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 1–4 of 4 · filtered from 100Clear filters
  1. 025What is the expected number of fair coin flips to see two heads in a row, and how does it compare to heads followed by tails?Stochastic processesIntermediatetechnicalQuant tradingQuant research

    Say this

    Six flips for HH and four for HT. They differ because HH can destroy its own progress: a tail after a single head sends you back to nothing, while for HT a head after a head keeps you one step from done.

    Then walk it

    1. Set up states for HH. Let A be the expected flips from scratch and B from having one head. A equals 1 plus half A plus half B. B equals 1 plus half times 0 plus half A.
    2. Substitute: B equals 1 plus A/2, so A equals 1 plus A/2 plus (1 plus A/2)/2, which gives A equals 1.5 plus 0.75A, so 0.25A equals 1.5 and A equals 6.
    3. Now HT. Let A be from scratch, B from having a head. A equals 1 plus half A plus half B. But B equals 1 plus half times 0 plus half B, because another head leaves you still in state B rather than resetting. So B equals 2.
    4. Then A equals 1 plus A/2 plus 1, so A/2 equals 2 and A equals 4.
    5. The lesson worth saying out loud: patterns with self-overlap take longer. Both patterns have probability 1/4 per pair of positions, yet the waiting times differ, and that is purely about overlap structure. It generalises: the expected wait for a pattern equals the sum of 2 to the power of the length of each of its self-overlapping prefixes. HH gives 4 plus 2 equals 6, HT gives 4 plus 0 equals 4.

    Where candidates lose it

    Assuming both answers are 4 because each two-flip pattern has probability a quarter. That is the intuition the question is designed to break. Set up the state equations explicitly and pay attention to where a failed attempt lands you, because that is the only difference between the two problems.

    Expect next

    • Now do HHH.
    • In a race between HH and HT, which appears first and with what probability?
    • Derive it with the martingale approach instead.
  2. 026You start with fifty dollars and bet a dollar on a fair coin each time. What is the probability you reach a hundred before going broke, and how does it change if the coin is slightly against you?Stochastic processesIntermediatetechnicalQuant tradingQuant research

    Say this

    In a fair game it is exactly one half, because your wealth is a martingale and the stopping value must average back to fifty. Tilt the odds slightly against you and the probability collapses, not linearly but exponentially in the number of steps.

    Then walk it

    1. Fair case: wealth is a martingale, so by optional stopping, 50 equals 100 times p plus 0 times (1 minus p), giving p equal to 0.5. In general starting at a with an upper barrier b, the probability is a over b.
    2. Biased case: with win probability q the hitting probability is (1 minus r to the a) over (1 minus r to the b) where r is (1-q)/q.
    3. Put a number on it. At q equal to 0.49, r is about 1.0408. With a equal to 50 and b equal to 100, the probability of reaching 100 drops to roughly 12 percent. A one percent edge against you turns a coin flip into a 1-in-8 shot.
    4. That sensitivity is the entire lesson. Expected value per bet is minus two cents, which sounds trivial, but over the hundreds of bets you need to walk the barrier it compounds into near certainty of ruin.
    5. And the practical version on a desk: expected time to absorption in the fair case is a times (b minus a), so 50 times 50 equals 2,500 bets. Casinos and market makers both live on this asymmetry. Small edge, high repetition, deep pockets.

    Where candidates lose it

    Giving a over b and stopping. The interesting content is how brutally the biased case differs, and candidates who cannot state the r to the power formula usually also guess that a one percent edge changes the answer by about one percent. It changes it from 50 percent to 12 percent. Put a number on it.

    Expect next

    • What is the expected number of bets until you stop?
    • What happens if you bet your whole stack each time instead?
    • How does this relate to a trader's drawdown limit?
  3. 027What is a martingale, and how would you use optional stopping to solve a problem?Stochastic processesHardtechnicalQuant researchQuant trading

    Say this

    A martingale is a process whose expected next value, given everything you know now, equals its current value. Optional stopping says that for a suitably bounded stopping time, the expected value at the stopping time equals the starting value, which is what turns a hard path-dependent question into one line of algebra.

    Then walk it

    1. Formally: E of X_{n+1} given the filtration F_n equals X_n. No drift, conditional on history. It is not the same as independence, and increments need not be identically distributed.
    2. Optional stopping needs a condition, and you should name one: bounded stopping time, or bounded increments plus finite expected stopping time, or uniform integrability. Without it the theorem fails, and the classic failure is the doubling strategy, where a stopping time that is finite with probability one still produces E of X_tau equal to 1 rather than 0.
    3. How I use it: find a quantity that is conserved in expectation, then evaluate it at the stopping time. Gambler's ruin falls out immediately from wealth being a martingale.
    4. A second example, expected time in a symmetric random walk: W_n squared minus n is a martingale, so E of tau equals E of W_tau squared. With barriers at 0 and b starting from a, that gives E of tau equal to a(b minus a) in a line.
    5. And the reason it matters beyond puzzles: risk-neutral pricing is exactly the statement that the discounted price is a martingale under the pricing measure. Delta hedging is the construction of that martingale. If you can say that connection, the puzzle answer becomes a conversation about derivatives.

    Where candidates lose it

    Defining a martingale as a fair game and stopping there, or applying optional stopping without checking the integrability condition. Interviewers at the good shops will hand you the doubling strategy specifically to see whether you know why the theorem does not apply. Name the condition before you use the theorem.

    Expect next

    • Why does optional stopping fail for the doubling strategy?
    • Is the square of a martingale a martingale?
    • Connect this to risk-neutral pricing.
  4. 028An ant walks randomly along the edges of a cube starting at one corner. What is the expected number of steps to reach the opposite corner?Stochastic processesHardsuperdayQuant tradingQuant research

    Say this

    Ten steps. Collapse the eight vertices into four states by distance from the start, then solve three linear equations. The symmetry reduction is the whole trick.

    Then walk it

    1. By symmetry, all that matters is your graph distance from the target: state 3 is the start, then 2, then 1, then 0 which is the target. Each vertex has three neighbours.
    2. From state 3 all three neighbours are at distance 2, so E3 equals 1 plus E2.
    3. From state 2, one neighbour is at distance 3 and two are at distance 1. So E2 equals 1 plus (1/3)E3 plus (2/3)E1.
    4. From state 1, one neighbour is the target and two are at distance 2. So E1 equals 1 plus (2/3)E2.
    5. Solve: substitute E3 equals 1 plus E2 into the second equation to get E2 equals 1 plus (1 plus E2)/3 plus (2/3)(1 plus (2/3)E2). That yields E2 equal to 9, so E1 equals 7 and E3 equals 10. Sanity check with the general theorem: for a random walk on a regular graph the expected return time to a vertex is the number of vertices, 8, which is the right order of magnitude for a 10-step commute across the diagonal.

    Where candidates lose it

    Trying to track all eight vertices individually and drowning in eight equations. Say the word symmetry, lump the states by distance, and you have three unknowns. The other error is miscounting neighbours in state 2, where it is one back and two forward, not two back and one forward.

    Expect next

    • What is the expected time to return to the starting corner?
    • Do it for a tetrahedron.
    • What if the walk is on a hypercube in n dimensions?

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.