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. 001There are two bags of stones and you do not know how many black or white are in each. You draw two stones and both are black. What is the probability the next one is black, and will you bet on it?ProbabilityHardtechnicalCitadelQuantitative Trading · New York · 2025

   Say this

   Higher than a half, and yes I would bet on black. Because I do not know the composition, the two black draws are evidence about the composition itself, so I update towards bags that are black-heavy. The draws are not independent trials, they are a sample that teaches me about the urn.

   Then walk it

   1. Set it up properly: put a prior over the unknown mixture, say the proportion of black p is uniform on zero to one, and the draws are conditionally independent given p.
   2. Then this is Laplace's rule of succession. With k blacks out of n draws the posterior predictive probability of another black is (k+1)/(n+2). Two blacks out of two gives 3/4.
   3. The intuition without algebra: seeing black twice shifts the posterior mass towards high p, and the predictive probability is the posterior mean of p, which is now above a half.
   4. Compare it with the alternative model. If I were told the bag was exactly 50/50 and I was drawing with replacement, the answer would be exactly a half and the history would be irrelevant. The whole question is which model you are in.
   5. On the betting half: I would take anything better than even money on black, and I would size it small because 3/4 is a function of my prior, not of data. Two draws is almost no information. If the prior were concentrated near a half the answer moves back towards a half.

   Where candidates lose it

   Saying one half because the draws are independent. They are only independent conditional on the unknown composition, and the composition is exactly what you are learning. The second failure is giving 3/4 with no mention of the prior, as if it were a fact rather than the output of a uniform prior you chose.

   Expect next

   • What if the prior were Beta(2,2) instead of uniform?
   • Now make me a market on the probability and I will trade it.
   • Same question but sampling without replacement from a bag of 10 stones. Does the answer move?

   Reported by candidates at Citadel (Quantitative Trading, New York, 2025). Source: Wall Street Oasis.

2. 012Four points are chosen at random on the surface of a sphere. What is the probability that the tetrahedron they form contains the centre?ProbabilityHardsuperdayOld Mission CapitalProp Trading · Chicago · 2018

   Say this

   One eighth. The clean argument: take three random points and their three antipodes, giving eight candidate tetrahedra from the eight sign choices, and exactly one of the eight contains the centre.

   Then walk it

   1. Build the construction. Draw three random points P1, P2, P3 and three random diameters through them. The fourth point is then the head or tail of an independent diameter, and by symmetry each of the eight sign combinations of the three diameters is equally likely as the configuration.
   2. For almost every set of three diameters, exactly one of the eight tetrahedra formed by choosing one endpoint from each diameter, plus the fourth point, contains the centre. So the probability is 1/8.
   3. Warm up with the two-dimensional version first if you are stuck. Three points on a circle contain the centre with probability 1/4, by the same argument with two diameters and four sign choices.
   4. The pattern generalises: n plus 1 points on the surface of an n-sphere contain the centre with probability 1 over 2 to the n. Two to the power n sign choices, one winner.
   5. Say the 2D case out loud before the 3D case. It is the same proof at half the cognitive load, and it shows the interviewer your method rather than a memorised number. The number alone is worthless here because the answer is famous.

   Where candidates lose it

   Attempting to integrate over solid angles. It is a five-line symmetry argument and any attempt at brute-force geometry will run out of time. The other trap is stating one eighth flatly, which reads as recall. Construct the antipodal argument, because with a famous answer the reasoning is all they can grade.

   Expect next

   • Do the circle case in two dimensions.
   • What is the expected volume of that tetrahedron?
   • Three random points on a circle: what is the probability the triangle is acute?

   Reported by candidates at Old Mission Capital (Prop Trading, Chicago, 2018). Source: Wall Street Oasis.

3. 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.

4. 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.

5. 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.

6. 023Monty Hall. Three doors, one car, you pick one, I open a door with a goat, do you switch?ProbabilityCorephone / first roundProp trading firmsQuant trading

   Say this

   Switch. Your original door wins one third of the time, so the other door wins two thirds. The host's choice is not random, and that is where the information comes from.

   Then walk it

   1. Condition on your first pick. One third of the time you picked the car, and switching loses. Two thirds of the time you picked a goat, the host is forced to reveal the only other goat, and switching wins.
   2. So switching wins two thirds. The Bayes calculation agrees: the likelihood of the host opening door 3 is 1/2 if the car is behind your door 1, and 1 if the car is behind door 2, which is what tilts the posterior two to one.
   3. The intuition people find convincing: extend it to a hundred doors. You pick one, the host opens 98 goats, and switching wins 99 times out of 100. The host did all the work of avoiding the car.
   4. The critical assumption, and this is what a quant interview is really checking: the host knows where the car is and always opens a goat. If the host opens a door at random and happens to show a goat, the posterior is fifty-fifty and switching gains nothing.
   5. So the honest answer is: switch, and the reason it works is that the host's constraint leaks information. Change the host's rule and the answer changes.

   Where candidates lose it

   Getting the right answer for the wrong reason, or failing to state the host's rule. Everyone knows the answer is switch, so the only thing being graded is whether you can name the assumption that makes it true. Say explicitly that the host knows and is forced to reveal a goat.

   Expect next

   • What if the host does not know where the car is?
   • What if the host only offers the switch when you picked the car?
   • Do it with a hundred doors.
7. 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?
8. 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?
9. 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?
10. 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.