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

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Type

AnyBrainteaserTechnicalCaseMarket viewFit

1. 010Same die game, but now you earn one dollar per dot and each re-roll costs you one dollar, with unlimited re-rolls. What is the value of the game and when do you stop?Expected valueHardsuperdayOld Mission CapitalFinance · New York · 2018

   Say this

   The game is worth 4 dollars if the first roll is free, and you stop on a 3 or better. The continuation value of choosing to roll again is exactly 3, so the acceptance threshold is 3, and the value of a free first roll is the average of 3, 3, 3, 4, 5 and 6, which is 4.

   Then walk it

   1. Set up the recursion. If you decide to roll, you pay 1, then with the threshold t you accept faces above or equal to t and otherwise roll again. So V equals minus 1 plus the average over faces of the max of the face value and V.
   2. Guess the threshold is 4, meaning V sits in the interval 3 to 4. Then faces 4, 5, 6 are accepted for 15 total, and faces 1, 2, 3 all continue at V each.
   3. V equals minus 1 plus (15 plus 3V)/6. Multiply through: 6V equals minus 6 plus 15 plus 3V, so 3V equals 9, V equals 3.
   4. Check consistency: V equal to 3 means you should accept anything at or above 3, not 4. Re-solve with threshold 3: accepted faces 3,4,5,6 sum to 18, continuing faces 1,2 give 2V. V equals minus 1 plus (18 plus 2V)/6 gives 6V equals 12 plus 2V, so V equals 3. Consistent, since 3 is in the interval 2 to 3 boundary case. So the value of choosing to roll is 3 and you stop on 3 or better.
   5. Check consistency, which is the step that matters: V equal to 3 means you accept anything at or above 3, so re-solve with threshold 3. Accepted faces 3, 4, 5, 6 sum to 18 and continuing faces 1 and 2 give 2V, so V equals minus 1 plus (18 plus 2V)/6, which gives 4V equals 12 and V equals 3. Now the assumed threshold and the solved value agree, so 3 is the answer. With a free first roll the game is worth the average of max(face, 3), which is 24 over 6, equals 4.

   Where candidates lose it

   Solving the fixed point once and not checking that the threshold you assumed is consistent with the value you found. That verification step is the whole exercise in an optimal-stopping problem, and skipping it is how candidates report a threshold of 4 with a value of 3 and never notice the contradiction.

   Expect next

   • What if the re-roll cost were 2 dollars instead?
   • At what cost per re-roll does the game become worthless?
   • How does this map to pricing an American option?

   Reported by candidates at Old Mission Capital (Finance, New York, 2018). 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. 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.