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. 006You want to draw a black card followed by a red card. One deck is a full 52-card deck, another has had some cards removed. Which deck do you choose and why?Coins, cards and gamesIntermediatetechnicalOld Mission CapitalQuantitative Research · New York · 2014

   Say this

   Write the probability down before you pick. For a deck with b blacks and r reds, drawing black then red is b/(b+r) times r/(b+r-1). Then just compare the candidate decks on that expression, and you will find you want the deck that is as balanced as possible and as small as possible.

   Then walk it

   1. Full deck: 26/52 times 25/51, which is 0.5 times 0.490, about 24.5 percent.
   2. Now try a tiny balanced deck, one black and one red. That is 1/2 times 1/1, which is 50 percent. Far better.
   3. So the direction is clear. Removing cards helps if it keeps the deck balanced, because the second draw's conditional probability improves once the black card you removed is a bigger fraction of a smaller deck.
   4. Unbalancing hurts. A deck of 26 blacks and 1 red gives 26/27 times 1/26, which is 1/27, about 3.7 percent. Almost all your probability mass dies on the second draw.
   5. So: balanced beats unbalanced, small beats large, and the extreme is one black plus one red at fifty percent. Say the formula first, then test the corners. That is faster and less error-prone than trying to reason about it verbally.

   Where candidates lose it

   Reasoning in words about whether removing cards helps or hurts, and getting tangled. Write b/(b+r) times r/(b+r-1) immediately, then plug in three corner cases. Also do not forget the minus one in the denominator, because sampling without replacement is the entire content of the question.

   Expect next

   • What deck maximises the probability of black then red then black?
   • What if you wanted two cards of the same colour instead?
   • Now make me a market on the probability for the standard deck.

   Reported by candidates at Old Mission Capital (Quantitative Research, New York, 2014). Source: Wall Street Oasis.

2. 007What is the probability of being dealt four of a kind in a five-card poker hand?Coins, cards and gamesIntermediatetechnicalOld Mission CapitalQuantitative Research · New York · 2014

   Say this

   624 hands out of 2,598,960, which is about 0.024 percent, or one in roughly 4,165. Thirteen choices of rank for the quad, times 48 remaining cards for the fifth card.

   Then walk it

   1. Denominator: 52 choose 5 is 2,598,960. Worth memorising, it comes up constantly.
   2. Numerator: pick the rank of the four of a kind, 13 ways. All four suits are forced. Then the fifth card is any of the 48 cards left, so 13 times 48 is 624.
   3. 624 over 2,598,960 simplifies to 1 over 4,165. Call it one in four thousand.
   4. The counting discipline that matters: the kicker is 48, not 12. If you write 13 times 12 you are counting ranks not cards, and you would be off by a factor of four.
   5. Quick cross-check against a fact you might already know: a full house is 3,744 hands and a straight flush is 40. Four of a kind sitting between them at 624 is consistent with the standard hand ranking, which is ordered by exactly this rarity.

   Where candidates lose it

   Double counting, or using 12 instead of 48 for the fifth card. The other classic error is dividing by 5 factorial somewhere by accident. Use combinations consistently in both numerator and denominator, and state the denominator before you start so the interviewer can follow.

   Expect next

   • Now do a full house.
   • What is the probability of a flush, excluding straight flushes?
   • How would that change in a seven-card game like Texas hold'em?

   Reported by candidates at Old Mission Capital (Quantitative Research, New York, 2014). Source: Wall Street Oasis.

3. 008What is the expected value of a roll of a fair six-sided die?Expected valueCorephone / first roundOld Mission CapitalTrading · Chicago · 2020

   Say this

   Three and a half. Each face has probability one sixth, and one through six sums to 21, so 21 over 6 is 3.5. Or faster: for a uniform run of integers the mean is the midpoint, which is (1 plus 6) over 2.

   Then walk it

   1. The sum one to n is n(n+1)/2, so the mean of a uniform die with n faces is (n+1)/2. For six faces that is 3.5.
   2. Variance is worth having ready too, because it is the immediate follow-up. E of X squared is (1+4+9+16+25+36)/6, which is 91/6, about 15.17. Subtract 3.5 squared, 12.25, and you get 35/12, about 2.917. Standard deviation is about 1.71.
   3. Also useful to know the general formula: the variance of a uniform die with n faces is (n squared minus 1) over 12. For n equal to 6 that is 35/12, matching.
   4. Two dice: expected sum is 7 by linearity, and variance is 35/6 because they are independent, so standard deviation is about 2.42.
   5. The reason a trading desk opens with this is speed and composure, not difficulty. Answer in under two seconds and have the variance ready before they ask, because the real question is the next one.

   Where candidates lose it

   Hesitating. This is a warm-up and any pause reads badly. The second trap is being caught flat-footed on variance, which follows more than half the time. Know 35/12 and know that the standard deviation of a die is about 1.71.

   Expect next

   • What is the variance?
   • What about the sum of two dice?
   • What is the expected value of the maximum of two dice?

   Reported by candidates at Old Mission Capital (Trading, Chicago, 2020). Source: Wall Street Oasis.

4. 009You roll a fair die and may choose to re-roll once, taking the second value if you do. What is the expected value of the game, and what is your strategy?Expected valueIntermediatetechnicalOld Mission CapitalFinance · New York · 2018

   Say this

   4.25. Re-roll on a 1, 2 or 3, keep a 4, 5 or 6. The continuation value is 3.5, so you keep anything strictly above 3.5 and re-roll anything below.

   Then walk it

   1. Work backwards. If you re-roll you face a plain die, worth 3.5. So the rule is: keep the first roll if it beats 3.5.
   2. With probability one half you roll 4, 5 or 6 and keep it. The conditional mean of those three is 5.
   3. With probability one half you roll 1, 2 or 3 and re-roll, collecting 3.5.
   4. So the value is 0.5 times 5 plus 0.5 times 3.5, which is 2.5 plus 1.75, equals 4.25.
   5. Sanity check the bounds before you commit: the answer must sit between 3.5, which is the value of no re-roll option, and 6, which is the value of a free choice of face. 4.25 sits sensibly in between, and the option to re-roll is therefore worth 0.75 to you.
   6. The general principle, and the thing they are actually testing: the threshold is the continuation value, always. This is the same logic as an American option's exercise boundary. Exercise when the intrinsic value exceeds the value of holding on.

   Where candidates lose it

   Taking the average of the two rolls, or re-rolling a 4 because it is below the maximum. The rule is compare against the continuation value, not against the best possible outcome. Also say the threshold out loud before computing, because the interviewer wants to hear the backward-induction step, not just the number.

   Expect next

   • Now allow two re-rolls. What is the value and the thresholds?
   • What if you get n re-rolls, as n goes to infinity?
   • What if the re-roll costs you a dollar?

   Reported by candidates at Old Mission Capital (Finance, New York, 2018). Source: Wall Street Oasis.

5. 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.

6. 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.

7. 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.