Derivatives Foundation puzzles, solved step by step

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

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034A card is drawn from a 52-card deck. You may bet Rs 100 at even money on red or on black. How much would you pay to be told first whether the card is a heart?Expected value and optimal stoppingCoreOptiverChicago · 2025
Try it first
What is the tip worth, at most?
Show the worked solution
Up to Rs 50. Without the tip the bet is a coin flip worth zero. With it: one time in four the card is a heart, you bet red and win Rs 100 for certain. Three times in four it is not a heart, which leaves 13 red and 26 black cards, so you bet black and win 100 with probability 2/3 and lose 100 with probability 1/3, an expected Rs 33.33. The average, 25 + 25, is Rs 50, and that is the most the information is worth.
Why is a tip about hearts worth anything for a bet on colour?
A friend who will tell you whether it is raining in one of the four districts of a city is not telling you the weather everywhere, but if you have to bet on whether it rains in the city at all, the tip changes your odds. The heart tip never names the colour, but every heart is red, so a yes makes red certain and a no tilts the remaining deck two to one towards black. Both answers leave you with a bet that has an edge, where before you had none. Information is worth the gap between what you can earn with it and what you could earn without it, and here that gap is the whole value, because without the tip the best you can do is zero.
Without the tip a bet on red or black has an expected value of zero, while with it a heart lets you win Rs 100 for certain and a non-heart lets you bet black at 26 to 13 for an expected Rs 33.33, so the tip is worth a quarter of 100 plus three quarters of 33.33, which is Rs 50.
How do you set the calculation out so it cannot go wrong?
Price the decision in each branch, then weight the branches by how likely each answer is. If the answer is yes, probability 13/52, the card is red: bet red, expected gain 100. If the answer is no, probability 39/52, there are 39 cards left of which 13 are red diamonds and 26 are black: bet black, expected gain 100 x 26/39 minus 100 x 13/39, which is 33.33. The value of the tip is the probability-weighted average of the best you can do after each answer, minus the best you could do with no answer at all. That is 1/4 x 100 + 3/4 x 33.33 = 50, minus zero. Pay less than Rs 50 and the deal is in your favour; pay exactly 50 and you are indifferent.
The relationship
$V = \tfrac{13}{52}\,(100) + \tfrac{39}{52}\left(100\cdot\tfrac{26}{39} - 100\cdot\tfrac{13}{39}\right) - 0 = 25 + 25 = 50$
13/52 the chance the card is a heart
100 a certain win on red once you know it is a heart
26/39, 13/39 the chance of black and of red among the 39 non-hearts
0 the value of the bet with no information
What it says in wordsWeight the best decision after each possible answer by the chance of that answer, and subtract what the bet was worth before.
What does this have to do with a trading desk?
The question is a small model of paying for data. Being told the colour outright would be worth Rs 100, the full value of perfect information; the heart tip, which answers a narrower question, is worth exactly half of that. A piece of information is worth what it changes in your best decision, not how interesting it sounds: a tip that the card is an ace would be worth nothing here, because it leaves red and black at even money. The limitation to say out loud is that the Rs 50 is an expected value: on any single hand you could pay 50 and lose 100, so a desk with a limited bankroll should pay less than the full value, by the same logic that sizes bets below the edge.
Where candidates lose it
The common loss is answering 25 by reasoning that hearts are a quarter of the deck, so the tip is right a quarter of the time. The tip is informative in both of its answers: a no still tilts the deck two to one.
The second is pricing only the yes branch and forgetting to re-count the deck after a no. Thirty-nine cards remain, thirteen of them red, and that recount is where the second Rs 25 lives.
What the interviewer asks next
How much is it worth to be told whether the card is an ace?
How much is it worth to be told whether the card is a face card or a heart?
You can bet Rs 100 on the suit at 3 to 1 instead. What is the heart tip worth now?
Why should a desk with a small bankroll pay less than Rs 50 for this tip?
Asked at Optiver, Quantitative Research, Chicago, 2025 (Wall Street Oasis): Valuing information, taking directional bets when not plus EV.

037X and Y are independent and uniform on 0 to 1. What is the density of X + Y, and what is the probability that X + Y is less than 1.5?Distributions and statisticsCoreCitadelChicago · 2025Citadel SecuritiesChicago · 2025

Try it first

What is P(X + Y < 1.5)?

Show the worked solution

The density is a triangle: f(s) = s for s between 0 and 1, and 2 - s for s between 1 and 2; P(X + Y < 1.5) = 7/8. The sum lands in a thin band near s only if X and Y fit inside the unit square along the line X + Y = s, and the length of that line segment grows until s = 1 and then shrinks. The sum exceeds 1.5 only in the top-right corner of the square, a triangle with legs 1/2 and area 1/8, so the probability of less than 1.5 is 0.875.

Why is the sum of two flat distributions a triangle?

Roll two ordinary dice: a total of 7 can be made six ways, a total of 2 or 12 only one way, so the totals pile up in the middle even though each die is flat. The sum of two independent uniforms is the continuous version: the density at s is the length of the diagonal line X + Y = s inside the unit square, which grows from 0 up to s = 1 and shrinks back to 0 at s = 2. Formally the density is the convolution, the integral over x of f(x) times g(s - x), and with both densities equal to 1 on the unit interval the integrand is 1 exactly where both x and s - x lie between 0 and 1. That interval has length s for s below 1 and length 2 - s above, which is the triangle.

The relationship

f_{X+Y}(s) = \int_0^1 \mathbf{1}\{0 \le s - x \le 1\}\,dx = \begin{cases} s & 0 \le s \le 1 \\ 2 - s & 1 \le s \le 2 \end{cases} \qquad P(X+Y < 1.5) = 1 - \int_{1.5}^{2} (2-s)\,ds = 1 - \tfrac18

f X+Y	the density of the sum
the indicator	equals 1 where both x and s - x lie in the unit interval, which is where both densities are non-zero
the integral from 1.5 to 2	the area of the small triangle in the tail, 1/8

What it says in wordsConvolve two flat densities and the overlap of the two intervals gives a triangle; the tail beyond 1.5 is one eighth of it.

On the unit square the event X + Y > 1.5 is the corner triangle with legs one half and area one eighth, and on the density plot the same one eighth is the tail of the triangle beyond s = 1.5, so P(X + Y < 1.5) is seven eighths.

How do you get 7/8 in your head?

Use the square, not the integral. The pair (X, Y) is a uniformly random point in the unit square, so any probability is an area. X + Y > 1.5 is the region above the line from (0.5, 1) to (1, 0.5), a right triangle with both legs equal to 1/2, whose area is half of a quarter, which is 1/8. Everything else is 7/8. The same picture answers the mirror question at once: X + Y < 0.5 is the bottom-left corner, also 1/8. And it checks the density: integrating 2 - s from 1.5 to 2 gives (0.5)^2 / 2 = 1/8, the same number from the other direction, which is what you say to show the two views agree.

What does the interviewer ask next, and why?

The follow-ups test whether you understood the mechanism or memorised the triangle. Add a third uniform and the density becomes three parabolic pieces, smooth at the joins, with P(X + Y + Z < 1) = 1/6, the volume of the corner tetrahedron, and in general the sum of n uniforms heads towards a normal with mean n/2 and variance n/12, which is the central limit theorem arriving. The mean of X + Y is 1 and the variance is 1/12 + 1/12 = 1/6, by independence. Each extra convolution smooths the density by one degree: flat to triangular to parabolic, which is the general fact that sums of independent variables are smoother than their parts. The limitation worth saying: this all relies on independence; with X and Y correlated, the point is no longer uniform over the square and the areas no longer give probabilities.

Where candidates lose it

The common loss is treating the sum as uniform on 0 to 2 and answering 0.75. The sum is a triangle, and the mass in the top quarter of its range is only 1/8, not 1/4.

The second is setting up the convolution integral correctly and then mishandling the limits, integrating over the wrong interval of x. Draw the square; the limits are the overlap of two unit intervals, and the picture makes them obvious.

What the interviewer asks next

What is P(X + Y + Z < 1) for three independent uniforms?
What is the density of X - Y, and of max(X, Y)?
What are the mean and variance of X + Y, and how do you get them without the density?
X and Y are now correlated with rho = 0.5. Can you still find P(X + Y < 1.5) by area?

Asked at Citadel, Quant Research Interview, Chicago, 2025 (Wall Street Oasis): He was asking some questions about the probability, especially on the convolution.
Asked at Citadel Securities, Quant Research Interview, Chicago, 2025 (Wall Street Oasis): He was asking some questions about the probability, especially on the convolution.

041Make me a market on the number of heads in 100 flips of a fair coin. How wide do you quote, and how does your quote change when I tell you the first 10 flips produced 8 heads?Market makingCoreDRWNew York · 2026

Try it first

After you hear that the first 10 flips gave 8 heads, where is the new mid?

Show the worked solution

Centre on 50 and quote something like 48 at 52; after 8 heads in 10, move to 51 at 55, around 53. The mid is the expected count, 100 x 1/2. The standard deviation is sqrt(100 x 1/4) = 5, which tells you how much one lot can swing. Nobody can know more than you about a fair coin, so the quote can be tight. Once 8 heads are banked, the 90 flips left average 45, so the mid is 53 and the sd is sqrt(22.5), about 4.74.

Where does the middle of the quote come from?

If a friend asks how many of the next 100 cars past a junction will be white, and one car in two is white, you say 50 and you are not embarrassed by it. The middle of a market is your expected value, and for 100 fair flips that is 100 x 1/2 = 50 heads; everything else in the answer is about the width. The spread of the outcome comes from the binomial: variance n x p x (1 - p) = 25, so the standard deviation is 5. The count lands between 45 and 55 about 72.9% of the time, and outside 40 to 60 only about 3% of the time. Say both numbers out loud; the interviewer is checking that you know the centre and the scale before you name a price.

The relationship

\mu = np = 100 \times \tfrac12 = 50, \qquad \sigma = \sqrt{np(1-p)} = \sqrt{25} = 5, \qquad \mu' = 8 + 90 \times \tfrac12 = 53, \quad \sigma' = \sqrt{90 \times \tfrac14} = 4.74

n	the number of flips still random
p	the chance of heads on each flip, one half
mu, sigma	the mean and standard deviation of the total before any flip
mu prime, sigma prime	the same after the first 10 flips are known: the 8 heads are fixed and only 90 flips remain random

What it says in wordsKnown flips add to the centre one for one, and only the flips still to come feed the standard deviation.

Before any flip the outcome is centred on 50 with standard deviation 5 and the quote sits at 48 bid, 52 offered, and after 8 heads in the first 10 flips the whole curve moves to 53 and narrows to 4.74, so the same four-wide quote moves up to 51 at 55.

How wide should the quote be, and why not plus or minus one standard deviation?

Width is what you charge for two risks: that the person trading with you knows something you do not, and that you have to carry a position whose outcome swings. On a fair coin nobody can know more than you, so the first risk is zero and the quote can be tight, a point or two either side of 50; the standard deviation of 5 sets how many lots you are willing to show, not how far apart your bid and offer are. A quote of 45 at 55 is not wrong, but it says you think the other side may be informed, and an interviewer will ask why. Open at 48 at 52, say you would tighten to 49 at 51 against someone who clearly has no edge, and say what would make you widen: hidden information, a size much bigger than you want to hold, or a coin you have not inspected.

What exactly changes when I tell you 8 of the first 10 were heads?

Two things, and candidates usually get only one. The centre moves to 8 plus the expected heads on the 90 flips left, 45, which is 53. The coin is still fair, so 80% is not the new rate; the 8 are simply banked. The uncertainty also falls, because 10 of the flips are no longer random: the standard deviation goes from 5 to sqrt(90 x 1/4) = 4.74, about 5% lower. Move the whole quote up by 3 to 51 at 55, and notice what the news did to your old offer: a counterparty who had seen those flips would have lifted your 52 and made 1 a lot on average. That is the lesson of the follow-up, and it is why market makers widen or pull quotes when they suspect the other side has seen something they have not. The limitation of the tidy answer is that it trusts the coin; after a run of 8 heads a careful trader at least asks whether the coin is fair.

Where candidates lose it

The common loss is re-centring on 80, as if 8 heads in 10 revealed the coin, or leaving the mid at 50 because the coin has no memory. The flips have no memory, but the total does: 8 heads are already in it, and the right mid is 53.

The second is quoting plus or minus one standard deviation, 45 at 55, without saying why. Width is a statement about information and risk. Give the centre and the standard deviation first, then defend the width you choose.

What the interviewer asks next

I tell you the coin may be biased, with heads anywhere between 40% and 60%. How does your market change?
You are lifted on your 52 offer three times in a row. What do you do?
Make a market on the number of heads in the last 50 flips only, given the same news about the first 10.
Make a market on the square of the number of heads.

Asked at DRW, Quantitative Trading, New York, 2026 (Wall Street Oasis): Make a market on the number of heads out of 100 coin flips.

048You may roll a fair die up to three times, stopping whenever you like, and you are paid the value of the last roll. What is the optimal stopping rule and the value of the game?Expected value and optimal stoppingCoreRBC Capital MarketsToronto · 2025
Try it first
On the first roll you get a 4. Do you stop?
Show the worked solution
Stop on 5 or 6 on the first roll, on 4 or more on the second, and take the third; the game is worth 14/3, about 4.67. Work backwards. One roll left is worth 3.5. With two left, keep anything above 3.5, so 4, 5 or 6, which makes the game worth 15/6 + 3/6 x 3.5 = 4.25. With three left, keep anything above 4.25, so 5 or 6, which gives 11/6 + 4/6 x 4.25 = 14/3.
Why start from the last roll?
If you are flat-hunting and can see three flats, one a week, you take the first only if it beats what you expect from the remaining two, and you can only know that by thinking about the last week first. Every stop-or-continue decision compares the number in hand with the value of continuing, and the value of continuing is only known once you have solved the rounds after it, so you solve from the end. With one roll left there is no choice: you take whatever comes, worth 3.5 on average. That number becomes the bar for the roll before it, and the value of that roll becomes the bar for the one before that.
Read from right to left, the last roll is worth 3.5, so with two rolls left you keep 4 or more and the game is worth 4.25, so with three rolls left you keep only 5 or 6 and the game is worth 14/3, about 4.67.
How do the two thresholds come out?
With two rolls left, you keep the first of them if it beats 3.5, so 4, 5 or 6 are kept, each with probability 1/6, and on 1, 2 or 3 you roll once more for 3.5. The value is (4 + 5 + 6)/6 + (3/6) x 3.5 = 2.5 + 1.75 = 4.25. With three rolls left, the bar rises to 4.25, so a 4 is no longer good enough: keep only 5 or 6, and the value is (5 + 6)/6 + (4/6) x 4.25 = 11/6 + 17/6 = 14/3. The thresholds rise as more rolls remain, because the option to keep rolling is worth more when there are more chances left. Notice that the threshold is a value, not a face: you keep a face only if it is strictly greater than the value of carrying on.
The relationship
$V_1 = 3.5, \qquad V_{n} = \frac{1}{6}\sum_{f=1}^{6} \max(f, V_{n-1}) \quad\Rightarrow\quad V_2 = 4.25, \quad V_3 = \tfrac{14}{3} \approx 4.67$
V n the value of the game with n rolls left, played optimally
max(f, V n-1) on rolling face f you keep it or carry on, whichever is worth more
the sum over f the average over the six equally likely faces
What it says in wordsThe value with n rolls left is the average, over the faces, of the better of keeping the face and rolling on.
What do the follow-ups test?
They test whether you can re-run the recursion. With more rolls the value climbs towards 6 but slowly: 4 rolls give 4.944, 6 give 5.275, 10 give 5.650. A cost per roll lowers each continuation value and drops the thresholds. Keeping a 4 on the first roll costs you: that rule is worth 4.625 against 4.667, a small gap that an interviewer will still ask you to explain. A simulation of 300,000 games of the optimal rule gives 4.666. The desk link is direct: an American option is a stopping problem of exactly this shape, exercise when the value in hand beats the value of holding on, and a binomial tree solves it by the same backward pass.
Where candidates lose it
The common loss is keeping a 4 on the first roll because it beats 3.5. The right comparison is with the value of continuing, which with two rolls left is 4.25, not 3.5.
The second is solving forwards, trying to guess the first-roll threshold before knowing what the later rolls are worth. Say the last roll is worth 3.5, then build up; the answer arrives in two lines.
What the interviewer asks next
What is the game worth with four rolls, and what are the thresholds?
Each roll after the first costs 0.25. How do the thresholds change?
You are paid the square of the final roll. Does the stopping rule change?
How is this related to exercising an American option?
Asked at RBC Capital Markets, Quantitative Trading, Toronto, 2025 (Wall Street Oasis): Best way to maximize EV across 3 chosen dice rolls (can choose to continue or not).
049A 3 x 3 x 3 cube is painted on the outside and cut into 27 small cubes. You pick one small cube at random and roll it like a die. What is the probability the top face is painted?Probability and countingCoreJane StreetNew York · 2026
Try it first
What is the chance the top face is painted?
Show the worked solution
1/3. Picking a cube at random and then a face at random makes every one of the 27 x 6 = 162 small faces equally likely to end up on top. The painted small faces are exactly the squares on the big cube's surface, 6 faces of 9 each, 54 in all. So the chance is 54/162 = 1/3. The breakdown by cube agrees: corners give 8 x 3, edges 12 x 2, face centres 6 x 1, the core 0, total 54.
Why count faces rather than cubes?
If a bag holds sweets of different sizes and you want the chance a random bite is chocolate, you count chocolate bites, not chocolate sweets. Every small cube is equally likely and every face of it is equally likely to land on top, so every one of the 162 small faces has the same chance, 1/162, and the answer is just the share of small faces that are painted. That share is easy, because the painted small faces are exactly the visible squares of the big cube: 6 faces with 9 squares each, 54. The answer, 54/162 = 1/3, comes in one line without classifying a single cube, which is what the interviewer hopes to see.
The 8 corner cubes carry 24 painted faces, the 12 edge cubes 24, the 6 face centres 6 and the core none, so 54 of the 162 small faces are painted and a random top face is painted with probability one third.
How does the cube-by-cube count confirm it?
Classify the 27 cubes by position. The 8 corners have 3 painted faces each, the 12 edge cubes have 2, the 6 face centres have 1, and the single core cube has none: 8 + 12 + 6 + 1 = 27. Weight each type by how often you pick it and by the chance its top is painted: 8/27 x 3/6 + 12/27 x 2/6 + 6/27 x 1/6 + 1/27 x 0 = (24 + 24 + 6) / 162 = 1/3, the same answer by the long road. The two methods are the law of total probability written two ways, once by cube and once by face, and saying that out loud shows you know why they must agree. For an n x n x n cube the face count gives 6n squared painted faces out of 6n cubed, so the answer is 1/n: 1/2 for a 2 x 2 x 2 cube, 1/10 for a 10 x 10 x 10.
The relationship
$P(\text{top painted}) = \frac{6 \times 3^2}{27 \times 6} = \frac{54}{162} = \frac13, \qquad \text{in general } \frac{6n^2}{6n^3} = \frac1n$
6 x 3 squared the painted small faces, the 9 squares on each of the 6 outer faces
27 x 6 all small faces, each equally likely to end on top
n the number of cuts along each edge
What it says in wordsThe chance is the painted share of all small faces, which for an n-cube is one over n.
What is the natural follow-up, and how do you answer it?
Turn it round: the top face is painted; what is the chance you picked a corner? That is Bayes on the same count. Of the 54 painted faces, 24 belong to corners, so the chance is 24/54 = 4/9, far above the 8/27 a corner has before you look. Seeing paint is evidence for the cubes with more paint, and the face count gives the posterior directly without a formula. A second follow-up asks for the chance that the picked cube has any paint at all, which is 26/27, and the gap between 26/27 and 1/3 is exactly the trap in the original question. The limitation is that the face count relies on every face being equally likely to land on top; a weighted cube, or a rule that picks cubes by size, would need the long route.
Where candidates lose it
The common loss is answering 26/27, the chance the cube has some paint. The question asks about the top face, and most painted cubes are painted on only a few of their six faces.
The second is miscounting the cube types, often 6 edges instead of 12, and then forcing the total to 27 with the core. Skip the classification: count the 54 visible squares, divide by 162, and use the breakdown only as a check.
What the interviewer asks next
The top face is painted. What is the probability the cube is a corner?
Do the same for a 4 x 4 x 4 cube. Is there a general formula?
You roll the chosen cube twice. What is the chance both tops are painted?
How many of the 27 cubes have exactly two painted faces, and for an n-cube?
Asked at Jane Street, Engineering, New York, 2026 (Wall Street Oasis): How you got to the answer matters even if you got the question right. Strawberry question + 3x3 cube question
053It rains on 20% of days. Two forecasters work independently: each calls rain on 90% of the days it actually rains, and also calls rain on 15% of the dry days. Both call rain for tomorrow. What is the probability it rains? And if one calls rain and the other does not?Conditional probability and BayesCoreJane StreetNew York · 2025
Try it first
Two independent forecasters both say rain. Instinct first: how likely is rain?
Show the worked solution
90% if both call rain, and 15% if they disagree. Over 1,000 days, 200 are rainy and 800 are dry. Both call rain on 200 x 0.9 x 0.9 = 162 rainy days and on 800 x 0.15 x 0.15 = 18 dry days, so rain is 162 of 180. A split call happens on 36 rainy days and 204 dry days, so rain is 36 of 240, which is below the 20% base rate.
Why count days instead of multiplying probabilities?
A doctor reading two test results does the same thing: imagine a thousand patients, split them by who is actually ill, then split each group by what the tests said. Natural frequencies keep the base rate in the picture, where the probability form hides it, so you never confuse the chance of the evidence given rain with the chance of rain given the evidence. The 200 rainy days produce two rain calls 162 times; the 800 dry days, despite each forecaster being wrong only 15% of the time, produce two rain calls 18 times because 800 is a big base. The ratio is 162 to 18, so 90%.
Of 1,000 days, 200 rainy days give 162 double rain calls, 36 split calls and 2 with no call, while 800 dry days give 18, 204 and 578, so two rain calls make rain 90% likely and a split call makes it only 15% likely.
Why does a split call push the chance below the base rate?
Because a forecaster saying dry is itself evidence, and it is strong evidence. Each forecaster calls dry on 85% of dry days but only 10% of rainy days, so a dry call divides the odds of rain by 8.5, while a rain call multiplies them by only 6. Put together, one rain call and one dry call multiply the odds by 6 x (1/8.5), about 0.71, so the odds fall from 1 to 4 to about 1 to 5.67, which is 15%. The common feeling that the two calls cancel is wrong because the two kinds of error are not symmetric.
The relationship
$\frac{P(R\mid\text{both})}{P(D\mid\text{both})} = \frac{0.2}{0.8}\times\frac{0.9}{0.15}\times\frac{0.9}{0.15} = \frac{1}{4}\times 36 = 9 \;\Rightarrow\; P(R\mid\text{both}) = \tfrac{9}{10}$
0.2/0.8 the prior odds of rain, 1 to 4
0.9/0.15 the likelihood ratio of one rain call, 6
9 the posterior odds of rain after two independent calls
What it says in wordsPrior odds of one to four, multiplied by six for each independent rain call, give odds of nine to one, which is ninety per cent.
What assumption does the answer lean on, and when would it fail?
Independence given the weather. If both forecasters read the same satellite feed, two rain calls are closer to one piece of evidence than two, and the odds should be multiplied by something well short of 36. On a desk the same mistake appears when three models trained on the same data all flag a trade: the signals agree because they share inputs, not because the evidence is three times as strong. The answer also treats the 90% and 15% rates as known; with rates estimated from a short record, the posterior is softer than it looks.
Where candidates lose it
The fast wrong answer is 81%, from 0.9 times 0.9, which is the likelihood of the evidence rather than the probability of rain. Candidates who write Bayes' formula from memory often lose the base rate and give the same number.
The second trap is the split call: most people say about 50% or go back to 20%. Say that a dry call is strong evidence, divide the odds by 8.5 and multiply by 6, and the 15% follows.
What the interviewer asks next
Both forecasters say dry. What is the chance of rain now?
The two forecasters share the same data and always agree. What does two rain calls tell you?
A third independent forecaster with the same accuracy calls rain. Update the 90%.
How accurate would one forecaster need to be for a single rain call to make rain more likely than not?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): a question about the probability of rain the next day that relied on a very in depth understanding of bayes theorem
054Give me two random variables that are uncorrelated but clearly dependent, and show me that the correlation is zero.Distributions and statisticsCoreTwo SigmaNew York · 2025Tower Research CapitalNew York · 2014
Try it first
X is a standard normal and Y = X squared. What is the correlation of X and Y?
Show the worked solution
Take X standard normal and Y = X squared. Y is completely determined by X, the strongest dependence there is, yet Cov(X, Y) = E[X cubed] - E[X] E[X squared] = 0 - 0 x 1 = 0, because a symmetric distribution has zero third moment. A discrete version is X uniform on minus 1, 0 and 1 with Y = X squared, where E[XY] = (minus 1 + 0 + 1)/3 = 0 and E[X] = 0.
What does correlation actually measure?
Think of a shop whose sales rise when it is very cold and when it is very hot, and sag in mild weather. Plot sales against temperature and you get a U. Ask a straight line to summarise the U and it comes out flat: the hot days pull it up on the right exactly as much as the cold days pull it up on the left. Correlation is the slope of the best straight line through the cloud, scaled to lie between minus 1 and 1, so it can only detect the part of a relationship that runs in one direction. A relationship that turns around is invisible to it.
Forty-one points on the parabola Y = X squared, with X spread like a standard normal, have a best-fit line that is exactly flat at height 0.97, so their correlation is zero while every Y is fixed by its X.
How do you prove the zero in two lines?
Write the covariance as E[XY] minus E[X] E[Y]. With Y = X squared, E[XY] is E[X cubed], and every distribution symmetric about zero has E[X cubed] = 0, because each positive value x cubed is cancelled by the equally likely minus x cubed. E[X] is also zero, so both terms vanish and the correlation, which is the covariance divided by two positive standard deviations, is zero. The same proof works for any even function of a symmetric X, such as the absolute value, and for any symmetric X, not just the normal.
The relationship
$\operatorname{Cov}(X, X^2) = E[X^3] - E[X]\,E[X^2] = 0 - 0 \times 1 = 0 \qquad \text{while } Y = X^2 \text{ exactly}$
E[X^3] the third moment, zero for any distribution symmetric about zero
E[X] the mean of X, zero
E[X^2] the variance of a standard normal, one
What it says in wordsThe covariance of a symmetric variable with its own square is zero, so correlation is zero even though the square is a function of the variable.
Why does a derivatives desk care about this distinction?
Because a hedge built on correlation only neutralises straight-line exposure. A stock's return and its squared return are close to uncorrelated, yet the squared return is exactly what a long option position pays for, so a delta hedge that zeroes the linear exposure leaves the gamma exposure untouched. The limitation runs the other way too: independence does imply zero correlation, so a measured correlation near zero rules nothing out, and a desk that reads zero correlation as no relationship will be surprised in both tails.
Where candidates lose it
Candidates reach for a definition, saying independence implies zero correlation but not the reverse, and then cannot produce an example. The interviewer wants the example first: X and X squared, said inside ten seconds.
The second loss is giving the example without the proof. Write the covariance as E[X cubed] minus E[X] E[X squared] and say why the third moment of a symmetric variable is zero.
What the interviewer asks next
Give a second example where neither variable is a function of the other.
X is uniform on 0 to 1 rather than symmetric. Is X still uncorrelated with X squared?
What single number would catch the dependence between X and X squared that correlation misses?
Asked at Two Sigma, Generalist, New York, 2025 (Wall Street Oasis): Come up with two uncorrelated but dependent variables. Lots of questions regarding regression
Asked at Tower Research Capital, Quantitative Research, New York, 2014 (Wall Street Oasis): He also asked a question about independence and correlation. I did not respond fast enough
063The gaps between trades in an illiquid option are exponentially distributed with an unknown rate. You observe gaps of 2, 3 and 7 seconds. Derive the maximum likelihood estimate of the rate, and tell me why you would not trust it much.Distributions and statisticsCoreAQR Capital ManagementTown of Greenwich · 2022
Try it first
Gaps of 2, 3 and 7 seconds. What is the maximum likelihood rate?
Show the worked solution
The maximum likelihood estimate is 3/12 = 0.25 trades per second, the count divided by the total time, and with three observations it is both noisy and biased high. The likelihood of gaps t_1, t_2, t_3 is lambda cubed times exp(minus lambda times 12). Its log, 3 ln lambda minus 12 lambda, has derivative 3/lambda minus 12, which is zero at lambda = 0.25. On average this estimator reads 1.5 times the true rate when n = 3.
What does the likelihood actually say, in words?
A shopkeeper who saw customers arrive 2, 3 and 7 minutes apart would say roughly one every four minutes, and the likelihood is the formal version of that. For each candidate rate, the likelihood is how probable the observed gaps would be under that rate; the maximum likelihood estimate is the rate that makes what you saw least surprising. With exponential gaps the density of each gap t is lambda times exp(minus lambda t), so three independent gaps multiply to lambda cubed times exp(minus lambda times their sum). The sum, 12 seconds, is all the data you need; the individual values 2, 3 and 7 drop out.
The likelihood of the gaps 2, 3 and 7 seconds, plotted against the trade rate, peaks at 3 divided by 12 = 0.25 trades per second, but stays above half its peak from about 0.11 to 0.47, so three observations pin the rate down only loosely, and the bias-corrected estimate of 0.167 sits well to the left.
How do you derive the peak in three lines?
Take logs first, because a product of exponentials becomes a sum. The log likelihood is n ln lambda minus lambda times the sum of the gaps, its derivative is n over lambda minus the sum, and setting that to zero gives lambda equal to n over the sum, the number of events divided by the time they took. Here that is 3 over 12. The second derivative, minus n over lambda squared, is negative everywhere, so the stationary point is a maximum. The same derivation gives the familiar result that the maximum likelihood rate is one over the sample mean gap, 1 over 4 seconds.
The relationship
$\ell(\lambda) = n\ln\lambda - \lambda\sum t_i, \qquad \ell'(\lambda) = \frac{n}{\lambda} - \sum t_i = 0 \;\Rightarrow\; \hat\lambda = \frac{n}{\sum t_i} = \frac{3}{12} = 0.25$
l(lambda) the log likelihood of the observed gaps as a function of the rate
n the number of gaps observed, three
sum of t_i the total time covered by the gaps, 12 seconds
What it says in wordsThe maximum likelihood rate is the number of trades divided by the total time between them, which is one over the average gap.
Why would you not trust 0.25, and what would you say instead?
Three reasons, in the order a desk cares about them. The estimate is built on three numbers, so the likelihood hill is wide and any rate from roughly 0.11 to 0.47 fits almost as well; one over a sample mean is biased upward, with an expected value of n over n minus 1 times the true rate, 1.5 times here, so the unbiased version is (n - 1) over the sum, 0.167; and nothing in three gaps tests the exponential assumption itself. Real trade arrivals cluster, with bursts after news and dead stretches overnight, so a single constant rate is a model you chose, not a fact you found. The honest statement is a rate near 0.25 with a wide interval and a flag that the model may be wrong.
Where candidates lose it
The common loss is a formula without a derivation: candidates say one over the mean and stop. The interviewer asked you to derive it, so write the likelihood, take the log, differentiate and check the sign of the second derivative.
The second loss is answering the trust question with only the sample size. Mention the bias, give the corrected estimate, and question the exponential assumption, because that is the part a desk actually gets wrong.
What the interviewer asks next
What is the maximum likelihood estimate of the mean gap, and is it biased?
Give an approximate 95% interval for the rate from these three gaps.
How would you test whether trade gaps are really exponential?
The fourth gap is 60 seconds. What happens to the estimate, and does that worry you?
Asked at AQR Capital Management, Trading, Town of Greenwich, 2022 (Wall Street Oasis): Derive the mle for some given distribution. Explain linear regression intuitively and derive the ols estimate.
070Make me a two-way price on the product of two dice rolls. I then show you that one of the dice is a 4. Requote.Market makingCoreOptiverChicago · 2026DRWChicago · 2025
Try it first
Fair value of the product of two dice, before anything is revealed?
Show the worked solution
First quote around 12.25: say 11.75 bid, 12.75 offer. After the 4 is shown, requote around 14: say 13.50 at 14.50. Two independent dice have an expected product of 3.5 x 3.5 = 12.25. Once one die is known to be a 4, the product is 4 times the other die, with expectation 4 x 3.5 = 14. The second market can be the same width or tighter, because only one die of uncertainty is left: the standard deviation of the outcome falls from about 8.9 to about 6.8.
Why does fair value come from multiplying the averages?
A canteen's daily takings are the number of customers times the average spend, and if the crowd size has nothing to do with how hungry people are, the average takings are the average crowd times the average spend. For two independent quantities the expectation of the product is the product of the expectations, so two dice thrown separately have an expected product of 3.5 x 3.5 = 12.25. You can check it on the grid: row i of the multiplication table averages 3.5 i, and the six row averages, 3.5 through 21, average 12.25. The product is skewed, with most of the 36 cells below the mean and a few large ones pulling it up.
Across the 36 equally likely products the mean is 3.5 x 3.5 = 12.25, so the opening market sits around 12.25; once one die is revealed as a 4 the outcome is 4 times a single die, with a mean of 14 and a standard deviation that falls from about 8.9 to about 6.8, so the requote moves up and can tighten.
How does the reveal change the quote, and why can the market tighten?
Replace the revealed die with its value and keep the other at its expectation. The product is now 4 times one unknown die, so fair value is 4 x 3.5 = 14 and the only uncertainty left is a single die, with a standard deviation of 4 times 1.71, about 6.8, against about 8.9 before. Less uncertainty means less risk per trade, so a market maker can quote the same width with more confidence or tighten it. What you must not do is anchor on the old 12.25: a quote that still straddles 12 after a 4 has been shown is a free trade for the other side, who lifts your offer and collects the difference.
The relationship
$E[XY] = E[X]\,E[Y] = 3.5 \times 3.5 = 12.25, \qquad E[XY \mid X = 4] = 4\,E[Y] = 14, \qquad \sigma_{XY\mid X=4} = 4\,\sigma_Y = 4\sqrt{35/12} \approx 6.83$
X, Y the two independent dice
E[XY | X = 4] the expected product once one die is known to be a 4
sigma_Y the standard deviation of one die, about 1.71
What it says in wordsBefore the reveal the fair product is twelve and a quarter; after seeing a four it is fourteen, with the uncertainty of one die rather than two.
What is the interviewer watching for in the requote?
Speed and direction first, then the width. A trader who says 14 inside a second, moves the market up without hesitation and gives a reason for the width is passing; one who recomputes from the grid, or leaves the old market up, is failing. The next step is usually a trade: if the interviewer lifts your 14.50 offer, you are short at above fair and should hold or edge the market up slightly rather than chase; if they hit your bid at 13.50, you are long below fair. The limitation is that this is a one-shot game with a known distribution; in a real market the reveal would itself be a signal about what else the counterparty knows.
Where candidates lose it
The first loss is the opening fair value: 18.5 from the midpoint of 1 and 36, or 15.17 from squaring one die. Say independent, say 3.5 times 3.5, and the grid check if asked.
The second loss is the requote. Candidates either freeze or adjust by a token amount. The product is now 4 times one die, fair value 14, and the market should move there immediately and may tighten.
What the interviewer asks next
Instead of showing a 4, I tell you the two dice are the same. Requote.
Now the revealed die is a 1. Where is your market, and how wide?
Make a market on the sum of the two dice, then on the sum given one is a 4.
I lift your 14.50 offer twice in a row. What do you do?
Asked at Optiver, Prop Trading, Chicago, 2026 (Wall Street Oasis): Market making game full simulation including fast mental math and quick ev/fair value calculation
Asked at DRW, Quantitative Trading, Chicago, 2025 (Wall Street Oasis): Market making and fermi estimation on random quantities
071You walk on a grid from (0,0) to (6,4), each step one unit right or one unit up. The point (3,2) is blocked. How many routes avoid it?Probability and countingCoreSusquehanna International GroupLondon · 2026
Try it first
Before the block: how many routes from (0,0) to (6,4) with right and up steps only?
Show the worked solution
110 routes. Without the block there are C(10,4) = 210 routes, one for each way of placing 4 ups among 10 moves. A route through (3,2) is a route from (0,0) to (3,2), C(5,2) = 10 ways, followed by a route from (3,2) to (6,4), another C(5,2) = 10 ways, so 100 routes pass through the block. Subtract: 210 - 100 = 110.
Why is a lattice route a choice of positions rather than a sequence of decisions?
Think of a delivery driver in a city laid out as a grid who only ever drives east or north. Whatever order the turns come in, the trip is six blocks east and four blocks north, and the only freedom is which of the ten blocks are the north ones. A monotone route is fully described by choosing which 4 of its 10 moves go up, so the number of routes is 10 choose 4, which is 210, and no decision tree is needed. The same logic prices any question that asks how many ways a count can reach a level in fixed-size steps.
Of the 210 monotone routes from (0,0) to (6,4), every route through (3,2) is one of 10 routes into the block followed by one of 10 routes out of it, so 100 routes pass through it and 110 avoid it.
Why does multiplying the two legs count each bad route exactly once?
Because a monotone route visits a given point at most once; it can never come back. A route through (3,2) splits uniquely into the part before the block and the part after, so the number of such routes is the product of the two leg counts, 10 x 10 = 100, with no double counting to correct. Each leg is three rights and two ups, so C(5,2) = 10. If there were two blocked points, the same idea works but needs inclusion and exclusion: subtract the routes through each, then add back the routes through both.
The relationship
$N = \binom{10}{4} - \binom{5}{2}\binom{5}{2} = 210 - 10\times 10 = 110$
C(10,4) all routes: 10 moves, choose which 4 go up
C(5,2) C(5,2) routes into the block times routes out of it
N the routes that never touch the blocked point
What it says in wordsCount every route, subtract the ones that pass through the blocked point, which are the product of the two legs.
How do you check 110 another way?
Fill the grid with counts. Each point's count is the sum of the counts to its left and below, with the blocked point set to zero, and the corner comes out at 110. That dynamic-programming check takes a minute on paper and catches arithmetic slips in the binomials. It also answers the probability version the interviewer sometimes asks: if each step is right or up with equal chance, the chance a random walk reaches (6,4) at all is not 1, because it can overshoot, so the probability of avoiding the block among routes that do arrive is 110/210, about 52%, which is a different question from the probability for a free walk.
Where candidates lose it
Candidates try to count the avoiding routes directly and get lost in cases. The move is to count the complement: all routes less the routes through the block, with the block routes as a product of two binomials.
The second loss is a wrong binomial, often C(10,6) confused with something else or C(5,2) miscounted as 20. Say the legs out loud: three rights and two ups, 5 choose 2, is 10.
What the interviewer asks next
Now both (3,2) and (2,3) are blocked. How many routes avoid both?
Each step is right or up with probability one half. What is the probability a random walk from (0,0) passes through (3,2) before leaving the grid?
How many routes from (0,0) to (6,4) pass through (3,2) or (4,1)?
What is the general formula for routes from (0,0) to (m,n) avoiding a single point (a,b)?
Asked at Susquehanna International Group, Quantitative Research, London, 2026 (Wall Street Oasis): Probability about crossing from (0,0) to (6,4). Some point in the middle cannot pass through

13/52	the chance the card is a heart
100	a certain win on red once you know it is a heart
26/39, 13/39	the chance of black and of red among the 39 non-hearts
0	the value of the bet with no information

0.2/0.8	the prior odds of rain, 1 to 4
0.9/0.15	the likelihood ratio of one rain call, 6
9	the posterior odds of rain after two independent calls

l(lambda)	the log likelihood of the observed gaps as a function of the rate
n	the number of gaps observed, three
sum of t_i	the total time covered by the gaps, 12 seconds

C(10,4)	all routes: 10 moves, choose which 4 go up
C(5,2) C(5,2)	routes into the block times routes out of it
N	the routes that never touch the blocked point

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Why is a tip about hearts worth anything for a bet on colour?

How do you set the calculation out so it cannot go wrong?

What does this have to do with a trading desk?

Where candidates lose it

What the interviewer asks next

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Why is the sum of two flat distributions a triangle?

How do you get 7/8 in your head?

What does the interviewer ask next, and why?

Where candidates lose it

What the interviewer asks next

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Where does the middle of the quote come from?

How wide should the quote be, and why not plus or minus one standard deviation?

What exactly changes when I tell you 8 of the first 10 were heads?

Where candidates lose it

What the interviewer asks next

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Why start from the last roll?

How do the two thresholds come out?

What do the follow-ups test?

Where candidates lose it

What the interviewer asks next

Try it first

Why count faces rather than cubes?

How does the cube-by-cube count confirm it?

What is the natural follow-up, and how do you answer it?

Where candidates lose it

What the interviewer asks next

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Why count days instead of multiplying probabilities?

Why does a split call push the chance below the base rate?

What assumption does the answer lean on, and when would it fail?

Where candidates lose it

What the interviewer asks next

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What does correlation actually measure?

How do you prove the zero in two lines?

Why does a derivatives desk care about this distinction?

Where candidates lose it

What the interviewer asks next

Try it first

What does the likelihood actually say, in words?

How do you derive the peak in three lines?

Why would you not trust 0.25, and what would you say instead?

Where candidates lose it

What the interviewer asks next

Try it first

Why does fair value come from multiplying the averages?

How does the reveal change the quote, and why can the market tighten?

What is the interviewer watching for in the requote?

Where candidates lose it

What the interviewer asks next

Try it first

Why is a lattice route a choice of positions rather than a sequence of decisions?

Why does multiplying the two legs count each bad route exactly once?

How do you check 110 another way?

Where candidates lose it

What the interviewer asks next