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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100 questions, mapped to the firms that asked them

Questions
100
Traced to a firm
53
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15
Updated
September 2026
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All firmsOld Mission Capital12Tower Research Capital10Jump Trading7Akuna Capital5Citadel4D.E. Shaw3Jane Street3AQR Capital Management2DRW2Millennium Management2Schonfeld2Squarepoint Capital2Susquehanna International Group2Belvedere Trading1Optiver1
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All topicsProbability10Coins, cards and games6Expected value8Statistics11Market making15Estimation and mental maths4Stochastic processes4Regression5Machine learning6Time series6Programming10Options and derivatives8Fit and motivation7
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  1. 005I shuffle a deck and turn cards face up one at a time. At any point you may say stop, and you win if the next card is red. What is your optimal strategy and what is your probability of winning?Coins, cards and gamesHardtechnicalJump TradingResearch · Chicago · 2018

    Say this

    Every strategy wins with probability exactly one half, so there is no optimal strategy. Stopping before the first card is as good as any clever rule based on the count.

    Then walk it

    1. The quick proof is a symmetry argument. Fix any stopping rule and imagine swapping the colour of every card in the deck. The rule's decisions are determined by cards already seen, and the swap turns every win into a loss and every loss into a win, so wins and losses are equally likely.
    2. The cleaner proof is a martingale. Let X be the fraction of red cards remaining. Before you see a card, the expected fraction of reds remaining after you see it is exactly the current fraction, because the card you turn is a uniform draw from what is left. So X is a martingale.
    3. Your win probability when you stop is X at the stopping time. Optional stopping says the expected value of a bounded martingale at any stopping time equals its starting value, which is 26/52, or a half.
    4. This is the whole lesson of the problem. Your information at the moment you stop is already priced into the state. There is no edge in a fair game no matter how you time it.
    5. One caveat that makes it a real problem rather than a trick: if you are forced to keep going to the last card, you still win a half, because the last card is red with probability a half. But the variance of the outcomes differs across strategies even though the mean does not, and if you had a utility function that is not linear you would care.

    Where candidates lose it

    Inventing a rule like wait until more blacks than reds have come out, and claiming it beats a half. That intuition feels right and it is wrong, because the situations in which the rule fires are exactly the situations where the deck was red-heavy from the start. Name the martingale and use optional stopping, or at minimum give the colour-swap symmetry argument.

    Expect next

    • Prove it with optional stopping, precisely.
    • Does the answer change if you can also bet on black?
    • Which strategy has the lowest variance of outcome?

    Reported by candidates at Jump Trading (Research, 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.