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

AnyCoreIntermediateHard

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. 018You have n cars, each with fuel for a thousand miles, and you can transfer petrol between them mid-journey. What is the maximum distance one car can travel, and what happens as n goes to infinity?Expected valueHardsuperdayMillennium ManagementInvestments · London · 2024

   Say this

   A thousand times the harmonic sum: 1000 times (1 plus 1/2 plus 1/3 up to 1/n). It diverges, so as n goes to infinity the distance is unbounded, but only logarithmically, which is the interesting part.

   Then walk it

   1. Think in stages, working from the start. With all n cars moving together, you burn n tanks per 1000 miles of travel, so you can go 1000/n miles before you can consolidate one car's worth of fuel out of the collective and abandon it.
   2. After that leg, n minus 1 cars carry on, each full, and you get 1000/(n-1) more miles before dropping the next. Continue until one car is left, which contributes 1000/1.
   3. Sum the legs: 1000 times the sum of 1/k for k from 1 to n. That is 1000 times H_n.
   4. H_n grows like the natural log of n plus gamma, about 0.577. So with 10 cars you get roughly 2,929 miles, with 100 cars about 5,187, and with a million cars only about 14,392.
   5. That is the point worth making: the distance is unbounded but painfully inefficient. To double your range from 100 cars you need about 100 squared cars. This is the same log scaling as the coupon collector problem, and it is a good example of a divergent series that is useless in practice.

   Where candidates lose it

   Getting the legs backwards, i.e. putting the long leg first. The many-car legs are short because you are burning fuel n times as fast. Also do not answer infinite and stop. The number they want is 1000 H_n with the log growth spelled out, because the divergence-but-barely is the whole insight.

   Expect next

   • How many cars to reach ten thousand miles?
   • What if the cars must all return to the start?
   • Where else does the harmonic series show up in probability?

   Reported by candidates at Millennium Management (Investments, London, 2024). 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.