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. 029There are n distinct types of card in cereal boxes, uniformly at random. How many boxes do you expect to buy to collect all n?Expected valueIntermediatetechnicalQuant tradingQuant research

   Say this

   n times the harmonic number H_n, which is roughly n times (ln n plus 0.577). For 50 cards that is about 225 boxes, so four and a half times the number of cards.

   Then walk it

   1. Decompose by waiting times. Once you hold k distinct cards, the chance the next box is new is (n minus k)/n, so the wait for the next new card is geometric with mean n/(n minus k).
   2. Sum over k from 0 to n minus 1: n times (1/n plus 1/(n-1) up to 1/1), which is n H_n.
   3. Numbers: n equal to 6 gives 14.7 boxes, n equal to 50 gives 224.9, n equal to 365 gives about 2,364. The last one is the expected days to see every birthday.
   4. The tail is where the cost is. Getting the first half of the set takes about 0.69n boxes; the last single card alone takes n boxes in expectation. Most of the pain is the final few.
   5. Variance is worth flagging: it is about n squared times pi squared over 6, so the standard deviation is roughly 1.28n. For n equal to 50 that is 64 boxes, which is enormous relative to the mean of 225. Quoting the mean without the spread would be misleading if you were budgeting for it.

   Where candidates lose it

   Trying to compute it by inclusion-exclusion over the whole collection. The decomposition into independent geometric waits plus linearity of expectation is the intended route and it takes twenty seconds. Also note the harmonic sum by name, because the log growth is the insight the interviewer wants.

   Expect next

   • What is the variance?
   • What if the cards are not equally likely?
   • How many boxes for a 90 percent chance of completing the set?
2. 031Break a stick at two uniformly random points. What is the probability the three pieces form a triangle?ProbabilityIntermediatetechnicalProp trading firmsQuant trading

   Say this

   One quarter. Let the cuts be x and y on a stick of length one. The triangle condition is that no piece exceeds one half, and that region is a quarter of the unit square.

   Then walk it

   1. The triangle inequality for three pieces summing to 1 reduces to a single condition: every piece must be strictly less than 1/2. If any piece is at least a half it is at least as long as the other two together.
   2. Draw the unit square in x and y. Take x less than y without loss of generality, which is the lower triangle of area 1/2. The three pieces are x, y minus x, and 1 minus y.
   3. The three conditions x less than 1/2, y minus x less than 1/2, and 1 minus y less than 1/2 carve out the middle triangle with vertices at (0, 1/2), (1/2, 1/2) and (1/2, 1). That has area 1/8.
   4. Double it for the other ordering and divide by the total area 1, giving 1/4.
   5. Different setup, different answer, and this is the part worth saying: if instead you break the stick once and then break the longer piece, the probability drops to 2 ln 2 minus 1, about 0.386. The phrase break at two random points must mean both cuts on the original stick, and you should confirm that reading before you compute.

   Where candidates lose it

   Not reducing the three triangle inequalities to the single condition no piece over a half. Candidates who try to handle three inequalities geometrically in one pass usually get 1/2 or 1/8. Also state the sampling scheme, because the sequential-break version has a completely different answer and interviewers use the ambiguity deliberately.

   Expect next

   • Now break the stick once and then break the longer piece.
   • What is the expected length of the longest piece?
   • What is the probability the triangle is obtuse?
3. 032How many people do you need in a room for a better than even chance that two share a birthday, and why is the answer so small?ProbabilityCorephone / first roundProp trading firmsQuant trading

   Say this

   Twenty-three. The reason it feels small is that you are counting pairs, not people. Twenty-three people generate 253 pairs, and each pair matches with probability 1/365, so you expect about 0.69 matches.

   Then walk it

   1. Compute the complement: the probability all birthdays differ is 365/365 times 364/365 times down to 343/365. At 23 people that product is about 0.493, so the match probability is about 0.507.
   2. The back-of-envelope version: the probability of no match is approximately exp of minus n(n-1)/(2 times 365). Set that to 0.5, so n squared over 730 equals ln 2, giving n about 22.5. Round up to 23.
   3. The pair-counting intuition is the answer to why. n choose 2 grows quadratically, so the number of chances grows fast while your intuition tracks n linearly.
   4. Contrast with the question people confuse it with: for someone to share your specific birthday you need about 253 people, because now you have only n pairs, not n squared over 2.
   5. Where this bites in real work: hash collisions and the birthday attack follow the same square-root law, and so does the chance that two of your supposedly independent signals are accidentally the same trade. You need about the square root of the space to get a collision, which is far fewer than people expect.

   Where candidates lose it

   Confusing it with the probability that someone shares your birthday, which needs 253 people. Also do not just recite 23. The gradeable part is the pair-counting argument and the exp of minus n squared over 730 approximation, which lets you answer variants like how many for a 99 percent chance without a calculator.

   Expect next

   • How many for a 99 percent chance?
   • How many to share a birthday with you specifically?
   • What is the connection to hash collisions?
4. 034How many Starbucks are there in New York City?Estimation and mental mathsCorephone / first roundTower Research CapitalProp Trading · New York · 2019

   Say this

   I would say roughly 250 to 350, and I would build it from demand rather than from geography. Eight million people, maybe one in ten buys a Starbucks on a given day, a store serves around a thousand cups a day, so 800,000 over 1,000 is about 800 store-days of demand, which I would then cut for the fact that Manhattan stores are much busier than a thousand cups.

   Then walk it

   1. Build two independent estimates and reconcile them. That is the actual skill being tested, not the number.
   2. Demand side: 8 million residents plus commuters and tourists, call it 9 million daytime people. Ten percent buy coffee from Starbucks on a given day gives 900,000 cups. A busy Manhattan store does 1,500 to 3,000 cups a day, so 900,000 over 2,500 is about 360 stores.
   3. Supply side: Manhattan has roughly 200 avenue-blocks of dense commercial frontage and you see a Starbucks every few blocks in midtown, which suggests 150 to 200 in Manhattan alone, plus maybe the same again across the four outer boroughs. That lands around 300.
   4. Both routes land in the same band, 250 to 400, which is the useful output. I would quote 300 as my point estimate with a range.
   5. Then state your uncertainty honestly and where it sits: the biggest lever is cups per store, which I could be wrong on by a factor of two. The population number I am confident in to ten percent. Naming which assumption dominates the error is what separates an estimate from a guess.

   Where candidates lose it

   Producing one chain of assumptions and asserting the answer with false precision. Build two independent routes, reconcile them, give a range, and say which assumption carries the error. Also do not freeze because you do not know the answer. Nobody knows it, and the interviewer is grading the structure and your composure, not the number.

   Expect next

   • Now make me a market on it and I will trade you.
   • How many coffee shops in total?
   • How would you check your estimate if you had the internet for thirty seconds?

   Reported by candidates at Tower Research Capital (Prop Trading, New York, 2019). Source: Wall Street Oasis.

5. 035How many golf balls fit in the Empire State Building?Estimation and mental mathsCoretechnicalTower Research CapitalAssistant Trader · New York · 2013

   Say this

   Order of a hundred billion. The building is roughly a hundred million cubic feet, a golf ball plus its packing waste takes about 0.0015 cubic feet, so 100 million over 0.0015 is about 70 billion. I would quote 50 to 100 billion.

   Then walk it

   1. Volume of the building: footprint about 200 by 400 feet, so 80,000 square feet, times 1,250 feet of height. That is 100 million cubic feet. Taper the tower and subtract structure and you might call it 80 million usable.
   2. Volume of a golf ball: diameter 1.68 inches, so radius 0.84 inches. Four thirds pi r cubed is about 2.5 cubic inches. There are 1,728 cubic inches in a cubic foot, so a ball is 0.00145 cubic feet.
   3. Packing efficiency: random close packing of spheres is about 64 percent, so effective volume per ball is 0.00145 over 0.64, about 0.00226 cubic feet.
   4. 80 million divided by 0.00226 gives about 35 billion. Using the full 100 million cubic feet gives 44 billion. So my range is tens of billions, call it 40 billion, and I would say 20 to 100 billion to be honest about the error bars.
   5. Say the two things you are least sure about: the usable fraction of the volume, and whether the question means the empty shell or the building with floors, furniture and lift shafts. Those swing the answer by a factor of two, and the packing fraction only matters at the 30 percent level.

   Where candidates lose it

   Forgetting the 1,728 cubic inches per cubic foot conversion, which throws you off by three orders of magnitude, or ignoring packing efficiency entirely. Also decide out loud whether you are filling the empty shell or the furnished building. And always sanity check the magnitude: if your answer is in millions or trillions, something went wrong by a factor of a thousand.

   Expect next

   • What is the packing efficiency of spheres and why?
   • How much would they weigh?
   • Now estimate the market value of that many golf balls.

   Reported by candidates at Tower Research Capital (Assistant Trader, New York, 2013). 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.