Derivatives Foundation puzzles, solved step by step

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

Showing 1–10 of 66 · filtered from 100Clear filters

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

003A logger stamps every event to the nanosecond, nine decimal places, and whenever the timestamp is missing it writes nine zeros instead. In 100,000 records you find 15 whose fractional part is exactly nine zeros. What is the probability that at least one of those 15 is a filled-in missing value?Conditional probability and BayesHardJump TradingAnonymous interview candidate in · 2022

Try it first

First instinct: roughly how many genuine timestamps, out of 100,000, should end in nine zeros by chance?

Show the worked solution

Essentially 1; the 15 are gaps. A genuine stamp ends in nine zeros with probability 10^-9, so in 100,000 records you expect 0.0001 such endings. The chance that 15 or more arise genuinely is about 10^-72, which no reasonable prior on missing data can overcome: even a missing rate of one in ten thousand would produce about 10 filled-in endings. So the probability that at least one of the 15 is a filled-in value is 1 to every decimal place you could print.

What is the question really asking you to compare?

A shopkeeper who finds 15 notes with the same serial number does not ask what the chance of a coincidence is; she asks which explanation makes 15 identical notes likely. This is a Bayes question in disguise: compare how likely 15 all-zero endings are if nothing is missing against how likely they are if some values are missing, then weight by a prior. The first likelihood is astronomically small; the second is ordinary. The prior would need to be more extreme than anything a real system justifies to change the answer.

Genuine nanosecond stamps should produce 0.0001 all-zero endings in 100,000 records, while 15 were observed, five orders of magnitude more, and the chance of 15 or more genuine ones is about 10^-72, so the observation is explained only by filled-in missing values.

How do you put a number on the genuine case?

Each of the 100,000 stamps ends in a specific nine-digit string with probability one in a billion, so the count of genuine all-zero endings is Poisson with mean 0.0001. The probability of exactly 15 is e^(-0.0001) times 0.0001^15 over 15 factorial, which is about 10^-72. You do not need the exact figure in the room; say that 0.0001 to the fifteenth power is 10^-60 before dividing by 15 factorial, and the interviewer has what they need. The point is to show you can set up the count, not to print 72 zeros.

The relationship

P(\geq 1 \text{ missing} \mid 15) = 1 - \frac{P(15 \mid \text{none}) \, P(\text{none})}{P(15)} \approx 1 - \frac{10^{-72} \, P(\text{none})}{P(15)} \approx 1

P(15 \| none)	the chance of 15 genuine all-zero endings when no value is missing, Poisson with mean 0.0001
P(none)	your prior that the data set has no missing values at all
P(15)	the overall chance of seeing 15, dominated by the missing-value explanation

What it says in wordsThe probability that none are missing is the genuine likelihood times its prior, divided by the total, and the genuine likelihood is so small that the result rounds to 1 whatever prior you hold.

What does the interviewer want to hear about the prior?

The honest answer is that the question is underspecified: without a prior on how often values go missing, you cannot write a single number. Say that, then show it does not matter: for the posterior to drop even to 99.9% you would need a prior of no missing values more than 10^69 times stronger than the alternative, and no logging system earns that confidence. For contrast, a modest missing rate of one record in ten thousand would give an expected 10 filled-in endings, right where the observed 15 sits. A limitation worth adding: the argument assumes the genuine fractional digits are uniform, which breaks if the clock quantises to microseconds and pads with zeros itself.

Where candidates lose it

Candidates reach for the binomial probability of 15 genuine zeros and stop, reporting a tiny number as if it were the answer. The question asks for the probability of a missing value given the data, which needs the comparison with the alternative, not a single likelihood.

The second loss is freezing because no prior is given. The strong move is to name the missing input, then show that the likelihood ratio is so lopsided that the prior cannot matter. That is what a desk wants: a conclusion that survives the unknown.

What the interviewer asks next

Now the logger stamps to the microsecond, six digits, and pads with three zeros. Does the argument survive?
Suppose only 1 record ends in nine zeros. What would you conclude then, and what would you need to know?
How would you check the data itself rather than reason about it?

Asked at Jump Trading, Prop Trading, Anonymous interview candidate in, 2022 (Wall Street Oasis): What is the probability of at least 1 missing value given that we see 15 data points with 0's in the end

006A trade surveillance system raises an alert on 95% of genuinely suspicious trades and, wrongly, on 2% of normal trades. One trade in a thousand is genuinely suspicious. An alert has just fired on a trade. What is the probability the trade is suspicious?Conditional probability and BayesWarm upCitadelMiami · 2022
Try it first
Gut answer before you count anything.
Show the worked solution
About 4.5%. Count 100,000 trades. One in a thousand is suspicious, so 100 are, and 95 of those alert. The other 99,900 are normal, and 2% of them, 1,998, alert anyway. Alerts total 2,093, of which 95 are genuine, so the probability that an alerted trade is suspicious is 95 over 2,093, about 4.5%. The 95% hit rate is not the answer; the base rate is what decides it.
Why does a 95% accurate system give a 4.5% answer?
A smoke alarm that goes off for 2% of toast is a fine alarm in a house that is never on fire; nearly every ring will be toast. When the thing you are looking for is rare, even a small false-alarm rate applied to the huge normal pile produces more alerts than the true cases produce. Here 2% of 99,900 normal trades is 1,998, twenty times the 95 genuine alerts. The system is not bad; the base rate is low, and that is what the question is testing.
Of 100,000 trades, 100 are suspicious and raise 95 true alerts, while the 99,900 normal trades raise 1,998 false ones, so alerts total 2,093 and a trade that alerts is genuinely suspicious only 4.5% of the time.
The relationship
$P(S \mid A) = \frac{P(A \mid S)\,P(S)}{P(A \mid S)\,P(S) + P(A \mid N)\,P(N)} = \frac{0.95 \times 0.001}{0.95 \times 0.001 + 0.02 \times 0.999} = \frac{95}{2{,}093} \approx 4.5\%$
S, N a suspicious trade, a normal trade
A an alert fires
P(A | S) = 0.95 the hit rate
P(A | N) = 0.02 the false-alarm rate
P(S) = 0.001 the base rate
What it says in wordsTrue alerts divided by all alerts, where all alerts are the true ones plus the false ones from the normal pile.
What is the fastest way to say it in the room?
Do not write Bayes' formula; count a round number of trades. Say: in 100,000 trades, 100 are suspicious and 95 alert; 99,900 are normal and 1,998 alert; 95 over 2,093 is about 4.5%. Three sentences, no algebra, and every number is checkable by the person listening. The odds form is just as quick: prior odds 1 to 999, likelihood ratio 0.95 over 0.02, about 47.5, so posterior odds 47.5 to 999, roughly 1 to 21.
What does the desk do with a 4.5% answer?
It decides what the alert is for. A 4.5% hit rate is fine for a filter that sends trades to a human for a second look, and useless for an automatic block, because 95% of blocked trades would be legitimate business. That is the trade-off every surveillance, fraud and risk-limit system lives with: a lower threshold catches more of the 100 but drags in more of the 99,900. Say the limitation too: the 2% and 95% are themselves estimates from past data, and a system tuned on last year's patterns can drift.
Where candidates lose it
The whole trap is answering 95%, confusing the probability of an alert given a suspicious trade with the probability of a suspicious trade given an alert. Interviewers ask this precisely because the two sound the same and are twenty times apart.
The second loss is reaching for the formula and tangling the denominator. Count 100,000 trades and the denominator builds itself: 95 plus 1,998.
What the interviewer asks next
The false-alarm rate is cut to 0.5%. What is the probability now?
Two independent systems both alert on the same trade. What is the probability it is suspicious?
What base rate would make an alert a coin flip, and what does that tell you about where surveillance is worth running?
Asked at Citadel, Sales and Trading, Miami, 2022 (Wall Street Oasis): I got a question about Bayes' theorem applied to a practical scenario, which I handled decently

009Three players each hold one hidden card from a standard deck, ace counting 1 up to king counting 13. You can see only your own card, a 10. You must make a two-way market on the total of all three cards. The player on your left lifts your offer, you requote, and he lifts again. What do you do now?Market makingHardOptiverChicago · 2025

Try it first

Before the second lift: what is a fair value for the total when all you know is your own 10?

Show the worked solution

Raise the market and widen it; do not sell a third time near the old level. With only your 10 known, the fair total is 24 and the two unknowns give a standard deviation of about 5.3. A player who sees his own card and lifts your offer is telling you his card is high: if it is 8 or more the fair total is 27.5; after a second lift, 11 or more, it is 29. You are short 2 at an average near 27 against a value near 29. Quote something like 28 at 32.

What does a lift tell you that the cards do not?

If you are selling a second-hand bike and the first viewer pays your asking price without haggling, you have probably priced it low; if he immediately asks to buy a second one, you certainly have. A counterparty who can see something you cannot and keeps buying is telling you the value is higher than your offer, and every trade you do with him before repricing is a trade you will regret. This is adverse selection, and a market-making game exists to see whether you notice it in time.

With your 10 the prior fair total is 24; one lift suggests the buyer's card is 8 or more and moves the fair value to 27.5; a second lift suggests 11 or more and moves it to 29, so the quote climbs from 23 at 25 to 28 at 32 and widens from 2 to 4 while you sit short 2.

The relationship

E[\text{total}] = 10 + E[L] + 7,\qquad E[L \mid L \ge 8] = 10.5,\quad E[L \mid L \ge 11] = 12 \;\Rightarrow\; 24 \to 27.5 \to 29

L	the card held by the player who keeps lifting
7	the mean of the third player's card, still unknown and uniform on 1 to 13
L >= 8, L >= 11	the rough information in a first and a second lift: his card is above what your offer implied

What it says in wordsEach lift raises your estimate of the lifter's card, and the fair total moves by exactly that amount.

How much do you move, and how much do you widen?

Move at least as far as the information says and widen because your uncertainty about his behaviour has grown. After one lift a quote of 26 at 29 sits around the new estimate of 27.5; after two lifts 28 at 32 sits around 29 and is twice as wide, because the next lift would mean his card is 12 or 13 and the total near 30 or more. The width is not a penalty on him; it is the price of your own blindness. A standard deviation of 5.3 on the total from the two unknown cards is the natural scale for the width before any lifts.

What about the position you already have?

You are short 2 at an average near 27 and the value is near 29, so you are losing about 4 on paper. Do not try to earn it back by selling more at a worse price, and do not flip to buying from the other player at any cost; skew your quote up so the next trade is more likely to reduce the short than add to it. Say your position out loud when the interviewer asks; the game checks whether you can hold the fair value, the quote and the inventory in your head at once. The limitation to state: the thresholds 8 and 11 are a rough model of his behaviour, and a player who bluffs changes the inference.

Where candidates lose it

The common failure is to keep quoting 23 at 25 after the first lift, and to sell a third unit at 25 after the second, because the cards have not changed. Your information has changed. Two lifts from a player who sees his own card are worth more than the deck statistics.

The second loss is overreacting: moving the quote to 35 at 40 after one lift. He may hold a 9. Move to what the evidence supports, widen for what it does not, and keep your position in mind.

What the interviewer asks next

Now the third player hits your bid at 28. How do you reprice?
What if the lifter can see your card as well as his own?
Make a market on the product of the three cards instead of the sum. What changes about the width?

Asked at Optiver, Quantitative Research, Chicago, 2025 (Wall Street Oasis): The next round was a poker style market making game as well as a separate behavioural interview

011n points are dropped at random on a circle of circumference 1. Each point colours in the arc between itself and its nearest neighbour. As n grows large, what fraction of the circle do you expect to be coloured?Probability and countingHardSusquehanna International GroupLondon · 2026

Try it first

First instinct: a gap between two neighbouring points stays blank when?

Show the worked solution

7/18 of the circle, about 38.9%. A gap between neighbouring points stays blank only when it is longer than both gaps beside it. For large n the gaps behave like independent exponentials with the same mean, and the expected length of the longest of three is 1 + 1/2 + 1/3 = 11/6 times the mean. Any one of the three is longest a third of the time, so the blank share of length is (11/6)/3 = 11/18, and the coloured share is 1 minus that, 7/18.

Why is the question about gaps and not about points?

Think of houses along a ring road where each household paints the stretch of road to its nearest neighbour. A stretch of road gets paint from the house at either end, so to find the unpainted road you ask which stretches are chosen by neither house. A gap is blank exactly when it is longer than both of its neighbouring gaps, because then each of its end points has a closer neighbour on the other side. That turns the problem into a question about one gap and its two neighbours, which is small enough to solve.

Fourteen random points cut the circle into fourteen gaps; the green gaps are coloured because each is shorter than at least one neighbour, the grey gaps are blank because each beats both neighbours, and in the limit the blank gaps carry 11/18 of the length and the coloured ones 7/18, about 38.9%.

Why 11/18 and not 1/3 for the blank share?

Each gap is the longest of its three with probability 1/3, by symmetry. But the question asks for length, not count, and the gaps that stay blank are the long ones. The blank share is the expected length of a gap that is the longest of three, divided by the mean gap, which for exponential gaps is (1 + 1/2 + 1/3)/3 = 11/18. Count and length give different answers because being blank is correlated with being long. That distinction is the whole difficulty of the question, and saying it out loud is most of the marks.

The relationship

E[\max(G_1, G_2, G_3)] = \mu\left(1 + \tfrac{1}{2} + \tfrac{1}{3}\right) = \tfrac{11}{6}\mu \quad\Rightarrow\quad \text{blank} = \frac{\tfrac{1}{3}\cdot\tfrac{11}{6}\mu}{\mu} = \frac{11}{18},\qquad \text{coloured} = \frac{7}{18}

G_1, G_2, G_3	a gap and its two neighbours, approximately independent exponentials for large n
mu	the mean gap, 1/n
1 + 1/2 + 1/3	the expected maximum of three unit exponentials, from the memoryless property

What it says in wordsA gap's expected blank length is a third of the expected longest of three gaps, and dividing by the mean gap gives the blank share of the circle.

Where does the 1 + 1/2 + 1/3 come from, and what are you assuming?

Three exponential clocks run together. The first to ring takes an expected 1/3 of the mean; then two remain, memoryless, and the next takes 1/2; the last takes a full mean. Adding gives 11/6 for the longest. The assumption is that neighbouring gaps are independent, which is exact in the limit of many points and only approximate for small n, where the gaps must sum to 1. A quick simulation with 2,000 points gives a coloured share of 0.388, against 7/18 = 0.389. The exact answer for any n differs slightly and settles to 7/18 as n grows.

Where candidates lose it

The common wrong answer is 2/3, from the count: each gap is the longest of three one time in three, so one third of the gaps are blank. The blank gaps are the long ones, so by length they carry more than a third, 11/18.

The second loss is trying to integrate over the joint distribution of n spacings. The limit is a three-gap problem with exponential gaps, and a desk wants the memoryless argument, not the integral.

What the interviewer asks next

What is the expected number of blank gaps when there are n points?
Now each point colours the arcs to both of its neighbours. What changes?
Why do the gaps between uniform points on a circle look exponential when n is large?

Asked at Susquehanna International Group, Quantitative Research, London, 2026 (Wall Street Oasis): if n points are placed on a circle and each point colours in the arc to its nearest neighbour, what is the expected length of coloured circumference

012You are making a market on a contract that settles at the sum of two dice. During the game you sell 5 at 7.5, buy 3 at 6.5 and sell 2 at 8. The dice are rolled and total 9. What is your final position, what is your profit or loss in rupees, and what was your expected profit at the moment you finished trading?Market makingWarm upOptiverAmsterdam · 2023
Try it first
Before the blotter: what is the fair value of the sum of two dice?
Show the worked solution
You finish short 4, you lose Rs 2 on the settlement, and your expected profit when you stopped trading was Rs 6. Sold 5, bought 3, sold 2 is a net short of 4. Cash is +37.5 - 19.5 + 16 = +34. Settling at 9 costs 4 x 9 = 36, so the result is 34 - 36 = -2. Against the fair value of 7 the short would have cost 28, leaving +6: the edge of 2.5 + 1.5 + 2 captured on the three trades.
Why keep three numbers in your head and not one?
A shopkeeper who sells umbrellas at a markup has a profit on each sale, a stock count, and a worry about whether it rains. Three separate things. A market maker tracks the same three: edge per trade against fair value, net position, and the exposure to the final number, and the game checks that you never let one of them slip. Say the fair value, 7, first. Then say each trade's edge as you do it: +2.5 on selling 5 at 7.5, +1.5 on buying 3 at 6.5, +2 on selling 2 at 8. Then say the position: short 4.
Selling 5 at 7.5, buying 3 at 6.5 and selling 2 at 8 leaves a short of 4 and cash of +34 with an edge of 2.5, 1.5 and 2 against fair value 7, so at the settlement of 9 the position costs 36 and the result is Rs 2 lost, while at fair value the same trades were worth Rs 6.
The relationship
$\text{P\&L} = \underbrace{5(7.5) - 3(6.5) + 2(8)}_{\text{cash} = 34} + \underbrace{(-4)}_{\text{position}} \times S, \qquad S = 9 \Rightarrow -2,\quad S = 7 \Rightarrow +6$
cash money received for sales minus money paid for purchases
position contracts bought minus contracts sold, here minus 4
S the settlement value of the contract, the dice total
What it says in wordsProfit is the cash already banked plus the position times the settlement, and replacing the settlement with the fair value gives the expected profit.
Was the loss a mistake?
No. Every trade was done at a better price than fair value, so the trading was right; the dice came in high. Expected profit of +6 is what you controlled, and the realised minus 2 is what the dice did; an interviewer wants to hear you separate the two without being asked. The sum of two dice has a standard deviation of about 2.4, so a short of 4 carries a one-standard-deviation swing of nearly 10, far larger than the 6 of edge. The real question is whether a short of 4 was more risk than you wanted to carry against 6 of edge.
What would you have done differently in the game?
Skewed the quote as the short grew. After selling 5, you are short 5 and should lower both your bid and your offer so that the next trade is more likely to be a buy that cuts the position, which is exactly what buying 3 at 6.5 did. Selling 2 more at 8 added to the short again; at that point a wider or higher quote would have protected you. The limitation of the puzzle: with two interviewers trading against you, their trades carry information about nothing, because the dice are not rolled yet, so here the only reason to skew is inventory, not adverse selection.
Where candidates lose it
Candidates lose the position count under pressure, saying short 6 or short 2 because they forget the buy of 3. State the running position after every trade, aloud, as the sample blotter does.
The second loss is reporting the minus 2 as if the trading was bad. The expected profit was plus 6 and the dice were unkind. Say both numbers and which one you controlled.
What the interviewer asks next
The dice settle at 5 instead. What is your P&L, and does your expected P&L change?
What is the standard deviation of the two-dice total, and how does it size the risk of being short 4?
After the first sale of 5 at 7.5, what market would you show next, and why?
Asked at Optiver, Prop Trading, Amsterdam, 2023 (Wall Street Oasis): some difficult trading games where you had to profit making a market whilst remembering your position and the position of two interviewers

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

S, N	a suspicious trade, a normal trade
A	an alert fires
P(A \| S) = 0.95	the hit rate
P(A \| N) = 0.02	the false-alarm rate
P(S) = 0.001	the base rate

cash	money received for sales minus money paid for purchases
position	contracts bought minus contracts sold, here minus 4
S	the settlement value of the contract, the dice total

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%

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

What is the question really asking you to compare?

How do you put a number on the genuine case?

What does the interviewer want to hear about the prior?

Where candidates lose it

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Why does a 95% accurate system give a 4.5% answer?

What is the fastest way to say it in the room?

What does the desk do with a 4.5% answer?

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Was the loss a mistake?

What would you have done differently in the game?

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Which assumption should you spend your time on?

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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