anything about the relationship between X and Z?ProbabilityIntermediatetechnicalTower Research CapitalProp Trading · New York · 2019

   Say this

   Nothing at all. Dependence is not transitive and it says nothing about a third variable you have not mentioned. X can be dependent on Y and completely independent of Z.

   Then walk it

   1. Trivial counterexample: let X and Y be the same fair coin and let Z be a separate independent coin. X and Y are maximally dependent, X and Z are independent.
   2. The deeper point is that even if X depends on Y and Y depends on Z, X need not depend on Z. Let Y be X plus Z with X and Z independent. Y is dependent on both, and X and Z remain independent of each other.
   3. Correlation is a bit more constrained than dependence because the correlation matrix must be positive semi-definite. If corr(X,Y) is 0.9 and corr(Y,Z) is 0.9, then corr(X,Z) is bounded below by about 0.62. So high correlations do restrict the third pair, but only through that PSD constraint, and dependence in general carries no such bound.
   4. The formula for the bound: rho_xz is at least rho_xy times rho_yz minus the square root of (1 minus rho_xy squared)(1 minus rho_yz squared). Plug in 0.9 and 0.9 and you get 0.81 minus 0.19, which is 0.62.
   5. Why this matters on a desk: people assume that if two assets both correlate with a factor they must correlate with each other. If the loadings are moderate, say 0.5 and 0.5, the bound is minus 0.5, so they can be strongly negatively correlated. That mistake shows up in risk models constantly.

   Where candidates lose it

   Answering yes because it feels like dependence should chain. Give the counterexample in one breath, then earn the extra credit with the correlation bound, because the interviewer's follow-up is almost always the correlation version. And be precise that zero correlation does not mean independence, only the converse holds.

   Expect next

   • Now with correlations. If corr(X,Y) is 0.9 and corr(Y,Z) is 0.9, what do you know about corr(X,Z)?
   • Give me an example of zero correlation with strong dependence.
   • What is conditional independence and why does it matter for factor models?

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

4. 039What are the differences between Lasso and Ridge regression?Machine learningIntermediatetechnicalTower Research CapitalTrading · Princeton · 2018

   Say this

   Both add a penalty on coefficient size to trade variance for bias. Ridge penalises the sum of squares and shrinks everything smoothly towards zero without eliminating anything. Lasso penalises the sum of absolute values and sets coefficients exactly to zero, so it selects features.

   Then walk it

   1. The geometry explains it. The L1 constraint region is a diamond with corners on the axes, so the solution tends to land on a corner, which means a zero coefficient. The L2 region is a ball with no corners, so solutions are interior and nothing is exactly zero.
   2. Ridge has a closed form, beta equals (X'X plus lambda I) inverse X'y, which is why it also fixes a singular X'X. Lasso has no closed form and needs coordinate descent or LARS.
   3. Correlated predictors behave very differently. Ridge splits the weight across a group of correlated features, which is stable. Lasso arbitrarily picks one and zeroes the rest, which is unstable across samples. Elastic net, which mixes both penalties, exists precisely to get sparsity without that instability.
   4. In a Bayesian reading, ridge is a Gaussian prior on the coefficients and lasso is a Laplace prior. The Laplace prior's spike at zero is what produces exact zeros.
   5. What I would say about which to use on financial data: predictors are usually highly correlated and the signal-to-noise ratio is awful, so ridge or elastic net typically beats pure lasso out of sample. Lasso is attractive when you need an interpretable short list of factors, but do not confuse the features it selected with the features that matter, because a slightly different sample gives you a different list.

   Where candidates lose it

   Stopping at L1 gives sparsity, L2 does not. Everyone says that. The differentiators are the diamond-versus-ball geometry, the behaviour under correlated predictors, and the Bayesian priors. Also always say that both require standardised features, because the penalty is scale-dependent and forgetting to standardise silently ruins the fit.

   Expect next

   • What is elastic net for?
   • How do you choose lambda?
   • Why do you have to standardise your features first?

   Reported by candidates at Tower Research Capital (Trading, Princeton, 2018). 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.