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

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Type

AnyBrainteaserTechnicalCaseMarket viewFit

1. 004Given a biased coin with probability p, how would you generate n independent fair coin tosses, and what is the lower bound on the number of biased tosses you need?Coins, cards and gamesHardsuperdayTower Research CapitalTrading · Princeton · 2018DED.E. ShawResearch · New York · 2026

   Say this

   The bound is information-theoretic. Each biased flip carries H(p) bits of entropy, where H(p) is minus p log2 p minus (1-p) log2 (1-p), and each fair flip you output consumes exactly one bit. So you need at least n divided by H(p) biased flips on average, and no scheme can beat that.

   Then walk it

   1. The argument is conservation of randomness. You cannot manufacture entropy, only repackage it, so expected input entropy must be at least expected output entropy.
   2. At p equal to a half, H is 1 and the bound is n flips, which is obviously right. At p equal to 0.1, H is about 0.47, so you need roughly 2.1 flips per fair bit at best.
   3. Von Neumann pairing achieves 1/(2p(1-p)) flips per bit, which at p equal to 0.1 is about 5.6. Miles off the 2.1 floor.
   4. To get close, you recycle the discarded information. Elias's and Peres's extractors take the sequence of discarded HH/TT outcomes and the positions of the successes, both of which still carry entropy, and feed them back in recursively. Peres's construction converges to the entropy bound as the block length grows.
   5. What I would actually say on a desk: for n fair bits I would use pairing because it is three lines of code and provably correct, and I would only reach for a Peres extractor if the biased source were expensive, which in practice it never is.

   Where candidates lose it

   Answering only with the construction and not the bound, or quoting the bound as n/H(p) without being able to say why entropy is the right currency. Also do not claim von Neumann is optimal. It is unbiased and simple, and it is provably wasteful, and saying so is the difference between having read the trick and understanding it.

   Expect next

   • Where exactly does the discarded entropy live in the von Neumann scheme?
   • Now the reverse problem: simulate a p-coin from fair coins.
   • What if p is unknown but you need to hit the entropy bound?

   Reported by candidates at Tower Research Capital (Trading, Princeton, 2018); D.E. Shaw (Research, New York, 2026). Source: Wall Street Oasis.

2. 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.

3. 017Five pirates must split a hundred gold coins. The most senior proposes a split, everyone votes, and if at least half agree it passes, otherwise he is thrown overboard and the next most senior proposes. How should the senior pirate split the coins to survive and maximise his take?Coins, cards and gamesHardsuperdayOld Mission CapitalProp Trading · New York · 2014

   Say this

   98 for himself, 0 to the second, 1 to the third, 0 to the fourth, 1 to the fifth. Solve it by backward induction from two pirates, because each pirate's vote depends only on what they would get if the current proposer dies.

   Then walk it

   1. Two pirates left: the senior of the two takes 100, votes for himself, and half of two is one vote, so it passes. Pirate 4 gets 100 and pirate 5 gets 0.
   2. Three left: pirate 3 needs one more vote. Pirate 5 gets nothing in the two-pirate world, so 1 coin buys him. Split is 99, 0, 1.
   3. Four left: pirate 2 needs one more vote out of four. He buys pirate 4, who gets 0 in the three-pirate world, for 1 coin. Split is 99, 0, 1, 0.
   4. Five left: pirate 1 needs two more votes. The pirates who get 0 under pirate 2's plan are 3 and 5, so he buys both for 1 coin each. That gives 98, 0, 1, 0, 1.
   5. The whole method is: work out what each pirate gets if the proposal fails, then pay each cheap vote exactly one coin more than that. The assumptions matter and you should state them: pirates are perfectly rational, prefer gold, prefer to live, and prefer fewer rivals if otherwise indifferent.

   Where candidates lose it

   Trying to reason forwards from five pirates, which is impossible. State that you are doing backward induction and start from the base case of two. The second trap is the tie rule. Half of an even number counts as passing here, and if you assume a strict majority the whole answer shifts, so say your reading of the rule out loud before you solve.

   Expect next

   • What happens with two hundred pirates and a hundred coins?
   • How does the answer change if a tie means the proposer dies?
   • What if pirates value killing above one extra coin?

   Reported by candidates at Old Mission Capital (Prop Trading, New York, 2014). 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.