Quant interview preparation

Prop market making and quantitative research, weighted the way the interviews actually are: probability and expected value, statistics and machine learning, market making logic, programming and options. Every question is either traced to a named firm from a public candidate report, or tagged at desk level when we could not trace it, and every probability answer shows the reasoning path rather than just the number.

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Question bank

100 questions, mapped to the firms that asked them

Questions
100
Traced to a firm
53
Firms
15
Updated
September 2026
Asked at
All firmsOld Mission Capital12Tower Research Capital10Jump Trading7Akuna Capital5Citadel4D.E. Shaw3Jane Street3AQR Capital Management2DRW2Millennium Management2Schonfeld2Squarepoint 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
Showing 1–2 of 2 · filtered from 100Clear filters
  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.

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.