Derivatives Foundation puzzles, solved step by step

Puzzles: 100
Traced to a firm: 66
Topics: 12
Hard: 29

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001How many golf balls fit inside a 100-storey office tower? Talk me through your thought process; I care more about the structure than the final number.Mental maths and estimationCoreTower Research CapitalNew York · 2013

Try it first

Before any arithmetic: which single assumption will move your answer the most, once the building's size is fixed?

Show the worked solution

About 12 billion, with an honest range of 10 to 14 billion. A floor plate of 4,000 square metres at 3.8 metres a floor gives 1.52 million cubic metres over 100 floors; take half as usable, 0.76 million. A golf ball is 4.27 centimetres across, about 41 cubic centimetres, so 24,500 would fit in a solid cubic metre. Spheres leave gaps: at a packing factor of 0.64 that is 15,700 per cubic metre, and 0.76 million times 15,700 is 11.9 billion.

Why build a chain instead of guessing a number?

If someone asks how much rice a storeroom holds, you do not guess a tonnage. You measure the room, decide how much of it is shelving, and know how much a sack holds. An estimation question is marked on whether each link in the chain is stated, sized and defensible, not on the final figure. Here the chain is floors, floor plate, floor height, usable fraction, ball volume, packing factor. Say the six links before you fill any of them in, so the interviewer can follow and correct a single link rather than the whole answer.

The chain runs from 100 floors, a 4,000 square metre plate and 3.8 metre floors to 1.52 million gross cubic metres and 0.76 million usable, then from a 40.8 cubic centimetre ball to 24,531 balls per solid cubic metre, and the packing factor turns that into 10.4, 11.9 or 13.8 billion balls, a spread of about a third from the packing assumption alone.

Where does the packing factor come from, and why does it matter so much?

Pour marbles into a jar and shake it: they settle at roughly 64% of the jar's volume filled, with the rest air. Stack them by hand in the tightest possible pattern and you reach about 74%. Tip them in gently without shaking and you can be as low as 56%. The floor plate and height are things you can look up or pace out, but the packing factor is a physical assumption you must own, and it moves the answer from 10.4 to 13.8 billion on its own. Say which one you are using and why: balls poured into a building are a shaken random pile, so 0.64 is the defensible middle.

The relationship

N = \frac{100 \times 4{,}000 \times 3.8 \times 0.5}{40.8 \times 10^{-6}} \times 0.64 \approx 1.19 \times 10^{10}

100 x 4,000 x 3.8	floors times floor plate times floor height, the gross volume in cubic metres
0.5	the usable fraction after the core, structure and services
40.8 x 10^-6	one golf ball in cubic metres
0.64	the packing factor for a shaken random pile of spheres

What it says in wordsUsable volume divided by the volume of one ball, scaled down for the gaps between balls.

What do you say when the interviewer pushes on the usable fraction?

Lifts, stairs, structural columns, ducts and ceiling voids take a large slice of a tower, and nobody knows it to the percent. Give a range and show which way it pushes the answer: 40% usable takes you to about 9.5 billion, 60% to about 14 billion, so the final figure is somewhere between 10 and 15 billion whatever you assume. A candidate who reports a single number to three figures has missed the point of the question. A candidate who says 12 billion, give or take a third, with the reasons, has answered it.

Where candidates lose it

The common failure is to fix on one number early, such as the number of balls in a room, and then multiply by a guessed room count without ever stating the building's volume. The interviewer cannot follow it, cannot correct it, and marks it as a guess.

The second loss is forgetting that spheres do not fill space. Dividing the volume by the ball's volume overstates the answer by about half. Saying the words packing factor, and a number for it, is what separates a trader's estimate from a schoolchild's.

What the interviewer asks next

Now the balls are tennis balls, 6.7 centimetres across. Roughly how does your answer change?
If I told you the real answer was 20 billion, which of your assumptions would you revisit first?
Make me a market on it: give me a bid and an offer in billions, and tell me how wide and why.

Asked at Tower Research Capital, Assistant Trader, New York, 2013 (Wall Street Oasis): How many golf balls fit in the empire state building? Explain thought process and detailed solution

013How many people work in a large bank's 45-storey London headquarters tower? Give me a number and the assumptions behind it.Mental maths and estimationCoreHSBCCentral · 2026

Try it first

Before building anything: which route gives an estimate the interviewer can check link by link?

Show the worked solution

About 8,400 people, with an honest range of roughly 5,558 to 12,994. Of 45 floors, take 40 as ordinary office floors after plant rooms, lobby and trading floors. A floor plate of 3,000 square metres with 70% usable gives 2,100 square metres of desk space per floor. At 10 square metres per person that is 210 people a floor, and 40 floors give 8,400. The widest assumption is the space per person, 8 to 12 square metres, which alone moves the answer by half.

Why build from the floor rather than from the bank?

If you wanted to know how many people a wedding hall holds, you would not guess from the size of the family; you would pace the hall and think about chairs per row. An estimate is only as good as the link the listener can check, and anyone who has worked on an office floor has a feel for how many desks it holds. So the chain is office floors, floor plate, usable share, square metres per person. Say the four links before any number, so the interviewer can argue with one of them rather than with the whole answer.

Forty office floors of 3,000 square metres at 70% usable and 10 square metres a person give 8,400 people, while taking every assumption at its low end gives 5,558 and at its high end 12,994, and the space per person is the link with the widest range, a ratio of 1.5 between its ends.

The relationship

N = \frac{\text{floors} \times \text{plate} \times \text{usable}}{\text{sq m per person}} = \frac{40 \times 3{,}000 \times 0.70}{10} = 8{,}400

floors	office floors out of 45, after plant, lobby and other uses
plate	gross floor area in square metres
usable	the share of a floor that holds desks rather than lifts, cores and meeting rooms
sq m per person	the desk plus a share of corridors and kitchens

What it says in wordsPeople equals total desk area divided by the area each person uses.

Which assumption should you spend your time on?

The one with the widest range. Floors run 38 to 42, a ratio of 1.11; the plate 2,700 to 3,300, a ratio of 1.22; the usable share 65% to 75%, 1.15; but space per person runs 8 to 12 square metres, a ratio of 1.5, so it moves the answer most. Check it against a floor you know: a trading floor packs people at 6 to 8 square metres, a floor of meeting rooms and offices spreads them at 15 or more. If the interviewer gives you one fact, ask for that one.

What would you add about occupancy?

That desks and people are different counts. With hot-desking, a floor of 210 desks might be home to 250 or 300 people who are not all in on the same day, so a question about who works in the building can give a larger answer than a question about who is in it. Say which one you are answering. The limitation to state: the usable share and the space per person are guesses from general experience, not measurements, and a real number would come from the building's floor plans and the badge-in data.

Where candidates lose it

The common failure is to answer with a bare number, often a round 10,000, and then be unable to defend any part of it. The number is not what is marked; the chain is.

The second loss is spending the time on floors and plate, which are tight, and waving at the space per person, which is loose. Put the effort where the range is.

What the interviewer asks next

Now estimate how many lifts the tower needs to get everyone in between 8 and 9 in the morning.
How many taxis operate in a city's central business district on a weekday morning? Build the chain.
If the building's badge data showed 6,000 entries a day, which assumption would you revisit first?

Asked at HSBC, Sales and Trading, Central, 2026 (Wall Street Oasis): How many employees in London hsbc building How many taxis are in HK central How many beds in the nyc hotel

014The sample variance with n minus 1 in the denominator is an unbiased estimate of the true variance. Is its square root an unbiased estimate of the standard deviation? If not, which way is it off, and does it matter when you estimate volatility from 20 daily returns?Distributions and statisticsCoreSquarepoint CapitalLondon · 2026
Try it first
Pick before you reason: the sample standard deviation on average is
Show the worked solution
No. The sample standard deviation is biased low, by about 1.3% at n = 20, and for a volatility estimate that is small next to the sampling noise. The square root is concave, so by Jensen's inequality the expected value of the root is below the root of the expected value. For normal returns the exact factor is c4(n): the expected sample standard deviation is 0.9869 times the true one at n = 20. On the same 20 points the estimate's own standard error is around 16%, so the bias is not what you should worry about.
Why does an unbiased variance give a biased standard deviation?
If you average the areas of several square plots and then take the square root, you do not get the average of their side lengths; the big plots pull the area average up more than they pull the side average. Unbiasedness is a statement about averages, and averages do not pass through a curved function: for a concave function like the square root, the average of the outputs is below the output of the average. That is Jensen's inequality, and it is the whole answer. The variance is unbiased; its root is not, and the direction is down.
Two sample variances of 0.1 and 1.9 times the truth average to 1.0, whose root is 1.00, but their roots average only 0.85, and for 20 normal returns the exact expected shortfall is c4(20) = 0.9869, so the sample standard deviation runs about 1.3% low on average.
The relationship
$E[s] = c_4(n)\,\sigma,\qquad c_4(n) = \sqrt{\frac{2}{n-1}}\;\frac{\Gamma(n/2)}{\Gamma((n-1)/2)},\qquad c_4(20) = 0.9869$
s the sample standard deviation, the root of the n minus 1 sample variance
sigma the true standard deviation
c4(n) the exact correction factor for normal data, always below 1 and rising to 1 as n grows
What it says in wordsFor normal data the sample standard deviation underestimates the true one by a known factor that depends only on the sample size.
n c4(n) shortfall
5 0.9400 6.0%
10 0.9727 2.7%
20 0.9869 1.3%
50 0.9949 0.5%
100 0.9975 0.2%
The shortfall of the sample standard deviation falls quickly with the sample size, from about 6% at five observations to about 1.3% at twenty and half a percent at fifty.
Does it matter for a 20-day volatility estimate?
Not much, and saying why is the second half of the marks. The bias is 1.3%, but the standard error of a standard deviation from 20 observations is roughly 1 over the root of 2 times 19, about 16% of the true value, so the noise is more than ten times the bias. You can multiply by 1/c4 to remove the bias if you like, but you cannot remove the noise without more data, and 20 daily returns is simply a short window. On a volatility desk the honest answer is that a 20-day estimate of 16% could easily have been 13% or 19%.
What assumption does the exact factor need?
Normal returns. The direction of the bias, downward, holds for any distribution, because it comes from concavity alone; the size of the factor c4 depends on the distribution, and fat-tailed daily returns change it. Say the general result first, Jensen, then the normal-case number, then the limitation. That order shows you know what is a theorem and what is a model.
Where candidates lose it
The fast wrong answer is yes, on the grounds that the variance is unbiased and the root is just a relabelling. The root is a curved function, and expectations do not commute with curved functions; the sign of the curvature gives the direction.
The second loss is stopping at biased low. The interviewer asked whether it matters, and the answer is a comparison of 1.3% of bias against roughly 16% of sampling noise. Numbers, not adjectives.
What the interviewer asks next
Derive the direction of the bias from Jensen's inequality in one line.
What is the standard error of the sample standard deviation for normal data with n observations, roughly?
Would you use the n or the n minus 1 version for a volatility estimate, and does the choice matter at n = 20?
Asked at Squarepoint Capital, Quantitative Research, London, 2026 (Wall Street Oasis): Is the square root of the unbiased estimator for sample variance unbiased for standard deviation?
015You have three six-sided dice. Red has the faces 2, 6, 7; green has 1, 5, 12; blue has 3, 4, 8, with each number on two faces. Two players each pick a die and roll; the higher number wins. Which die would you choose to play with?Games and logicCoreBelvedere TradingChicago · 2022
Try it first
Before working the pairs: green has the highest average face, 6 against 5 for red and blue. Does that make green the best die?
Show the worked solution
Let the other player choose first, then take the die that beats theirs. Red beats green 5 times in 9, green beats blue 5 in 9, and blue beats red 5 in 9. The three dice form a cycle like rock, paper, scissors, so no die is best on its own; the advantage belongs to whoever picks second. If you must pick first, no choice does better than 4 in 9 against a wise opponent, and you should say so rather than pretend one die is stronger.
How can three dice with the same average not have a best one?
Three cricket teams can each beat one of the others and lose to the third; a league table would show them level, and still no team is the best. Winning a roll depends only on which die shows the higher face, pair by pair, and pairwise comparisons do not have to line up in a single order the way averages do. Red and blue average 5 and green averages 6, and yet green loses to red 5 times in 9: every head-to-head is lopsided, 5 to 4, in a circle, and the highest average sits inside it. Green's 12 wins by a mile and its 1 loses by a mile, and a roll pays nothing for the margin. The question tests whether you check the comparison that matters instead of the summary that does not.
Red beats green in 5 of the 9 equally likely face pairs, green beats blue in 5 of 9 and blue beats red in 5 of 9, so the three dice form a cycle with no best die and the player who picks second always holds a 5 in 9 edge.
How do you check a pair quickly in the room?
Write one die's faces across and the other's down and count the cells where the first is higher. Red against green: 2 beats only the 1; 6 beats 1 and 5; 7 beats 1 and 5; that is 1 + 2 + 2 = 5 of 9. Green against blue: 1 beats nothing, 5 beats 3 and 4, 12 beats everything, again 5 of 9. Blue against red: 3 and 4 each beat the 2, and 8 beats 2, 6 and 7, again 5 of 9. Three counts, under a minute, and the cycle appears.
The relationship
$P(R > G) = \frac{1 + 2 + 2}{9} = \frac{5}{9},\qquad P(G > B) = \frac{0 + 2 + 3}{9} = \frac{5}{9},\qquad P(B > R) = \frac{1 + 1 + 3}{9} = \frac{5}{9}$
R, G, B the face shown by the red, green and blue die
9 the number of equally likely face pairs, three distinct faces on each die
5/9 each die's edge over the next one around the cycle
What it says in wordsCount the winning face pairs out of nine for each ordered pair, and the three results form a cycle.
What is the trading lesson the interviewer is after?
That the order of moves can be worth more than the thing being chosen. The second mover has a guaranteed 5 in 9; the first mover, against someone who knows the cycle, has at best 4 in 9, so you should pay to move second and never volunteer to move first. That is the same instinct as quoting after you have seen the other side's interest rather than before. The limitation to state: the edge is only 5 to 4, so over a few rolls luck dominates, and a one-roll bet on it is a small edge with a large variance.
Where candidates lose it
The common answer is green, because it has the biggest face and the highest average. Both facts are true and both are irrelevant: a roll pays for being higher, not for being higher by a lot, and the pairwise count is the only thing that decides it.
The second loss is finding the cycle and still naming a die. The answer to which die is a question back: which one is the other player taking? Say that you want to choose second, and why.
What the interviewer asks next
Each player rolls their die twice and the totals are compared. Does the cycle survive, and does it change direction?
Design a fourth die that beats all three of these more often than not, or show that none exists.
Where on a trading desk does moving second carry an edge, and where does it cost you?
Asked at Belvedere Trading, Prop Trading, Chicago, 2022 (Wall Street Oasis): You have 3 dice: red has 2, 6, 7; green has 1, 5, 12; blue has 3, 4, 8. Highest number wins the game. Which one would you choose to play with?
017Two friends agree to meet at a cafe between 1 pm and 2 pm. Each arrives at a uniformly random time within that hour, independently of the other, and waits 20 minutes for the other before leaving (or until 2 pm, whichever is sooner). What is the probability they meet?Probability and countingCoreJane StreetNew York · 2026
Try it first
Pick before you draw anything: the chance the two friends meet is
Show the worked solution
5/9, about 55.6%. Put A's arrival time across and B's up a 60 by 60 square; every pair of times is a point, all equally likely, so probability is area. They meet when the times are within 20 minutes, the band either side of the diagonal. They miss in two corner triangles, each with legs of 40 minutes and area 800 of 3,600, which is 2/9. So the meeting chance is 1 minus 4/9 = 5/9.
Why turn two arrival times into a square?
Throw a dart at a square board without aiming and the chance it lands in any patch is just that patch's share of the board. Two independent arrival times, each spread evenly over the hour, behave exactly like that dart: A's time picks a position across, B's time picks a position up. With two independent uniform times, every pair of arrivals is a point in a 60 by 60 square, equally likely anywhere in it, so a probability becomes an area you can see. The event they meet is the set of points where the two times differ by less than 20 minutes, a band hugging the diagonal.
In the 60 by 60 square of arrival times the friends meet in the band within 20 minutes of the diagonal and miss in two corner triangles with legs of 40 minutes, each 2/9 of the area, so the meeting probability is 1 minus 4/9, which is 5/9 or about 55.6%.
Why is it easier to compute where they miss?
The band is an awkward six-sided shape; the regions outside it are two clean triangles. In the top-left triangle B arrives more than 20 minutes after A, so A has gone; in the bottom-right one, A is the late one. Each triangle has legs of 60 minus 20 = 40 minutes, so its area is 40 x 40 / 2 = 800 square minutes out of 3,600, which is 2/9, and the two together are 4/9. The clause about leaving at 2 pm changes nothing, because no one can arrive after 2 pm anyway; it only stops the question from looking ambiguous.
The relationship
$P(\text{meet}) = 1 - \left(1 - \frac{w}{60}\right)^2 = 1 - \left(\frac{40}{60}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9}$
w the waiting time in minutes, here 20
60 the length of the window in minutes
(1 - w/60)^2 the two miss triangles together, which fit into one square of side 1 - w/60
What it says in wordsThe chance of meeting is one minus the square of the share of the hour that falls outside the waiting time.
How does the answer move with the waiting time?
Not in a straight line, and that is a common follow-up. Doubling the wait from 10 to 20 minutes takes the meeting chance from about 31% to about 56%, not from one third to two thirds, because the miss region shrinks as a square. The table runs the formula for four waits. The trading version is two orders that must arrive within a latency window to match: halving the gap you can tolerate does more than halve the matches, and a picture of the square is the fastest way to see by how much. The limitation to state is the uniform assumption; real arrivals bunch near the hour, which raises the meeting chance.
Wait (minutes) Miss region Meet probability
10 25/36 11/36, 30.6%
20 4/9 5/9, 55.6%
30 1/4 3/4, 75.0%
40 1/9 8/9, 88.9%
The meeting probability rises faster than the waiting time at first and then flattens, because the miss region is the square of the share of the hour outside the wait.
Where candidates lose it
The fast wrong answer is one third, from reading the 20 minutes as a share of the hour. It forgets that either friend can be the late one and that the window is cut off at both ends of the hour. Without a picture, people also land on two thirds by doubling the window.
The second loss is drawing the square and then computing the band directly, with a hexagon and several pieces. The interviewer is watching for the complement: two identical triangles, one line of arithmetic, done in under a minute.
What the interviewer asks next
Each friend now waits 20 minutes but B always arrives in the second half hour. What is the probability they meet?
Three friends, each waiting 20 minutes. What is the chance all three are there at once?
What waiting time gives a meeting chance of exactly one half?
Asked at Jane Street, Technology, New York, 2026 (Wall Street Oasis): two people arrive at a location uniform random time within an hour, each wait 20min, what's the prob they meet
019You roll a fair die again and again until the first six appears. What is the expected total of all the numbers rolled, including the final six?Expected value and optimal stoppingCoreJane StreetNew York · 2026
Try it first
Commit to a number first: the expected total, six included, is
Show the worked solution
21. A six comes up one time in six, so you expect six rolls, and each roll averages 3.5; by Wald's identity the expected total is 6 x 3.5 = 21. From the other side: the five expected rolls before the six can only be 1 to 5, so each averages 3, giving 15, and the closing six makes 21. The tempting 5 x 3.5 + 6 = 23.5 forgets that the earlier rolls cannot be sixes.
Why is 23.5 wrong when every roll averages 3.5?
Count people walking through a door until the first one in a red shirt arrives. The people before that one are, by the way you counted, not in red; nobody would estimate their shirts from the whole crowd. Conditioning on a roll not being a six changes its average from 3.5 to 3, and every roll before the stopping roll carries that condition. So the honest split is five expected rolls at 3 each, which is 15, plus the six that ends the game. The slip to 23.5 counts the sixes twice: once in the 3.5 of the earlier rolls, and again as the final roll.
The five expected rolls before the first six can only show 1 to 5, so they average 3 and total 15, and with the closing six the expected total is 21, the same as six expected rolls at 3.5 each, while the tempting 23.5 wrongly lets the earlier rolls average 3.5; a seeded simulation of 200,000 games gives 20.92.
Why does 6 x 3.5 still give the right answer?
Because of Wald's identity, and it is worth naming in the room. If the decision to stop depends only on rolls you have already seen, the expected total equals the expected number of rolls times the average roll. Before any roll is made, each one is a fresh die with average 3.5; the stopping rule only decides how many of them you take, and it cannot peek ahead. The two routes agree because the missing sixes in the early rolls are exactly balanced by the guaranteed six at the end. A one-line check also works: the first roll averages 3.5, and five times in six the game restarts.
The relationship
$E[S] = 3.5 + \tfrac{5}{6}\,E[S] \;\Rightarrow\; E[S] = 21,\qquad E[S] = E[N]\,E[X] = 6 \times 3.5 = 21$
S the total of all rolls, six included
N the number of rolls until the first six, with E[N] = 6
X one roll of the die, with E[X] = 3.5
5/6 the chance the first roll is not a six and the game starts afresh
What it says in wordsConditioning on the first roll gives the same 21 as multiplying the expected number of rolls by the average roll.
Where does this reasoning show up on a desk?
Any rule of the form trade until something happens. A rule such as take profit once up 2% cannot change the expected P&L of a fair bet, for the same reason: a stopping rule that only looks backwards changes when you stop, not the average of what you collect per step. The limitation to state is that Wald needs the expected number of steps to be finite; a rule like keep going until you are ahead can break it. A sharp follow-up the interviewer may use: given that every roll was even, what is the expected number of rolls? The answer is 1.5, not 3, and the same conditioning idea explains it.
Where candidates lose it
The common answer is 23.5: five rolls at the familiar 3.5, plus the six. It sounds careful because it treats the last roll separately, and it is wrong because the earlier rolls are conditioned on not being sixes.
The second loss is getting 21 from 6 x 3.5 and not being able to say why it is allowed. The interviewer will push: the number of rolls depends on the rolls, so why can you multiply? The answer is that the stop depends only on the past, which is Wald's identity.
What the interviewer asks next
Given that every roll in the game was even, what is the expected number of rolls?
What is the expected total if you stop at the first roll of 5 or 6, counting that roll?
You are paid the total but must pay Rs 4 per roll. Would you play, and at what cost per roll is the game fair?
Asked at Jane Street, Investment Operations, New York, 2026 (Wall Street Oasis): First interview was testing simple math brainteasers (e.g. expected value of dice throws, etc.)
022Make me a market on the total number of Olympic gold medals the United States has ever won, Summer and Winter Games together. You cannot look anything up.Market makingCoreOld Mission CapitalBoston · 2024
Try it first
Once the structure is written down, which input deserves most of your quote's width?
Show the worked solution
About 1,096 in the middle, so quote 1,000 bid, 1,200 offered, and say the share of golds is where you are least sure. Summer: about 29 Games attended, times about 190 events, times about 18% won by the United States, gives about 992. Winter: 24 Games, about 62 events, about 7%, adds about 104. The honest range is 756 to 1,491; the quote sits around the middle and moves as soon as someone trades on it.
What is the interviewer actually marking?
If a friend asks how much your monthly grocery bill is, you do not recall a number; you think of trips a week, spend a trip, four weeks, and say a figure with a range. A market on an unknown quantity is marked on the structure behind the middle, the honesty of the range, and what you do when someone trades with you, not on knowing the answer. The interviewer knows the real count. Treat every input below as an assumption to be checked against the published medal table, and say so; inventing precision is worse than a wide range.
Summer Games contribute about 29 x 190 x 18% = 992 golds and Winter Games about 24 x 62 x 7% = 104, a total of about 1,096 with an honest range of 756 to 1,491, and the share of golds is the widest link, so the quote of 1,000 at 1,200 sits around the middle.
How do you build the middle?
Split it into Summer and Winter, because they differ by an order of magnitude. Modern Summer Games run every four years from 1896, less three wartime cancellations, about 30 Games, and take one off for the boycotted 1980 Games, 29. Events grew from a few dozen to over three hundred, so call the average about 190, with 170 to 210 as the bracket. The share of golds is the input you own: the United States wins more than any other country, so something like a sixth to a fifth of all golds, 18% in the middle, is defensible but soft. Winter adds 24 Games at about 62 events and a smaller share, about 7%.
The relationship
$G \approx 29 \times 190 \times 0.18 + 24 \times 62 \times 0.07 \approx 992 + 104 \approx 1{,}096$
29, 24 Summer and Winter Games the United States took part in, as estimated
190, 62 average events per Games, one gold each
0.18, 0.07 the assumed United States share of golds in each
What it says in wordsGolds are the number of Games times events per Games times the share the United States wins, added across Summer and Winter.
How wide should the quote be, and what happens when someone trades?
Running all lows together and all highs together gives 756 to 1,491, but every input is unlikely to sit at its extreme at once, so the quote can be much tighter than that range. Quote about 10% either side of the middle, 1,000 bid and 1,200 offered, and treat the first trade as information from someone who may know the answer. If the interviewer pays 1,200, move both sides up, perhaps to 1,150 at 1,350, and ask yourself which link you underestimated; the share is the first suspect. The limitation to admit is that a quote built from memory of rough facts can be confidently off-centre, which is exactly why the market moves when it is traded.
Where candidates lose it
The common failure is to blurt one number, or a market as wide as 500 at 2,000, with no structure behind it. A number with no structure cannot be defended, and a market that wide is a refusal to quote dressed up as caution.
The second loss is holding the quote after the interviewer lifts it twice. A counterparty who keeps buying is telling you the answer is higher. Move up, widen if you must, and say which assumption you are revising.
What the interviewer asks next
I buy 10 at your offer. Where is your new market?
Now make a market on the total golds won by all countries ever. Which of your inputs carries over?
Your structure gave about 1,100. If the true answer were 1,500, which link would you suspect, and why?
Asked at Old Mission Capital, Equities, Boston, 2024 (Wall Street Oasis): One was on making a market on the total number of Olympic gold medals the US has won.
028How many zeros are at the end of 1000! (1000 factorial)?Mental maths and estimationCoreJump TradingChicago · 2013
Try it first
Pick the count before you work it out.
Show the worked solution
249 trailing zeros. A zero at the end is a factor of 10, and a 10 is a 2 times a 5. Among 1 to 1,000 there are 994 factors of 2 but only 249 factors of 5, so the 5s decide. Multiples of 5 contribute 200, multiples of 25 add a second 5 each for 40 more, multiples of 125 add 8, and 625 adds 1: 200 + 40 + 8 + 1 = 249.
Why count 5s and not 10s?
Imagine packing gift boxes that each need one lid and one base, from a pile of 994 lids and 249 bases. You can make 249 boxes, and the extra lids are useless. A trailing zero is a box: it needs one factor of 2 and one factor of 5, and in the product 1 x 2 x 3 x ... x 1,000 the 5s run out long before the 2s. Counting multiples of 10 misses every zero that comes from pairing a 5 in one number with a 2 in another; 4 x 5 is 20, which ends in a zero though neither factor is a multiple of 10. So the question is simply: how many times does 5 divide into 1,000 factorial?
Multiples of 5 up to 1,000 contribute 200 factors of 5, multiples of 25 a further 40, multiples of 125 another 8 and 625 one more, which add to 249, while the 994 factors of 2 are never the constraint.
How do you count the 5s without missing the doubled ones?
Count in layers. Every multiple of 5 carries at least one 5: 1,000 divided by 5 is 200. Every multiple of 25 carries a second 5 that the first layer did not see: 1,000 over 25 is 40. Multiples of 125 carry a third, 8 of them, and 625 carries a fourth, once. Each layer counts only the extra 5 that the layer before it missed, which is why you add the plain quotients and never multiply. The sum is 249. The same method gives 24 zeros for 100 factorial, 20 plus 4, and the layering stops as soon as the power of 5 exceeds the number.
The relationship
$Z(n) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \left\lfloor \frac{n}{625} \right\rfloor = 200 + 40 + 8 + 1 = 249$
Z(n) the number of trailing zeros of n factorial
floor of n over 5^k how many numbers up to n carry at least k factors of 5
What it says in wordsAdd the whole-number quotients of n by each power of 5 until the power is bigger than n.
What does the interviewer learn from how you say it?
The answer is a test of whether you decompose a question into its prime factors rather than reach for the obvious count. Say the pairing argument first, then the layered count, then check the shape: the answer is a little under n over 4, because 200 + 40 + 8 + 1 is a geometric series that sums towards 1,000 over 4. That check is useful on the follow-ups. For the number of factors of 2 the same layering gives 994; for 3s it gives 498, so in base 6, where a trailing zero needs a 2 and a 3, 1,000 factorial ends in 498 zeros. The limitation is worth a sentence: this counts zeros at the end, not zeros anywhere in the 2,568-digit number, which is a different and much harder question.
Where candidates lose it
The fast wrong answer is 100, one per multiple of 10, or 200, one per multiple of 5. The first misses every zero made by pairing a 5 with a 2 from a different number; the second misses the extra 5s inside 25, 125 and 625.
The other slip is counting 625 as four and 125 as three in the final sum, which double counts. Each layer adds only one new 5 per number, so the layers are added as plain quotients.
What the interviewer asks next
How many trailing zeros does 100! have?
How many trailing zeros does 1000! have when written in base 6? In base 12?
What is the smallest n for which n! ends in exactly 100 zeros, and is there an n that ends in exactly 5 zeros?
Asked at Jump Trading, Trading, Chicago, 2013 (Wall Street Oasis): How many 0's are in 1000! (factorial)?
029A point is dropped uniformly at random inside a unit square. What is the expected distance from the point to the nearest edge?Expected value and optimal stoppingCoreHudson River TradingNew York · 2024
Try it first
Pick the expected distance before you integrate.
Show the worked solution
One sixth. The distance to the nearest edge exceeds t only when the point lands inside the inner square of side 1 minus 2t, which has area (1 - 2t)^2. The expected value of a non-negative quantity is the integral of its tail probability, so E[D] is the integral of (1 - 2t)^2 from 0 to 1/2, which is 1/6, about 0.167. One axis alone would give 1/4; the second axis trims it.
Why work with the chance of being further than t, rather than the distance itself?
Suppose you want the average waiting time at a counter, and the only thing you can observe is, for each t, the fraction of people still waiting after t minutes. That is enough: add up the survival fractions over all t and you have the average wait. For any non-negative quantity, the expected value equals the area under its tail curve, and here the tail curve is easy to see: the point is further than t from every edge exactly when it lands inside the inner square of side 1 minus 2t. That inner square has area (1 - 2t)^2, and it shrinks to nothing at t equal one half, the centre of the square. Writing the density of the minimum of four dependent distances directly is far messier.
A random point is further than t from every edge only when it lands in the inner square of side 1 minus 2t, so the tail probability is (1 - 2t)^2, and the area under that curve from 0 to 1/2 is the expected distance of one sixth, confirmed by a 0.1667 Monte Carlo estimate.
How does the integral come out to exactly 1/6?
Substitute u = 1 - 2t. As t runs from 0 to 1/2, u runs from 1 down to 0, and dt is minus du over 2. The integral becomes one half of the integral of u squared from 0 to 1, which is one half times one third. The answer is 1/6 because a squared tail integrates to a third and the half-width of the square halves it again. A sanity check: the median distance is where (1 - 2t)^2 equals a half, which is t = 0.146, a little below the mean, as you expect for a distribution with a long right tail reaching 0.5.
The relationship
$E[D] = \int_0^{1/2} P(D > t)\,dt = \int_0^{1/2} (1-2t)^2\,dt = \frac{1}{2}\int_0^{1} u^2\,du = \frac{1}{6}$
D the distance from the point to the nearest edge
(1 - 2t)^2 the area of the inner square where every edge is further than t away
u the side of that inner square, 1 - 2t
What it says in wordsThe expected distance is the area under the tail curve, and the tail curve is the area of a shrinking inner square.
What is the fast cross-check that shows you understand the structure?
Do one axis first. The distance to the nearer of the left and right edges is min(x, 1 - x), a triangle-shaped quantity with mean 0.25. The nearest edge of the square is the smaller of two such independent quantities, one per axis, and taking the smaller of two pulls the mean down from 1/4 to 1/6. Say 1/4 for one axis, then 1/6 for two, and the interviewer hears that you see the minimum of independent pieces rather than a formula. The same structure gives 1/8 for a unit cube, where the tail is (1 - 2t)^3, and it shows the limitation of the method: it works because the inner region stays the same shape as it shrinks, which fails for a disc, where the nearest point on the boundary is not along an axis.
Where candidates lose it
The common loss is answering 1/4, the one-axis answer, forgetting that the nearest of four edges is smaller on average than the nearest of two. The second axis is independent and matters.
The other is trying to write the density of the minimum directly and running out of time. The tail curve is one line, (1 - 2t)^2, and the expected value is its integral. Learn that identity; it solves half the expected value questions a desk asks.
What the interviewer asks next
What is the expected distance to the nearest face for a point dropped in a unit cube?
What is the expected distance to the nearest edge in a 2 by 1 rectangle?
What is the expected distance from the point to the centre of the square? Why is that a different kind of integral?
Two points are dropped. What is the chance that both are further than 0.1 from every edge?
Asked at Hudson River Trading, Campus Algo Dev Interview, New York, 2024 (Wall Street Oasis): expected value question involving the expected value among distance to an edge, with a randomly placed object

n	c4(n)	shortfall
5	0.9400	6.0%
10	0.9727	2.7%
20	0.9869	1.3%
50	0.9949	0.5%
100	0.9975	0.2%

Wait (minutes)	Miss region	Meet probability
10	25/36	11/36, 30.6%
20	4/9	5/9, 55.6%
30	1/4	3/4, 75.0%
40	1/9	8/9, 88.9%

030A coin falls your way 60% of the time and pays even money. What fraction of your capital should you bet each time, and what growth rate does that give?Betting and sizingCoreOptiverAustin · 2025

Try it first

What fraction of your capital goes on each flip?

Show the worked solution

Bet 20% of your capital each time, for a growth rate of about 2.0% per bet. The Kelly fraction at even money is the edge, p minus q, which is 0.6 minus 0.4. The expected log growth is 0.6 x ln(1.2) + 0.4 x ln(0.8) = 0.0201, so capital typically grows by a factor of about 7.5 over 100 bets. Bet twice that, 40%, and the growth turns slightly negative, -0.24% a bet: anything beyond about 39% loses money despite the edge.

Why not bet everything on a coin that favours you?

A street vendor with a stall that makes money six days in seven does not spend the whole float on stock each morning, because the seventh day would end the business. Expected value is the right guide for a single bet with money you can replace, but when you must survive to play again, what matters is the growth of your capital over many bets, and that is governed by the average of the logarithm, not the average of the rupees. Betting all of it has the highest expected value and a certainty of ruin: one loss and there is nothing left to compound. The logarithm punishes that loss infinitely, which is the mathematical way of saying you cannot come back from zero.

The relationship

g(f) = p\ln(1+f) + q\ln(1-f), \qquad f^* = p - q = 0.2, \qquad g(0.2) = 0.6\ln 1.2 + 0.4\ln 0.8 \approx 0.0201

f	the fraction of capital staked on each bet
p, q	the chance of winning and of losing, 0.6 and 0.4
f star	the fraction that maximises long-run growth; at even money it is the edge
g	the expected log growth per bet, about 2% here

What it says in wordsMaximise the average log of your wealth after one bet, and at even money the best stake is simply the edge.

Expected log growth per bet rises from zero to a peak of 2.01% at a 20% stake and falls back through zero at about 39%, so half Kelly keeps three quarters of the growth while double Kelly, just past the crossing, loses a little on every bet.

Where does the 20% come from, and what if the odds are not even?

Differentiate g with respect to f and set it to zero: p over (1 + f) equals q over (1 - f), which gives f = p - q. At even money that is the edge, 0.2. For a bet that pays b to 1 the formula becomes (bp - q) over b, edge over odds. Say the general form, because the interviewer will often move the payout: at 2 to 1 on the same coin the stake becomes (1.2 - 0.4) / 2 = 40%, and at odds where bp is below q the right stake is zero. The growth at 20% is 0.0201 per bet, which compounds to a factor of e to the 2.01, about 7.5 times, over 100 bets on the typical path. That is the median outcome, not the average: a few lucky paths do far better and pull the mean up.

Why do desks bet less than Kelly?

The curve is flat on the left and steep on the right. Half Kelly, 10%, still delivers 1.50% a bet, three quarters of the peak, with far smaller swings. Double Kelly, 40%, delivers -0.24%, already below zero, with wild swings: the crossing point is about 39%, and anything beyond it shrinks capital in the long run even though every single bet has positive expected value. Over-betting is punished much harder than under-betting, and in real trading the 60% is an estimate rather than a fact, so sizing at a fraction of Kelly is the standard discipline. The limitation to state: Kelly assumes you know p, that bets are independent and repeated, and that you can resize freely; on a desk, position limits and the uncertainty in your edge usually bind before the formula does.

Where candidates lose it

The common loss is answering with the expected value: 60% of the time you win, so bet big, or even bet everything. The interviewer wants to hear that growth is about the log, that ruin ends the game, and that the stake is the edge, 20%.

The second is giving 20% but no growth rate, or quoting the 20% expected return per rupee as the growth rate. The growth per bet is about 2%, far below the 20% edge, because the losses bite on a shrinking base.

What the interviewer asks next

The coin pays 2 to 1 instead of even money. What is the stake now?
Why does half Kelly give three quarters of the growth? What is the general relationship?
You are only 60% sure the coin is 60% biased. How does that change the stake?
What is the probability that a full-Kelly bettor halves their capital at some point before doubling it?

Asked at Optiver, Quantitative Research, Austin, 2025 (Wall Street Oasis): Use Kelly Criterion

s	the sample standard deviation, the root of the n minus 1 sample variance
sigma	the true standard deviation
c4(n)	the exact correction factor for normal data, always below 1 and rising to 1 as n grows

R, G, B	the face shown by the red, green and blue die
9	the number of equally likely face pairs, three distinct faces on each die
5/9	each die's edge over the next one around the cycle

w	the waiting time in minutes, here 20
60	the length of the window in minutes
(1 - w/60)^2	the two miss triangles together, which fit into one square of side 1 - w/60

29, 24	Summer and Winter Games the United States took part in, as estimated
190, 62	average events per Games, one gold each
0.18, 0.07	the assumed United States share of golds in each

Try it first

Why build a chain instead of guessing a number?

Where does the packing factor come from, and why does it matter so much?

What do you say when the interviewer pushes on the usable fraction?

Where candidates lose it

What the interviewer asks next

Try it first

Why build from the floor rather than from the bank?

Which assumption should you spend your time on?

What would you add about occupancy?

Where candidates lose it

What the interviewer asks next

Try it first

Why does an unbiased variance give a biased standard deviation?

Does it matter for a 20-day volatility estimate?

What assumption does the exact factor need?

Where candidates lose it

What the interviewer asks next

Try it first

How can three dice with the same average not have a best one?

How do you check a pair quickly in the room?

What is the trading lesson the interviewer is after?

Where candidates lose it

What the interviewer asks next

Try it first

Why turn two arrival times into a square?

Why is it easier to compute where they miss?

How does the answer move with the waiting time?

Where candidates lose it

What the interviewer asks next

Try it first

Why is 23.5 wrong when every roll averages 3.5?

Why does 6 x 3.5 still give the right answer?

Where does this reasoning show up on a desk?

Where candidates lose it

What the interviewer asks next

Try it first

What is the interviewer actually marking?

How do you build the middle?

How wide should the quote be, and what happens when someone trades?

Where candidates lose it

What the interviewer asks next

Try it first

Why count 5s and not 10s?

How do you count the 5s without missing the doubled ones?

What does the interviewer learn from how you say it?

Where candidates lose it

What the interviewer asks next

Try it first

Why work with the chance of being further than t, rather than the distance itself?

How does the integral come out to exactly 1/6?

What is the fast cross-check that shows you understand the structure?

Where candidates lose it

What the interviewer asks next

Try it first

Why not bet everything on a coin that favours you?

Where does the 20% come from, and what if the odds are not even?

Why do desks bet less than Kelly?

Where candidates lose it

What the interviewer asks next