Derivatives Foundation puzzles, solved step by step

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

Showing 11–20 of 100

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?
016You walk into a casino with Rs 63,000 and bet Rs 1,000 on red at even money, where red comes up 48% of the time. Every time you lose, you double the bet. You stop at the first win, or when you cannot cover the next bet. What is the chance you lose everything, and what is your expected result?Betting and sizingCoreRisk managementProp trading firms
Try it first
Before any arithmetic: the plan ends a session up Rs 1,000 about 98 times in 100. What is its expected result per session?
Show the worked solution
You lose everything about 2.0% of the time, 0.52 to the sixth power, and the expected result is about minus Rs 265. Rs 63,000 covers exactly six bets: 1, 2, 4, 8, 16 and 32 thousand. A win at any of them recovers every earlier loss and nets Rs 1,000, which happens 98.0% of the time. Six losses in a row cost all Rs 63,000. Weighted, 980 of expected winnings against 1,246 of expected loss leaves minus Rs 265.
Why does a plan that wins 98 times in 100 still lose money?
Picture a friend who sells phone insurance to classmates for Rs 50 a month. Month after month nobody drops a phone, and the Rs 50 notes pile up; it feels like free money until the month three phones go into a pond. A win rate tells you how often you are paid, not how much you are paid against how much you can lose, and the expected value needs both. Doubling after every loss builds exactly that shape: Rs 1,000 collected almost every time, and Rs 63,000 handed back rarely. The rare branch is 63 times the size of the common one, so a 2% chance of it more than cancels a 98% chance of the small win.
The doubling plan ends a session up Rs 1,000 with probability 98.0% and down Rs 63,000 with probability 2.0%, and weighting the two gives plus 980 against minus 1,246, an expected result of minus Rs 265, which is also 4% of the Rs 6,633 the plan expects to stake.
How do you lay out the six bets in the room?
Write the ladder down before computing anything. The stakes are 1, 2, 4, 8, 16 and 32 thousand, which add to 63 thousand exactly, so the seventh bet of 64 thousand can never be placed. If the first win comes at bet k, it pays 2 to the power k minus 1 thousand, and the losses before it add to one thousand less than that, so every winning session nets exactly plus Rs 1,000. There are only two outcomes, and the table shows how quickly the chance of reaching each rung falls: by the sixth bet you are staking Rs 32,000 to recover Rs 31,000 of losses and win one more thousand.
Bet Stake (Rs) Lost before it (Rs) Chance of reaching it
1 1,000 0 100.0%
2 2,000 1,000 52.0%
3 4,000 3,000 27.0%
4 8,000 7,000 14.1%
5 16,000 15,000 7.3%
6 32,000 31,000 3.8%
Each rung doubles the stake while the chance of reaching it falls by a factor of 0.52, and the chance of losing the sixth bet as well is 1.98%, the probability of ruin.
The relationship
$E = (1 - 0.52^6)(+1{,}000) + 0.52^6(-63{,}000) = 0.9802(1{,}000) - 0.0198(63{,}000) \approx -265$
0.52^6 the chance of six losses in a row, about 2%
+1,000 the net result of any session that wins before the money runs out
-63,000 the whole bankroll, lost when all six bets lose
What it says in wordsThe expected result is the frequent small win times its probability plus the rare total loss times its probability, and the second term is larger.
Is there a faster way to see the sign without the ladder?
Yes, and it is the one a trader reaches for first. Every rupee placed on red loses 4 paise on average, whatever happened on the previous spin, because the wheel has no memory. The expected result of any staking plan is the edge per rupee times the expected total amount staked: here minus 4% of Rs 6,633, which is minus Rs 265, the same figure as the ladder. Doubling raises the amount you put down when you are losing; it cannot change the sign of the edge. On a fair 50/50 wheel the same plan has an expected value of exactly zero, with the same lopsided shape.
Why does a desk interviewer care about a roulette plan?
Because the shape is the shape of selling far out-of-the-money options, or of adding to a losing position to get back to flat. Both produce a long run of small gains and a rare large loss, and a good-looking track record says almost nothing about the tail. Repetition makes the rare branch common: play 50 sessions and the chance of at least one ruin is 1 minus 0.98 to the 50th, about 63%. The limitation to state is that the plan assumes no table limit; a casino maximum bet cuts the ladder short and makes ruin more likely, not less.
Where candidates lose it
The common answer is that the plan wins, because it almost always wins. Candidates quote the 98% and stop, never weighing it against the size of the 2% branch. A probability without a payoff is half an expected value.
The second loss is the opposite slip: computing minus 4% of the Rs 63,000 bankroll, about minus Rs 2,520. The edge applies to rupees actually staked, and most sessions stake only Rs 1,000 or Rs 3,000 before the first win. Expected stake, Rs 6,633, is the base.
What the interviewer asks next
The wheel is fair, 50/50. What is the expected result now, and what is the chance of ruin?
You have unlimited money but the table caps any single bet at Rs 16,000. How does the picture change?
Name a trading strategy with the same payoff shape, and say how you would size it.
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
018A stock trades at Rs 1,000 and its options are priced at 16% implied volatility. You buy an at-the-money option and delta-hedge it every day. Roughly how large a daily move does the stock need to make for you to break even?Option pricing intuitionWarm upVolatility tradingMarket making
Try it first
Answer in your head before reading on: the breakeven daily move is about
Show the worked solution
About 1% a day, Rs 10. With roughly 256 trading days in a year and volatility growing with the square root of time, daily volatility is annual volatility divided by 16, so 16% a year is 1% a day. A delta-hedged long option earns half its gamma times the square of each day's move and pays theta every day; the two cancel when the move equals the implied daily move. Using 252 days gives 1.008%, still Rs 10.
Where does dividing by 16 come from?
Walk randomly on a straight road, one step forward or back each second, and after 100 seconds you are typically about 10 steps from where you began, not 100, because most steps undo each other. Price moves add up the same way. Volatility scales with the square root of time, and the square root of 256 trading days is 16, so an annual volatility divided by 16 is the standard deviation of one day's move. Traders call this the rule of 16. It makes 16% implied volatility the cleanest number on the screen: 1% a day, Rs 10 on this stock.
The relationship
$\sigma_{\text{day}} = \frac{\sigma_{\text{year}}}{\sqrt{256}} = \frac{16\%}{16} = 1\%,\qquad 1\% \times 1{,}000 = \text{Rs } 10$
sigma year the implied volatility, quoted per year
256 trading days in a year, rounded so the square root is a whole number
sigma day the standard deviation of one day's percentage move
What it says in wordsOne day's typical move is the annual volatility divided by the square root of the number of trading days.
Why does that daily move decide whether the hedged option makes money?
Once the delta is hedged, the option's daily P&L is two pieces. Gamma pays you half gamma times the square of the move, in either direction; theta charges you a fixed amount for the day passing. For a one-month at-the-money option here, gamma is 0.0087 per rupee and theta is 0.435 a day. A Rs 10 move earns 0.5 x 0.0087 x 100 = 0.435, exactly the theta, because the option's price was built so that theta pays for a move of one implied standard deviation. A flat day loses 0.435; a Rs 20 day makes 1.31.
A delta-hedged long at-the-money option loses its theta of 0.435 on a flat day, breaks even when the stock moves Rs 10 either way, and makes 1.31 on a Rs 20 move, because the gain grows with the square of the move while theta is fixed, and Rs 10 is 16% divided by 16.
What does the quick answer leave out?
Three things, and naming one earns the follow-up. First, the breakeven is on the average squared move, not the average move. If the stock's moves are normal with a standard deviation of Rs 10, its average absolute move is only about Rs 7.98, so a stock that typically moves Rs 8 a day is already moving enough to break even. Second, gamma changes as the stock drifts away from the strike and as expiry nears, so the Rs 10 holds for an at-the-money option on the day you measure it. Third, hedging once a day adds noise to the P&L even when realised volatility exactly matches implied. The rule of 16 is a desk shortcut, not a pricing model.
Where candidates lose it
The common slip is dividing 16% by the number of trading days, or by 365, and quoting a breakeven of a few paise. Volatility adds in squares, so time enters under a square root; dividing by days is the mistake the question is built to catch.
The second loss is quoting Rs 10 as an average move to expect every day. It is a standard deviation: plenty of days will move Rs 2 and a few will move Rs 25, and the hedged option breaks even only if the average of the squared moves matches 100.
What the interviewer asks next
The same stock's options are priced at 32% volatility. What is the breakeven move, and what is the theta in terms of gamma?
Over a week the stock moves 5, minus 12, 3, 15 and minus 9. Did a delta-hedged long option make or lose money, roughly?
Why might a trader quote 252 days rather than 256, and when does the difference matter?
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.)
020One glass holds 100 ml of wine and another holds 100 ml of water. You take a spoonful of wine, tip it into the water and stir. Then you take a spoonful of the mixture and tip it back into the wine glass. Is there now more wine in the water glass, or more water in the wine glass?Games and logicWarm upProp trading firms
Try it first
Decide before any arithmetic: after the two spoonfuls,
Show the worked solution
Exactly the same. Each glass ends with 100 ml, so whatever wine is missing from the wine glass has been replaced, millilitre for millilitre, by water, and the missing wine can only be in the water glass. With a 10 ml spoon and a thorough stir, the return spoon carries back 0.91 ml of wine and 9.09 ml of water, leaving 9.09 ml of water in the wine and 9.09 ml of wine in the water.
Why does the first spoon feel like it settles the question?
Because it is pure wine going one way and a diluted mixture coming back, so it feels as though more wine travelled. Think instead of two cricket teams of eleven who swap some players and still field eleven each. Each glass ends with exactly 100 ml, so every millilitre of wine that left the wine glass and did not come back has been replaced by a millilitre of water: the two foreign amounts must be equal. The number of team A players now in team B is the number of team B players now in team A, however the swaps were done.
With a 10 ml spoon, the wine glass goes from 100 ml of wine to 90 ml and then back to 100 ml holding 9.09 ml of water, while the water glass goes to 110 ml and back to 100 ml holding 9.09 ml of wine, so the two foreign amounts are equal.
What do the millilitres actually look like?
Take a 10 ml spoon. After the first transfer the water glass holds 100 ml of water and 10 ml of wine, 110 ml in all, so a stirred spoonful from it is 10/110 wine. The return spoon carries 0.91 ml of wine and 9.09 ml of water, so 9.09 ml of wine stays behind in the water glass and 9.09 ml of water arrives in the wine glass. The arithmetic confirms the argument, but the argument came first and did not need the spoon size, the stirring or any division.
The relationship
$\text{water in wine} = 10 \times \frac{100}{110} = 9.09,\qquad \text{wine in water} = 10 - 10 \times \frac{10}{110} = 9.09$
10 the spoon, in millilitres
100/110 the share of water in the stirred water glass after the first transfer
10/110 the share of wine in that glass
What it says in wordsThe water carried into the wine glass equals the wine left behind in the water glass, both 9.09 ml for a 10 ml spoon.
Why do the interviewer's variations not change the answer?
Interviewers vary the story: no stirring, five spoonfuls back and forth, a ladle instead of a spoon. As long as both glasses end at their starting volume, the answer is equal, because the argument uses only the totals. The limitation to say out loud: if the return spoon is a different size from the first, the glasses end at different volumes and the amounts differ, so check the volumes before using the shortcut. The desk lesson is the bookkeeper's: in a closed system, look at the totals before tracking every transfer, the same way a net position check catches a booking error faster than replaying every ticket.
Stage Wine glass Water glass
Start 100 wine 100 water
After spoon 1 90 wine 100 water + 10 wine
After spoon 2 90.91 wine + 9.09 water 90.91 water + 9.09 wine
Tracking a 10 ml spoon through both transfers leaves each glass at 100 ml with 9.09 ml of the other liquid, which is what the conservation argument predicted without any arithmetic.
Where candidates lose it
The common answer is more wine in the water, because the first spoon was undiluted. It anchors on one transfer and forgets that the second spoon also took some of that wine back.
The second loss is reaching the right answer by long arithmetic and then failing the follow-up, such as an unstirred glass or several transfers, because there was no argument underneath. Give the volume argument first and use the numbers only as a check.
What the interviewer asks next
The return spoon is 5 ml instead of 10 ml. Which glass now holds more of the other liquid, and by how much?
You repeat the two-spoon swap many times. What do both glasses converge to?
Where on a trading desk does checking a total first save you from tracking every transfer?

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

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

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%

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

Try it first

Why is the question about gaps and not about points?

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

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

Where candidates lose it

What the interviewer asks next

Try it first

Why keep three numbers in your head and not one?

Was the loss a mistake?

What would you have done differently in the game?

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 does a plan that wins 98 times in 100 still lose money?

How do you lay out the six bets in the room?

Is there a faster way to see the sign without the ladder?

Why does a desk interviewer care about a roulette plan?

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

Where does dividing by 16 come from?

Why does that daily move decide whether the hedged option makes money?

What does the quick answer leave out?

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

Why does the first spoon feel like it settles the question?

What do the millilitres actually look like?

Why do the interviewer's variations not change the answer?

Where candidates lose it

What the interviewer asks next