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. 013You have a feed of a hundred thousand data points and you know fifteen of them are missing, recorded as zeros at the end. If you pull a window, what is the probability of at least one missing value?ProbabilityIntermediatetechnicalJump TradingProp Trading · Remote · 2022

   Say this

   Use the complement. For a sample of n points drawn without replacement from 100,000 of which 15 are bad, the probability of at least one bad is one minus the hypergeometric probability of none, which is one minus the product over i of (99,985 minus i)/(100,000 minus i). For small n that is well approximated by one minus (1 minus 0.00015) to the n.

   Then walk it

   1. Always compute at least one as one minus none. Summing the cases is the slow road and it invites double counting.
   2. The exact object is hypergeometric: choose n from 99,985 good over choose n from 100,000. For n much smaller than 100,000 the with and without replacement answers agree to several decimals.
   3. Numbers give it life. p is 15 over 100,000, which is 0.00015. For a window of 1,000 points, one minus 0.99985 to the 1000 is about 13.9 percent. For a window of 100 it is about 1.5 percent. So this is a real problem, not a rounding issue.
   4. Useful shortcut: for small p and moderate n the answer is roughly n times p, capped by 1. A thousand times 0.00015 is 0.15, close to the exact 0.139, and the Poisson approximation 1 minus e to the minus 0.15 gives 0.1393, which is very close.
   5. The thing I would say next on a desk, because it is the real question: they are at the end of the series, which is not random at all. If they are the most recent 15 points, then any window containing the tail hits all 15 with certainty and every other window hits none. Position matters more than the count.

   Where candidates lose it

   Treating the missing points as randomly scattered when the question says they sit at the end. That is the detail being tested. Give the hypergeometric answer for the random case, then flag the structural point: trailing zeros are usually a feed-truncation artefact, so the right fix is to detect and drop the tail, not to price the probability.

   Expect next

   • How would you detect that the zeros are missing values rather than genuine zeros?
   • What is the Poisson approximation and when does it break?
   • How do you handle those points in a model without leaking future information?

   Reported by candidates at Jump Trading (Prop Trading, Remote, 2022). Source: Wall Street Oasis.

2. 019I owe you pi dollars, and we can only exchange whole cents. How do you settle the debt fairly?ProbabilityIntermediatetechnicalMillennium ManagementQuantitative Research · Hong Kong · 2025

   Say this

   Randomise the last cent. Pi is 3.14159 and change, so pay 3.14 with probability 1 minus 0.159 and 3.15 with probability 0.159. The expected payment is exactly pi, so the settlement is unbiased even though every individual payment is wrong.

   Then walk it

   1. The general rule: to pay an amount x on a grid, pay floor(x) with probability 1 minus the fractional part and floor(x) plus one tick with probability equal to the fractional part.
   2. Check it: 3.14 times 0.841 plus 3.15 times 0.159 equals 3.141590 to six places. Unbiased by construction.
   3. How do you generate the 0.159? Flip a fair coin repeatedly and read off binary digits until the number you have built is decisively above or below 0.159. That terminates with probability one and needs about two flips on average.
   4. This is called randomised rounding and it is not a party trick. It is exactly how you settle fractional share allocations, how you break ties in sub-penny pricing, and how stochastic rounding keeps low-precision numerics from accumulating drift.
   5. The limitation, said before they ask: unbiasedness buys you a variance of about a quarter of a cent squared per payment. If we settle once, I have injected noise to remove bias. If we settle a thousand times, the bias from always rounding down would be 1.59 dollars while the randomised total is within a few cents of correct. So the method pays off under repetition, not on a single trade.

   Where candidates lose it

   Answering just round to 3.14, which is the boring answer and is systematically biased against one party. Also do not overreach into give me an IOU, which dodges the question. The interviewer wants unbiased-in-expectation, and wants to hear you construct the random bit from fair coins.

   Expect next

   • How many fair coin flips do you need to generate that probability?
   • What is the variance of your payment?
   • Where does randomised rounding matter in a real trading system?

   Reported by candidates at Millennium Management (Quantitative Research, Hong Kong, 2025). 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. 024I have two children and at least one is a boy. What is the probability both are boys?ProbabilityIntermediatetechnicalProp trading firmsQuant trading

   Say this

   One third, if the information came from a statement about the pair. The sample space is BB, BG, GB, GG, the condition kills GG, and one of the three survivors is BB. But the answer becomes a half if you learned it by meeting one specific child.

   Then walk it

   1. Equally likely and independent births give four ordered outcomes, each one quarter. Conditioning on at least one boy leaves three, of which one is BB. So one third.
   2. Now the version that makes it a real question. Suppose instead I introduce you to my elder child and he is a boy. Now you have conditioned on the elder being a boy, which leaves BB and BG, so the answer is one half.
   3. Same words in English, different conditioning event, different answer. The phrase at least one is a boy is a statement about the pair; this is my son is a statement about a position.
   4. The famous extension is the Tuesday boy: at least one is a boy born on a Tuesday. Now the answer is 13/27, because the extra detail changes how many pairs satisfy the condition and it breaks the symmetry between the two children.
   5. What I would actually say in an interview: the answer is one third under the standard reading, and then immediately name the ambiguity, because the entire point of the question is whether you notice that the conditioning event is underspecified.

   Where candidates lose it

   Answering one half on instinct, or answering one third and stopping. Both are half answers. Give one third with the sample space, then say precisely which conditioning event gives a half, because a quant interviewer is testing whether you can spot an ill-posed conditioning statement, which is a daily hazard in real data work.

   Expect next

   • Now: at least one is a boy born on a Tuesday.
   • What if I tell you my eldest is a boy?
   • How does this relate to survivorship bias in a dataset?
5. 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?
6. 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?

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.