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.

036Three assets all have the same pairwise correlation rho. What are the eigenvalues of the correlation matrix, and how low can rho go?Volatility and correlationHardJump TradingPudong Xinqu · 2023

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How negative can the common correlation be?

Show the worked solution

The eigenvalues are 1 + 2 rho, once, and 1 - rho, twice; rho can go no lower than -1/2. The vector (1, 1, 1) is an eigenvector with eigenvalue 1 + 2 rho, and any vector whose entries sum to zero is an eigenvector with eigenvalue 1 - rho, which gives a two-dimensional space and hence a double root. A correlation matrix must have no negative eigenvalue, since each eigenvalue is a portfolio variance, so 1 + 2 rho is at least 0 and rho is at least -1/2.

Why can three assets not all be strongly negatively correlated?

Three friends cannot all sit opposite each other at a table: if A faces B and B faces C, then A and C are on the same side. Correlation has the same constraint. If asset A moves against B and B moves against C, A and C are pushed towards moving together, so there is a floor on how negative a common correlation can be, and for three assets that floor is -1/2. You can see the floor without any algebra by holding an equal-weight portfolio of the three, each with unit variance: its variance is (3 + 6 rho) / 9, which is (1 + 2 rho) / 3, and a variance cannot be negative. At rho = -1/2 the portfolio has zero variance; it is perfectly hedged, and nothing below that is possible.

Plotted against rho, the eigenvalue 1 + 2 rho rises steeply and crosses zero at rho = -1/2, while the double eigenvalue 1 - rho falls gently to zero at rho = 1, so the matrix is a valid correlation matrix only between those two points.

How do you find the eigenvalues without expanding a determinant?

Write the matrix as (1 - rho) times the identity plus rho times the all-ones matrix J. The identity leaves every vector alone, so you only need the eigenvalues of J, and J is easy: it maps (1, 1, 1) to (3, 3, 3), eigenvalue 3, and it maps any vector whose entries sum to zero to the zero vector, eigenvalue 0, with a two-dimensional space of such vectors. Shifting and scaling by (1 - rho) turns those into 1 - rho + 3 rho = 1 + 2 rho for the market direction and 1 - rho for the two spread directions. Check with the trace: the eigenvalues add to 1 + 2 rho + 2(1 - rho) = 3, the sum of the diagonal, as they must. At rho = 0.3 they are 1.6, 0.7 and 0.7.

The relationship

\Sigma = (1-\rho)I + \rho J, \qquad \lambda_1 = 1 + 2\rho \;\text{on}\;(1,1,1), \qquad \lambda_{2,3} = 1 - \rho \;\text{on}\;\{v : v_1+v_2+v_3 = 0\}, \qquad \rho \ge -\tfrac12

I	the identity matrix
J	the matrix of all ones, whose eigenvalues are 3 (once) and 0 (twice)
lambda 1	the eigenvalue of the common or market direction
lambda 2, 3	the double eigenvalue of the two directions that net to zero, the spread trades

What it says in wordsThe matrix is a stretch of the all-ones matrix, so the market direction gets 1 + 2 rho and every spread direction gets 1 - rho.

What does the structure tell a risk or trading desk?

The eigenvectors are the principal components. The (1, 1, 1) direction is the market factor, and its eigenvalue over the trace, (1 + 2 rho) / 3, is the share of total variance it explains: 53% at rho = 0.3 and 80% at rho = 0.7. The two spread directions carry the rest, equally. A long-short book that nets to zero across the three assets lives entirely in the 1 - rho directions, which is why pairs trades get calmer as correlation rises and why a correlation of 1 collapses them to nothing. For n assets the same argument gives eigenvalues 1 + (n - 1) rho and 1 - rho, so the floor is -1 / (n - 1): -1/3 for four assets and -1/9 for ten. Say the limitation as well: a historical correlation matrix estimated from more assets than observations is only barely positive semi-definite, and a hand-edited one, where a trader overrides a few pairs, can fail the test entirely, which is exactly the fault a risk system is built to catch.

Where candidates lose it

The common loss is answering -1, because a correlation can be -1. For a pair it can; for three assets pairwise, it cannot, and the interviewer wants the reason: a negative eigenvalue is a negative portfolio variance.

The second is expanding the characteristic polynomial by hand and getting lost. Spot the all-ones structure, name the eigenvector (1, 1, 1), and the rest is one line.

What the interviewer asks next

What is the floor on rho for n equally correlated assets?
The three assets have correlations 0.9, 0.9 and -0.9. Is that a valid correlation matrix?
What is the variance of the equal-weight portfolio at rho = -1/2, and what does that portfolio look like?
How would you repair an estimated correlation matrix that has a small negative eigenvalue?

Asked at Jump Trading, Prop Trading, Pudong Xinqu, 2023 (Wall Street Oasis): Some very difficult linear algebra questions about PCA and eigenvalues

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.

039You start with Rs 2 and bet Rs 1 at a time on a coin that falls your way 60% of the time. You stop when you reach Rs 5 or go broke. What is the probability you reach Rs 5?Random walks and Markov chainsHardTwo SigmaNew York · 2023
Try it first
Roughly how likely are you to reach Rs 5 before going broke?
Show the worked solution
135/211, about 64.0%. Let r be q over p, which is 0.4 / 0.6 = 2/3. The probability of reaching N from a stake of i is (1 - r^i) / (1 - r^N). With i = 2 and N = 5 that is (1 - 4/9) / (1 - 32/243) = (5/9) x (243/211) = 135/211. A fair coin would give 2/5 = 40%; the 60% edge lifts it to 64.0%. The game lasts about 6.0 bets on average.
Why is the answer not simply 2 out of 5?
With a fair coin the answer is 2/5, because a fair game cannot create or destroy expected money: you start with Rs 2, you finish with Rs 5 or Rs 0, so the chance of Rs 5 must be 2/5 to keep the average at 2. With a 60% coin each bet gains you Rs 0.20 on average, so the walk drifts upward and the chance of hitting the top is higher than the fair-coin fraction; what you need is a quantity that is still conserved under the biased coin. That quantity is (q/p) to the power of your stake. A win multiplies it by q/p, a loss by p/q, and weighted by their probabilities the two moves cancel: p x (q/p) + q x (p/q) = q + p = 1. Because that quantity is conserved, its starting value must equal its average finishing value, and that one line gives the formula.
The relationship
$r = \frac{q}{p} = \frac{2}{3}, \qquad P_i = \frac{1 - r^i}{1 - r^N} = \frac{1 - (2/3)^2}{1 - (2/3)^5} = \frac{5/9}{211/243} = \frac{135}{211} \approx 0.640$
r the loss probability over the win probability; below 1 when the coin favours you
i the starting stake, Rs 2
N the target, Rs 5
P i the chance of reaching the target before going broke
What it says in wordsSet the conserved quantity r to the stake equal to its average at the end, and solve for the chance of reaching the target.
The chance of reaching Rs 5 before Rs 0 rises along a curve above the fair-coin straight line, reaching 0.640 from a starting stake of Rs 2 against 0.4 for a fair coin, because each rupee of stake multiplies the odds of ruin by q over p, two thirds.
How do you derive it from the states if you forget the formula?
Write P_i for the chance of reaching 5 from a stake of i. Then P_0 = 0, P_5 = 1, and in between P_i = 0.6 P_(i+1) + 0.4 P_(i-1), one equation per state. That is a second-order linear recurrence whose solutions are of the form A + B r^i with r = q/p, and the two boundary conditions fix A and B. Solving the five equations directly gives P_1 = 81/211, P_2 = 135/211, P_3 = 171/211 and P_4 = 195/211, and the recurrence is the thing to write on the whiteboard first, because it works for any rule change. A simulation of 200,000 games gives 0.639, agreeing with the fraction to three places. The same system with a 1 on the right-hand side of each interior equation gives the expected duration, about 6.0 bets from Rs 2.
What does the biased formula tell you about trading with an edge?
Let the target go to infinity. With a fair coin the chance of never going broke is zero: any finite stake is eventually lost. With the 60% coin it is 1 minus r to the stake, which from Rs 2 is 1 - 4/9 = 5/9, about 56%, and from Rs 10 it is above 98%. An edge does not protect a thin stake: with Rs 2 behind a 60% coin you still go broke 44% of the time, and the cure is not a better coin but a bigger stake relative to the bet. That is why a desk with a genuine edge still caps position size, and why the question sits next to the Kelly one. The limitation is that the bets here are of fixed size; once you can resize the bet with your capital, the ruin arithmetic changes completely.
Where candidates lose it
The common loss is answering 2/5, the fair-coin answer, or guessing that 60% means roughly 60%. The edge changes the structure, and the interviewer wants to hear q over p.
The second is writing the formula with p/q instead of q/p, which gives a number below 40% for a coin that favours you. Sanity check the direction: an edge in your favour must raise the chance above the fair-coin 2/5.
What the interviewer asks next
What is the chance of reaching Rs 5 from Rs 2 with a fair coin, and why is it exactly 2/5?
The target is Rs 10 instead of Rs 5. What is the chance now?
There is no target: you play until you go broke or forever. What is the chance you never go broke?
How long does the game last on average from Rs 2?
Asked at Two Sigma, Research, New York, 2023 (Wall Street Oasis): Biased gamblers ruin problems; Markov Chain problems; sampling uniformly from triangle

040Five rational pirates, A the most senior down to E, split 100 gold coins. The most senior proposes a split; it passes if at least half of the pirates, his own vote included, agree, otherwise he is thrown overboard and the next most senior proposes. Each pirate wants to survive, then to get the most coins, then to see others thrown overboard. How should A split the coins to survive and keep the most?Games and logicHardOld Mission CapitalNew York · 2014

Try it first

How many coins can A keep?

Show the worked solution

A proposes 98 for himself, 0 for B, 1 for C, 0 for D and 1 for E. Work backwards. With two pirates, D keeps 100, since his own vote is half. With three, C keeps 99 and gives E 1, because E gets nothing if C dies. With four, B keeps 99 and gives D 1. With five, A needs two other votes and buys the cheapest: C and E, who get nothing under B's plan, accept one coin each. A's vote plus two makes three of five.

Why does the answer come from the end of the game, not the start?

When you negotiate a price you think about what the other side does if you walk away; what they do next determines what they will accept now. Each pirate compares the offer in front of him with what he would get under the next proposal, so to know what A must offer you first need to know what B would do, which needs C, which needs D with two pirates left. With two pirates, D proposes 100 for himself; his own vote is one of two, which is half, so it passes and E gets nothing. That is the anchor. Every earlier step is a question of who gets zero in the next round, because those are the pirates whose votes are cheapest to buy.

Working up from two pirates, D keeps 100, C keeps 99 by paying E one coin, B keeps 99 by paying D one coin, and A keeps 98 by paying C and E one coin each, the two pirates who would receive nothing under B's proposal.

How does each proposer decide whose vote to buy?

He buys exactly the number of extra votes he needs, from the pirates who are cheapest, and a pirate is cheap if the next proposal gives him nothing. With three pirates, C needs one more vote; under D's plan E gets zero, so C offers E one coin and keeps 99, and E accepts because one is better than zero. With four, B needs one more; under C's plan D gets zero, so B pays D one coin. With five, A needs two more votes, and under B's plan the pirates with nothing are C and E, so A pays them one coin each and keeps 98; offering anything to B or D is wasted, because B would get 99 and D one coin without A, and neither can be bought for less. The alternation is the pattern to say out loud: each round, the pirates who were paid last time are the ones left out this time.

The relationship

\text{2: }(100, 0) \;\to\; \text{3: }(99, 0, 1) \;\to\; \text{4: }(99, 0, 1, 0) \;\to\; \text{5: }(98, 0, 1, 0, 1)

each tuple	coins to the proposer first, then to the others in order of seniority
the 1s	votes bought for one coin from pirates who would get zero in the next round
votes needed	half of the pirates, rounded up, including the proposer: 1 of 2, 2 of 3, 2 of 4, 3 of 5

What it says in wordsEach proposer keeps everything except one coin for each extra vote he needs, bought from whoever the next round would leave empty-handed.

What assumptions is the answer resting on, and what happens when they change?

Three assumptions, and the interviewer will test at least one. Pirates are perfectly rational and know the others are. A pirate who is indifferent between two outcomes prefers the one where a rival is thrown overboard, which is why a bought pirate must be paid one coin and not zero. And the voting rule is at least half, with the proposer voting. Change the rule to a strict majority of all votes and the two-pirate case flips, D cannot pass anything and dies, so E would reject everything in the three-pirate round, and the whole ladder shifts. With a strict majority of the other pirates' votes the ladder shifts again, and a candidate who re-derives it from the two-pirate anchor is the one who gets hired. The limitation is that real negotiators are not this rational; the puzzle is a model of backward induction, not of pirates.

Where candidates lose it

The common loss is reasoning forward from fairness, offering 20 each or 50 for A, and being unable to say why those numbers rather than others. Fairness is not in the rules; survival and coins are.

The second is paying the wrong pirates. B and D would do well without A and cannot be bought cheaply; C and E would get nothing. Candidates who pay B and D have not worked out the four-pirate round.

What the interviewer asks next

The proposal now needs a strict majority of all votes. What does A propose?
The proposal needs a majority of the other pirates' votes, excluding the proposer. Does A survive?
There are 6 pirates. How does the pattern continue?
What if a pirate who is indifferent prefers to keep the proposer alive?

Asked at Old Mission Capital, Prop Trading, New York, 2014 (Wall Street Oasis): There are 5 pirates and they are trying to split 100 gold coins in a rational way.

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.

043In your head, no paper: convert 3/32 to a decimal, work out 38 x 42, and give 1/7 to four decimal places.Mental maths and estimationWarm upBelvedere TradingChicago · 2021
Try it first
What is 38 x 42?
Show the worked solution
0.09375, 1,596 and 0.1429. For 3/32, halve 1 five times to get 1/32 = 0.03125 and multiply by 3. For 38 x 42, both sit 2 away from 40, so the product is 40 squared minus 2 squared, 1,600 - 4. For 1/7, the repeating block is 142857, because 7 x 142857 = 999,999, so 1/7 = 0.142857... which rounds to 0.1429. Each one is a known anchor plus one step.
Why does the interviewer ask three small sums in a row?
A shopkeeper who totals a bill in his head is not doing long addition; he rounds to the nearest hundred and fixes the difference. Trading desks ask quick sums to see whether you reach for an anchor you already know, because that is how prices get checked in the two seconds before someone else trades. The three questions here each have a short route: fractions with a power of two below them are halvings, products of numbers either side of a round number are a difference of squares, and sevenths are one repeating block of six digits. The speed comes from knowing which route fits, and the interviewer listens to the route as much as to the number.
Halving 1 five times gives 1/32 = 0.03125 and three of those make 0.09375, a 40 by 40 square with a 2 by 2 corner removed shows 38 x 42 = 1,596, and the six-digit cycle 142857 gives 1/7 = 0.1429 to four places.
What is the route for each one?
For 3/32, keep halving from a half: 0.5, 0.25, 0.125, 0.0625, 0.03125. Every halving adds a digit, and 32 is five halvings, so 1/32 is 0.03125 and three of them are 0.09375. Bond traders do this all day, because US Treasury prices are quoted in 32nds. For 38 x 42, notice that both numbers sit 2 from 40, so the product is 40 squared minus 2 squared, 1,596, a trick that works for any pair placed evenly around a round number. The same trick gives 47 x 53 = 2,500 - 9 = 2,491. For 1/7, remember that 7 x 142857 = 999,999. So 1/7 is 142857 divided by 999,999, which is 0.142857 repeating, and the fifth decimal is 5, so it rounds up to 0.1429.
The relationship
$\frac{3}{32} = 3 \times \frac{1}{2^5} = 3 \times 0.03125, \qquad (a - b)(a + b) = a^2 - b^2, \qquad \frac17 = \frac{142857}{999999} = 0.\overline{142857}$
2 to the 5 32, five halvings of 1
a, b the round midpoint, 40, and the distance to each number, 2
the bar the block of six digits that repeats forever
What it says in wordsTurn each sum into a fact you already know plus one small step.
What does a strong candidate add after the answers?
Add the sanity check and the neighbours. A product of two numbers around a centre is always a little below the centre squared, so 1,596 must be under 1,600. Three 32nds must be just under a tenth, since 3.2 of them would make exactly 0.1. The sevenths share the same six digits in rotation, 2/7 = 0.285714 and 3/7 = 0.428571, so knowing one seventh gives you all six. The limitation is honest too: anchors cover the common cases, and for an awkward product like 37 x 46 you fall back on splitting, 37 x 46 = 37 x 50 - 37 x 4 = 1,850 - 148 = 1,702. Saying when you switch method sounds better than pretending one trick does everything.
Where candidates lose it
The common loss is starting long multiplication for 38 x 42 aloud, carrying digits and losing track. The interviewer gives you a pair either side of 40 on purpose; a candidate who does not spot it looks slow even if the answer is right.
The second is 1/7 as 0.1428, truncating instead of rounding. The next digit is 5, so the fourth decimal rounds up. Say the repeating block first, then round, and the slip disappears.
What the interviewer asks next
What is 5/16 as a decimal, and 7/64?
Work out 67 x 73 and 96 x 104 the same way.
What is 5/7 to four places?
Estimate 1/13 to three decimal places.
Asked at Belvedere Trading, Trading, Chicago, 2021 (Wall Street Oasis): 3/32 mental math, 38*42, crossing the bridge in the shortest amount of time

045Two players each ante Rs 1 and receive one card from an ace, a king and a queen. Player one may bet Rs 1 or check, and a check goes straight to showdown. Facing a bet, player two may call or fold. How often should player one bluff with the queen, and how often should player two call with the king?Games and logicHardOld Mission CapitalNew York · 2022

Try it first

How often should player one bet the queen?

Show the worked solution

Player one bets the ace always, checks the king, and bluffs the queen one time in three; player two calls with the ace, folds the queen, and calls with the king one time in three. The bluff rate makes one bet in four a bluff, which is the break-even for a call of 1 into a pot of 3. The king's call rate makes player two fold one time in three overall, the break-even for a bluff of 1 into a pot of 2. Player one gains 1/18 of a rupee a hand.

Which decisions are obvious, and which one is the real question?

Start by removing the choices nobody would make. Player one always bets the ace, since it wins every showdown, and player two always calls with the ace and folds with the queen, since those cards win or lose every time; the only real decisions are player one's queen and player two's king. Player one's king should check: if it bets, player two calls only with the ace and folds the queen, so the bet loses 2 half the time and wins 1 half the time, an average of -0.5, worse than the 0 a showdown gives. That leaves two numbers to find, the queen's bluff frequency and the king's calling frequency, and each is set to make the other player's choice a matter of indifference.

How do you find the bluffing frequency?

A goalkeeper who always dives left is easy to beat; a penalty taker mixes his side so the keeper gains nothing by guessing. Poker equilibrium works the same way. Player one bluffs just often enough that player two's king gains nothing by calling over folding, and player two calls just often enough that player one's queen gains nothing by bluffing over checking. Facing a bet with the king, folding loses the ante, -1. Calling wins 2 against a bluff and loses 2 against the ace. With the ace always bet and the queen bet a fraction b of the time, a bet is a bluff with probability b / (1 + b). Setting the call's value equal to -1 gives b = 1/3, so one bet in four is a bluff.

The relationship

\text{King indifferent: } \tfrac{b}{1+b}(+2) + \tfrac{1}{1+b}(-2) = -1 \Rightarrow b = \tfrac13, \qquad \text{Queen indifferent: } \tfrac12(-2) + \tfrac12\big(c(-2) + (1-c)(+1)\big) = -1 \Rightarrow c = \tfrac13

b	how often player one bets the queen
c	how often player two calls with the king
-1	the value of the alternative: folding the king, or checking the queen, each loses the ante
1/2	given the queen, player two holds the ace or the king with equal chance

What it says in wordsEach player's frequency is chosen so that the other player's two choices are worth the same.

Player one bets the ace, checks the king and bluffs the queen one time in three, while player two calls with the ace, folds the queen and calls with the king one time in three, because those frequencies leave each player's marginal choice worth exactly the same either way.

How do you sanity check the frequencies, and who wins the game?

Use pot odds as the check. Player two calls 1 to win a pot of 3, two antes and the bet, so a call breaks even when bluffs are 1 / (3 + 1) of bets, a quarter, which is what b = 1/3 delivers. Player one bluffs 1 to win the pot of 2, so a bluff breaks even when player two folds 1 / (2 + 1) of the time; he folds the queen always and the king two times in three, which averages to one in three. Average over the six deals and player one gains 1/18 of a rupee a hand, because only he can bet; neither player can improve on that by changing his own frequency. Against a player two who never calls with the king, bluffing every queen would earn 1/6 of a rupee a hand instead of 1/18. The limitation is that equilibrium is a defence, not the most profitable play against a predictable opponent; a desk uses it as the baseline and then leans towards the other side's mistakes.

Where candidates lose it

The common loss is saying player one should never bluff because the queen cannot win a showdown. Without bluffs player two simply folds his king to every bet, and player one's ace stops getting paid.

The second is giving the bluff rate as a quarter. A quarter is the share of bets that are bluffs; because the ace is always bet, that needs the queen bet one time in three. Keep the two fractions apart and say which one you mean.

What the interviewer asks next

Player two may also bet after player one checks. How does the equilibrium change?
The bet size doubles to Rs 2. What are the new bluffing and calling frequencies?
Player two never calls with the king. What is player one's best response, and what does it earn?
Why does player one never want to bet the king?

Asked at Old Mission Capital, Quantitative Research, New York, 2022 (Wall Street Oasis): Asking to find the game theory optimal strategy in a simplified poker game

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

2 to the 5	32, five halvings of 1
a, b	the round midpoint, 40, and the distance to each number, 2
the bar	the block of six digits that repeats forever

Try it first

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

Try it first

Why can three assets not all be strongly negatively correlated?

How do you find the eigenvalues without expanding a determinant?

What does the structure tell a risk or trading desk?

Where candidates lose it

What the interviewer asks next

Try it first

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

Try it first

Why is the answer not simply 2 out of 5?

How do you derive it from the states if you forget the formula?

What does the biased formula tell you about trading with an edge?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the answer come from the end of the game, not the start?

How does each proposer decide whose vote to buy?

What assumptions is the answer resting on, and what happens when they change?

Where candidates lose it

What the interviewer asks next

Try it first

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

Try it first

Why does the interviewer ask three small sums in a row?

What is the route for each one?

What does a strong candidate add after the answers?

Where candidates lose it

What the interviewer asks next

Try it first

Which decisions are obvious, and which one is the real question?

How do you find the bluffing frequency?

How do you sanity check the frequencies, and who wins the game?

Where candidates lose it

What the interviewer asks next

Try it first

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