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. 031Break a stick at two uniformly random points. What is the probability the three pieces form a triangle?ProbabilityIntermediatetechnicalProp trading firmsQuant trading

   Say this

   One quarter. Let the cuts be x and y on a stick of length one. The triangle condition is that no piece exceeds one half, and that region is a quarter of the unit square.

   Then walk it

   1. The triangle inequality for three pieces summing to 1 reduces to a single condition: every piece must be strictly less than 1/2. If any piece is at least a half it is at least as long as the other two together.
   2. Draw the unit square in x and y. Take x less than y without loss of generality, which is the lower triangle of area 1/2. The three pieces are x, y minus x, and 1 minus y.
   3. The three conditions x less than 1/2, y minus x less than 1/2, and 1 minus y less than 1/2 carve out the middle triangle with vertices at (0, 1/2), (1/2, 1/2) and (1/2, 1). That has area 1/8.
   4. Double it for the other ordering and divide by the total area 1, giving 1/4.
   5. Different setup, different answer, and this is the part worth saying: if instead you break the stick once and then break the longer piece, the probability drops to 2 ln 2 minus 1, about 0.386. The phrase break at two random points must mean both cuts on the original stick, and you should confirm that reading before you compute.

   Where candidates lose it

   Not reducing the three triangle inequalities to the single condition no piece over a half. Candidates who try to handle three inequalities geometrically in one pass usually get 1/2 or 1/8. Also state the sampling scheme, because the sequential-break version has a completely different answer and interviewers use the ambiguity deliberately.

   Expect next

   • Now break the stick once and then break the longer piece.
   • What is the expected length of the longest piece?
   • What is the probability the triangle is obtuse?
2. 032How many people do you need in a room for a better than even chance that two share a birthday, and why is the answer so small?ProbabilityCorephone / first roundProp trading firmsQuant trading

   Say this

   Twenty-three. The reason it feels small is that you are counting pairs, not people. Twenty-three people generate 253 pairs, and each pair matches with probability 1/365, so you expect about 0.69 matches.

   Then walk it

   1. Compute the complement: the probability all birthdays differ is 365/365 times 364/365 times down to 343/365. At 23 people that product is about 0.493, so the match probability is about 0.507.
   2. The back-of-envelope version: the probability of no match is approximately exp of minus n(n-1)/(2 times 365). Set that to 0.5, so n squared over 730 equals ln 2, giving n about 22.5. Round up to 23.
   3. The pair-counting intuition is the answer to why. n choose 2 grows quadratically, so the number of chances grows fast while your intuition tracks n linearly.
   4. Contrast with the question people confuse it with: for someone to share your specific birthday you need about 253 people, because now you have only n pairs, not n squared over 2.
   5. Where this bites in real work: hash collisions and the birthday attack follow the same square-root law, and so does the chance that two of your supposedly independent signals are accidentally the same trade. You need about the square root of the space to get a collision, which is far fewer than people expect.

   Where candidates lose it

   Confusing it with the probability that someone shares your birthday, which needs 253 people. Also do not just recite 23. The gradeable part is the pair-counting argument and the exp of minus n squared over 730 approximation, which lets you answer variants like how many for a 99 percent chance without a calculator.

   Expect next

   • How many for a 99 percent chance?
   • How many to share a birthday with you specifically?
   • What is the connection to hash collisions?
3. 033A test for a disease is 99 percent accurate and the disease affects one in ten thousand people. You test positive. What is the probability you have it?ProbabilityIntermediatephone / first roundQuant researchQuant trading

   Say this

   About one percent. Out of a million people, 100 are sick and about 99 of them test positive, while 999,900 are healthy and about 9,999 of them test positive falsely. So 99 out of roughly 10,098 positives are real, which is 0.98 percent.

   Then walk it

   1. Do it in counts, not Bayes notation. A population of a million makes the arithmetic trivial and the answer intuitive.
   2. The formula check: P(sick given positive) equals 0.0001 times 0.99 divided by (0.0001 times 0.99 plus 0.9999 times 0.01), which is 0.000099 over 0.010098, about 0.0098.
   3. The driver is base rate. False positives from the huge healthy population swamp the true positives from the tiny sick population. At a prevalence of 1 in 10,000 and a 1 percent false positive rate, you get a hundred false positives for every true one before adjusting for sensitivity.
   4. So the useful quantity is the likelihood ratio: 0.99 over 0.01 equals 99. It multiplies your prior odds of 1 in 9,999 into posterior odds of about 99 in 9,999, which is 1 percent. Thinking in odds and likelihood ratios is far faster than the fraction form.
   5. Where this shows up in trading: any rare-event detector, from fraud flags to regime-change signals to strategy alerts. A signal with 99 percent accuracy on a one-in-ten-thousand event fires 99 false alarms per real one, which is why alert systems get ignored.

   Where candidates lose it

   Answering 99 percent. The second trap is being sloppy about what 99 percent accurate means, since sensitivity and specificity need not be equal. State your reading, do it in counts per million, and name base rate neglect as the reason the intuitive answer is wrong by two orders of magnitude.

   Expect next

   • What prevalence would make the positive predictive value fifty percent?
   • You test positive twice. Now what?
   • How does this apply to a trading signal that fires rarely?
4. 034How many Starbucks are there in New York City?Estimation and mental mathsCorephone / first roundTower Research CapitalProp Trading · New York · 2019

   Say this

   I would say roughly 250 to 350, and I would build it from demand rather than from geography. Eight million people, maybe one in ten buys a Starbucks on a given day, a store serves around a thousand cups a day, so 800,000 over 1,000 is about 800 store-days of demand, which I would then cut for the fact that Manhattan stores are much busier than a thousand cups.

   Then walk it

   1. Build two independent estimates and reconcile them. That is the actual skill being tested, not the number.
   2. Demand side: 8 million residents plus commuters and tourists, call it 9 million daytime people. Ten percent buy coffee from Starbucks on a given day gives 900,000 cups. A busy Manhattan store does 1,500 to 3,000 cups a day, so 900,000 over 2,500 is about 360 stores.
   3. Supply side: Manhattan has roughly 200 avenue-blocks of dense commercial frontage and you see a Starbucks every few blocks in midtown, which suggests 150 to 200 in Manhattan alone, plus maybe the same again across the four outer boroughs. That lands around 300.
   4. Both routes land in the same band, 250 to 400, which is the useful output. I would quote 300 as my point estimate with a range.
   5. Then state your uncertainty honestly and where it sits: the biggest lever is cups per store, which I could be wrong on by a factor of two. The population number I am confident in to ten percent. Naming which assumption dominates the error is what separates an estimate from a guess.

   Where candidates lose it

   Producing one chain of assumptions and asserting the answer with false precision. Build two independent routes, reconcile them, give a range, and say which assumption carries the error. Also do not freeze because you do not know the answer. Nobody knows it, and the interviewer is grading the structure and your composure, not the number.

   Expect next

   • Now make me a market on it and I will trade you.
   • How many coffee shops in total?
   • How would you check your estimate if you had the internet for thirty seconds?

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

5. 035How many golf balls fit in the Empire State Building?Estimation and mental mathsCoretechnicalTower Research CapitalAssistant Trader · New York · 2013

   Say this

   Order of a hundred billion. The building is roughly a hundred million cubic feet, a golf ball plus its packing waste takes about 0.0015 cubic feet, so 100 million over 0.0015 is about 70 billion. I would quote 50 to 100 billion.

   Then walk it

   1. Volume of the building: footprint about 200 by 400 feet, so 80,000 square feet, times 1,250 feet of height. That is 100 million cubic feet. Taper the tower and subtract structure and you might call it 80 million usable.
   2. Volume of a golf ball: diameter 1.68 inches, so radius 0.84 inches. Four thirds pi r cubed is about 2.5 cubic inches. There are 1,728 cubic inches in a cubic foot, so a ball is 0.00145 cubic feet.
   3. Packing efficiency: random close packing of spheres is about 64 percent, so effective volume per ball is 0.00145 over 0.64, about 0.00226 cubic feet.
   4. 80 million divided by 0.00226 gives about 35 billion. Using the full 100 million cubic feet gives 44 billion. So my range is tens of billions, call it 40 billion, and I would say 20 to 100 billion to be honest about the error bars.
   5. Say the two things you are least sure about: the usable fraction of the volume, and whether the question means the empty shell or the building with floors, furniture and lift shafts. Those swing the answer by a factor of two, and the packing fraction only matters at the 30 percent level.

   Where candidates lose it

   Forgetting the 1,728 cubic inches per cubic foot conversion, which throws you off by three orders of magnitude, or ignoring packing efficiency entirely. Also decide out loud whether you are filling the empty shell or the furnished building. And always sanity check the magnitude: if your answer is in millions or trillions, something went wrong by a factor of a thousand.

   Expect next

   • What is the packing efficiency of spheres and why?
   • How much would they weigh?
   • Now estimate the market value of that many golf balls.

   Reported by candidates at Tower Research Capital (Assistant Trader, New York, 2013). Source: Wall Street Oasis.

6. 036The sample variance with the n minus one correction is unbiased. Is its square root an unbiased estimator of the standard deviation?StatisticsHardtechnicalSCSquarepoint CapitalQuantitative Research · London · 2026

   Say this

   No. The square root is concave, so by Jensen's inequality the expected square root is strictly less than the square root of the expected value. The sample standard deviation is biased downwards, always, for any distribution with positive variance.

   Then walk it

   1. Jensen: for a strictly concave g, E of g(X) is less than g of E of X unless X is degenerate. With g the square root and X the unbiased sample variance, E of s is less than sigma.
   2. Size the bias for normal data. E of s equals c4(n) times sigma, where c4 is a known constant involving gamma functions. At n equal to 2, c4 is about 0.798, so you understate sigma by 20 percent. At n equal to 10 it is 0.9727, a 2.7 percent understatement. At n equal to 30 it is 0.9914.
   3. So the bias is order 1/(4n) and it vanishes as n grows. It is a real problem for short samples and irrelevant for long ones.
   4. Unbiasedness is also not preserved under any nonlinear transform, which is the general lesson. The unbiased estimator of sigma squared does not give you an unbiased estimator of sigma, or of 1/sigma, or of log sigma.
   5. Where this bites on a desk: annualised volatility estimated from a few weeks of data, and any Sharpe ratio, since the Sharpe divides by s. Understating s inflates the Sharpe, so short-sample Sharpes are biased upwards. That is worth saying because it connects a textbook Jensen question to a live problem in strategy evaluation.

   Where candidates lose it

   Saying yes because the variance estimator is unbiased. Unbiasedness does not survive a nonlinear function. Name Jensen explicitly, give the direction of the bias, and quantify it with c4 for at least one small n. The follow-up about Sharpe ratios is where the real conversation is, so get there yourself.

   Expect next

   • How would you correct it?
   • What does that imply for a Sharpe ratio estimated on a short sample?
   • Is the sample correlation coefficient unbiased?

   Reported by candidates at Squarepoint Capital (Quantitative Research, London, 2026). Source: Wall Street Oasis.

7. 037You stand on a road and watch cars drive past. How would you estimate the parameter of the underlying distribution?StatisticsHardtechnicalJump TradingQuantitative Research · Chicago · 2018

   Say this

   First I would state the model: arrivals as a Poisson process with rate lambda, so inter-arrival times are exponential with mean 1 over lambda. Then the maximum likelihood estimate of lambda is just the count divided by the observation time, and its standard error is lambda over the square root of the count.

   Then walk it

   1. Model choice first, and justify it: independent arrivals at a constant rate with no memory gives a Poisson process. That is reasonable on a quiet road, and clearly wrong near a traffic light where cars arrive in platoons.
   2. MLE: for n arrivals in time T, lambda hat is n over T. It is unbiased, and the variance is lambda over T, so the relative standard error is 1 over the square root of n. Twenty-five cars gives you a 20 percent standard error, a hundred cars gives 10 percent.
   3. That tells you the sample size you need before you open your mouth about precision. If someone wants the rate to five percent, you need 400 cars.
   4. Now the diagnostics, which are what a research interview is actually about. Plot the inter-arrival times and check whether they look exponential. Over-dispersion, meaning variance above the mean of the counts, tells you arrivals are clustered and Poisson is wrong. Then I would go to a Cox process or a Hawkes process with self-excitation.
   5. And I would flag the estimation trap: if instead I sampled by picking a random moment and measuring the gap I happened to land in, I would oversample long gaps. That is the inspection paradox, and it biases the mean gap upward by a factor of one plus the squared coefficient of variation. It is the same bias that makes waiting times feel longer than the timetable says.

   Where candidates lose it

   Jumping to a formula without stating the model or checking it. The interviewer wants model, estimator, standard error, then diagnostics. The specific failure mode they are hunting is the inspection paradox, so mention length-biased sampling unprompted. Hawkes processes are the right answer for clustered arrivals and they are also how trade arrivals actually behave in markets.

   Expect next

   • How would you test whether the Poisson assumption holds?
   • What if the cars arrive in clusters?
   • How long do you need to watch to get the rate within five percent?

   Reported by candidates at Jump Trading (Quantitative Research, Chicago, 2018). Source: Wall Street Oasis.

8. 038How would you fix violations of the OLS assumptions?RegressionIntermediatetechnicalACAQR Capital ManagementInvestments · Greenwich · 2022

   Say this

   Depends which assumption breaks, and the fixes fall into two very different classes: violations that only break your standard errors, and violations that break the coefficients themselves. The first class you patch; the second class you have to re-specify the model.

   Then walk it

   1. Heteroskedasticity and autocorrelated errors: coefficients stay unbiased, only inference is wrong. Fix with White or Newey-West robust standard errors, or clustered errors if the dependence is by group. Cheap fix, always worth doing on financial data.
   2. Endogeneity, meaning a regressor correlated with the error, whether from omitted variables, simultaneity or measurement error: this biases the coefficients and no standard error fix helps. You need an instrument, a control for the omitted factor, a fixed effect, or a different design.
   3. Multicollinearity: coefficients are still unbiased but the variances explode and the signs flip sample to sample. Drop or combine the collinear regressors, use ridge, or work with principal components. And check the variance inflation factors before you interpret anything.
   4. Non-normal or fat-tailed errors: inference is still fine asymptotically thanks to the CLT, but outliers dominate the fit because OLS minimises squares. Use robust regression, Huber loss or quantile regression, and always look at the influence diagnostics.
   5. Non-linearity: add the relevant transform or interaction rather than pretending it away. And I would say the order I actually work in on real data: plot residuals against fitted values and against time first, because most violations announce themselves visually before any test does.

   Where candidates lose it

   Listing fixes without separating what biases the coefficients from what only biases the standard errors. That distinction is the question. Slapping Newey-West errors on an endogenous regression is a common and useless move, and an interviewer at a research shop will push on exactly that.

   Expect next

   • Which of those actually biases your coefficients?
   • How do you detect endogeneity if you have no instrument?
   • What do you do when the residuals are fat-tailed and autocorrelated at the same time?

   Reported by candidates at AQR Capital Management (Investments, Greenwich, 2022). Source: Wall Street Oasis.

9. 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.

10. 040Here is a dataset. Analyse it using probability metrics and tell me what you find.StatisticsHardcase studyJane StreetCredit Risk · London · 2025

    Say this

    I would spend the first third of the time on the data itself before any modelling: shape, missingness, duplicates, timestamps, and the univariate distributions. Then state a hypothesis, test it, and report the effect size with an honest uncertainty. Narrate every step, because the interviewer is grading the process, not the punchline.

    Then walk it

    1. Start with the boring checks, out loud. Row count, date range, obvious duplicates, missing values and whether they are missing at random, and whether any column is a leak of the outcome. Most real findings in interviews of this kind are data artefacts.
    2. Then univariates: mean, median, standard deviation, skew, kurtosis, and the tails. Plot histograms and the empirical CDF. If a column is heavy-tailed or bimodal, say so, because it changes every subsequent choice.
    3. Then the relationship you were asked about. Give a point estimate plus a confidence interval, and prefer a plot to a coefficient. If the data are time-ordered, check for autocorrelation and regime change before quoting any p-value, because serial dependence inflates significance badly.
    4. Then the discipline: state your null, say what result would change your mind, and count how many hypotheses you have looked at. If you tested twenty things, say so and adjust.
    5. Close with what the data cannot tell you. A credit dataset with survivors only cannot tell you about defaults. Ending on the limitation is what separates an analyst from someone producing numbers, and in a live exercise it is the cheapest way to sound senior.

    Where candidates lose it

    Going straight to a model. Almost every candidate opens a regression and never looks at a histogram, then reports a spurious result driven by three outliers or a broken timestamp. Talk through the data integrity checks first, and say your uncertainty on every number you quote.

    Expect next

    • What would you check before trusting that correlation?
    • How many hypotheses did you test, and how does that change your p-value?
    • What would you want that is not in this dataset?

    Reported by candidates at Jane Street (Credit Risk, London, 2025). 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.