Derivatives Foundation puzzles, solved step by step

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

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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
056In your head, no paper: 998 x 1,003. Then 2.5% of 4,860. Then 99 squared.Mental maths and estimationWarm upOld Mission CapitalChicago · 2015Old Mission CapitalChicago · 2015
Try it first
998 x 1,003. Say it inside five seconds.
Show the worked solution
1,000,994, then 121.5, then 9,801. 998 x 1,003 is (1,000 - 2)(1,000 + 3) = 1,000,000 + 1,000 - 6. For 2.5% of 4,860, take 10%, which is 486, and then a quarter of it, 121.5. For 99 squared, (100 - 1) squared is 10,000 - 200 + 1 = 9,801. In every case the move is the same: go to the nearest round number and fix the small error afterwards.
Why is the round number always the first move?
Buying seven items at Rs 99 each, nobody multiplies 99 by 7; you take Rs 700 and hand back Rs 7. Multiplying by a round number is free, and the correction is a small product you can hold in your head, so every mental multiplication is a round-number product plus or minus a correction. For 998 x 1,003 the round number is 1,000 for both, so the product is 1,000,000, the two cross terms are + 3 x 1,000 and - 2 x 1,000, which net to + 1,000, and the only real work is the sign of 2 x 3 at the end: a minus times a plus is a minus, so subtract 6.
Each calculation becomes a round-number product with a correction: 998 x 1,003 is a million plus 1,000 minus 6, which is 1,000,994; 2.5% of 4,860 is a tenth, 486, then a quarter, 121.5; and 99 squared is 10,000 minus 200 plus 1, which is 9,801.
How do you handle percentages that are not round?
Build the percentage out of ones you can do instantly. 2.5% is a quarter of 10%, so take a tenth of 4,860, which is 486, and quarter it: half is 243, half again is 121.5. The same habit covers 7.5% as 10% minus a quarter of 10%, 15% as 10% plus half of that, and 12.5% as an eighth. On a desk this is how you convert a basis-point move to rupees while someone is still talking: 25 basis points on Rs 4,860 crore of notional is the same 121.5, in crore.
The relationship
$(a-b)(a+c) = a^2 + a(c-b) - bc \qquad (a-b)^2 = a^2 - 2ab + b^2$
a the nearest round number, here 1,000 or 100
b, c the small distances from the round number, here 2 and 3, or 1
bc the small product that carries the sign most people get wrong
What it says in wordsA product near a round number is the round square plus the round number times the net offset, minus the product of the two offsets.
What does the interviewer listen for beyond the answer?
Speed, but also the sanity check. 998 x 1,003 must be a little above a million because 998 x 1,003 is roughly 1,000 x 1,001, and 99 squared must end in 1 because 9 x 9 does, so a candidate who says 1,000,994 and 9,801 without pausing shows both the trick and the check. Where this breaks down is numbers that are not near anything round, such as 47 x 83; there the move is to split one factor, 47 x 80 + 47 x 3, and accept that it takes two beats rather than one.
Where candidates lose it
The slip in 998 x 1,003 is the sign of the last term: candidates add 6 and say 1,001,006, or drop the cross terms and say 999,994. Say the expansion out loud, plus 1,000 then minus 6, and the sign looks after itself.
The slip in 99 squared is forgetting the plus 1 and answering 9,800. The check is the last digit: 9 times 9 ends in 1, so 9,800 cannot be right.
What the interviewer asks next
1,002 x 997, same method.
What is 12.5% of 4,860?
101 squared minus 99 squared, without squaring either.
A bond moves 35 basis points on Rs 2,400 crore. How many crore is that?
Asked at Old Mission Capital, Prop Trading, Chicago, 2015 (Wall Street Oasis): multiple questions based on arithmetic operations involving multiple digits. These questions are supposed to answered without consulting a calculator
Asked at Old Mission Capital, Prop Trading, Chicago, 2015 (Wall Street Oasis): Phone interview included arithmetic operations that was supposed to be done without pen or paper
057The correlation of X and Y is 0.2 and the correlation of Y and Z is 0.5. What range of values can the correlation of X and Z take?Volatility and correlationHardTower Research CapitalNew York · 2019
Try it first
Two correlations are given. Before any algebra: is the third one free to be anything between minus 1 and 1?
Show the worked solution
Between about -0.75 and 0.95. The three correlations must form a positive semi-definite matrix, and for three variables that means its determinant 1 - a^2 - b^2 - c^2 + 2abc cannot be negative. Solving for c with a = 0.2 and b = 0.5 gives c = ab plus or minus sqrt((1 - a^2)(1 - b^2)) = 0.1 plus or minus sqrt(0.96 x 0.75) = 0.1 plus or minus 0.849.
Why can the third correlation not be anything it likes?
If Amit is tall when Bela is tall, and Bela is tall when Chitra is tall, then Amit and Chitra cannot be perfectly opposite; the middle person ties them together. Three pairwise correlations describe one joint set of three variables, and a joint set has a variance for every combination of them that cannot be negative, which is the positive semi-definite condition on the correlation matrix. With weak links, 0.2 and 0.5, the tie through Y is loose and the band is wide; with links of 0.9 and 0.9, X and Z would be forced to be strongly positively correlated.
The correlation of X and Z can only sit in the green band from -0.75 to 0.95, centred on 0.2 x 0.5 = 0.1, because outside it the determinant of the 3 x 3 correlation matrix, plotted below the axis, turns negative and no three variables can have those correlations.
How do you get the band without writing out eigenvalues?
Use the determinant. For a 3 x 3 correlation matrix with off-diagonal entries a, b and c, the determinant is 1 - a^2 - b^2 - c^2 + 2abc, and the matrix is valid exactly when that is at least zero. Treat it as a quadratic in the unknown c: c^2 - 2ab c + (a^2 + b^2 - 1) must be at most zero, so c lies between the two roots ab plus or minus sqrt((1 - a^2)(1 - b^2)). With a = 0.2 and b = 0.5 the centre is 0.1 and the half-width is sqrt(0.96 x 0.75) = sqrt(0.72), about 0.849. The geometric reading is the same thing: think of each variable as a unit vector, correlations as cosines of the angles between them, and the third angle is bounded by the sum and difference of the first two.
The relationship
$\rho_{XZ} \in \left[\rho_{XY}\rho_{YZ} - \sqrt{(1-\rho_{XY}^2)(1-\rho_{YZ}^2)},\; \rho_{XY}\rho_{YZ} + \sqrt{(1-\rho_{XY}^2)(1-\rho_{YZ}^2)}\right] = [0.1 - 0.849,\; 0.1 + 0.849]$
rho_XY, rho_YZ the two given correlations, 0.2 and 0.5
rho_XZ the correlation being bounded
the square root term the half-width of the feasible band, which shrinks as the given correlations strengthen
What it says in wordsThe third correlation sits within a band centred on the product of the other two, with a width that vanishes only when the other two are plus or minus one.
Where does this show up on a desk?
In any correlation matrix that was assembled by hand or from mismatched histories. A desk that marks the stock-to-index correlation at 0.9, the index-to-sector correlation at 0.9, and the stock-to-sector correlation at 0.3 has written down a matrix that no world can produce, and a basket option priced from it will give nonsense, often a negative variance for some combination. The fix is to project the matrix back to the nearest valid one before pricing. The limitation of the puzzle is that it covers only three variables; with more, every principal sub-matrix must pass, and the feasible set has no simple closed form.
Where candidates lose it
The common answer is that the third correlation is unconstrained, or that it must equal 0.1. Both miss that three variables share one joint distribution, and that its correlation matrix must be valid.
The second loss is knowing the condition and failing to turn it into a number. Have the determinant formula ready, solve the quadratic in the unknown, and quote the band to two decimals.
What the interviewer asks next
Now the two given correlations are 0.9 and 0.9. What is the band?
Can the X Z correlation be negative if X Y and Y Z are both 0.8?
Explain the geometric picture using angles between unit vectors.
What does a desk do when its estimated correlation matrix fails this test?
Asked at Tower Research Capital, Prop Trading, New York, 2019 (Wall Street Oasis): What if the correlation between X and Y is 0.2 and the correlation between Y and Z is 0.5. What is the range for the correlation of X and Z?
058We roll two fair dice and you win if the total is exactly 10. How much would you be willing to risk to win Rs 100?Betting and sizingWarm upAkuna CapitalChicago · 2025
Try it first
First, how many of the 36 outcomes total 10?
Show the worked solution
At most Rs 100/11, about Rs 9.09. A total of 10 comes from (4, 6), (5, 5) and (6, 4), three of the 36 equally likely outcomes, so the chance is 1/12 and the odds are 11 to 1 against. A fair bet matches the odds: risk 1 to win 11, so to win Rs 100 the fair stake is 100/11. Risk Rs 10 and you lose about Rs 0.83 per game on average.
Why is the answer odds, not a probability?
A friend offers to pay you Rs 100 if your cricket team wins a match it wins one time in twelve. Paying Rs 20 to enter is a bad deal, and the way you know is by comparing what you risk to what you get: you lose Rs 20 eleven times and win Rs 100 once. A bet is fair when the stake and the prize are in the same ratio as the losing and winning chances, which is what odds express directly. One in twelve means eleven losses per win, so a fair stake is one eleventh of the prize: Rs 100/11, which is Rs 9.09. Anything above that is paying for the privilege of playing.
Of the 36 ordered outcomes of two dice, exactly three total 10, so the chance is 1/12, the odds are 11 to 1 against, and the fair stake to win Rs 100 is Rs 100/11, about Rs 9.09; staking Rs 10 loses about Rs 0.83 per game on average.
How do you check the stake with expected value?
Set the expected profit to zero. With stake s, you win Rs 100 with chance 1/12 and lose s with chance 11/12, so the expectation is 100/12 - 11s/12, which is zero when s = 100/11. The expected value route and the odds route must agree, and saying both in the room shows the interviewer you can move between them. Then say the direction: a lower stake is a profitable bet for you, a higher one is profitable for the house, and at Rs 10 the house edge is 8.3% of your stake, which is roughly the edge a casino takes on its better games.
The relationship
$E[\text{profit}] = \tfrac{1}{12}\times 100 - \tfrac{11}{12}\times s = 0 \;\Rightarrow\; s = \tfrac{100}{11} \approx 9.09$
1/12 the chance of a total of 10, three cells out of 36
s the stake you risk
100/11 the fair stake, where the bet has zero expected value
What it says in wordsThe fair stake is the prize divided by the odds against, which makes the expected profit zero.
What would a trader add after the number?
That the fair stake is a ceiling, not an offer. A trader asked to risk money quotes below fair, say Rs 8, and explains the gap as the edge for taking the other side, because a bet at exactly fair value makes nothing over many plays. The limitation is that 1/12 is the chance only if the dice are fair and the counting is right; if the interviewer changes the target to 7, the chance rises to 6/36 and the fair stake jumps to Rs 20, so the first thing to re-check whenever the rules change is the count of winning cells.
Where candidates lose it
The common error is in the count: giving 2 winning outcomes by treating 4-6 and 6-4 as the same, or 4 by counting 5-5 twice. Lay out the ordered pairs and say three of 36.
The second loss is quoting Rs 8.33, which is 100/12. That is the expected prize, not the fair stake. Odds of 11 to 1 mean the stake is 100/11.
What the interviewer asks next
Same game but you win on a total of 7. What is the fair stake now?
You are offered the bet at a stake of Rs 8. What is your expected profit per game?
How would you change the stake if the dice might be loaded?
You can play 100 times at Rs 8. What is the chance you end up behind?
Asked at Akuna Capital, Trading, Chicago, 2025 (Wall Street Oasis): We are playing a game where we roll two dice. You win if you roll a 10, how much would you risk to win $100?
060A bowl holds 100 cooked noodles. You repeatedly pick two free ends at random and tie them together, until no free ends remain. What is the expected number of loops in the bowl?Probability and countingHardD.E. ShawNew York · 2026
Try it first
100 noodles, 100 ties. Roughly how many loops do you expect at the end?
Show the worked solution
About 3.28 loops. Each tie either closes a loop or joins two strands into one longer strand, so after every tie there is one strand fewer. With n strands left there are 2n free ends; pick one, and of the other 2n - 1 ends exactly one belongs to the same strand, so that tie closes a loop with chance 1/(2n - 1). Linearity of expectation adds those chances over n = 100 down to 1: 1/199 + 1/197 + ... + 1/3 + 1.
Why does every tie either close a loop or shorten the list by one strand?
Think of 100 pieces of string on a table. Tie two ends from different pieces and you now have 99 pieces, one of them longer. Tie the two ends of the same piece and you have a ring and 99 pieces left as well. Whatever happens, the number of strands with free ends falls by exactly one per tie, so there are exactly 100 ties, and the only question at each tie is whether it closed a loop. That makes the count of loops a sum of 100 indicator events, which is the signal to use linearity of expectation rather than to enumerate outcomes.
The chance that a tie closes a loop is 1/(2n - 1) with n strands left, so it is 1/199 at the first tie and stays near zero until the last handful, reaching 1/3 and then 1; the running expected total climbs slowly and reaches 3.28 only because the last ten ties alone contribute 2.13.
Where does 1/(2n - 1) come from?
With n strands there are 2n free ends. Pick the first end; it belongs to some strand. Of the remaining 2n - 1 ends, exactly one is the other end of that same strand, and all are equally likely, so the tie closes a loop with chance 1/(2n - 1). It does not matter how long the strands have become or how many loops already sit in the bowl, because closed loops have no free ends and are out of the picture. The expectation is therefore the sum of 1/(2n - 1) for n from 100 down to 1, which is the odd-denominator half of the harmonic series.
The relationship
$E[\text{loops}] = \sum_{n=1}^{100} \frac{1}{2n-1} = 1 + \tfrac{1}{3} + \tfrac{1}{5} + \cdots + \tfrac{1}{199} \approx 3.284 \approx \tfrac{1}{2}\ln(4n) + \tfrac{\gamma}{2}$
n the number of strands with free ends before a tie
1/(2n - 1) the chance that tie closes a loop
gamma the Euler constant, about 0.577, in the logarithmic approximation
What it says in wordsThe expected number of loops is the sum of the odd reciprocals up to 1/199, which grows only like half the logarithm of the number of noodles.
What is the approximation, and why is the answer so small?
The sum of odd reciprocals up to 1/(2n - 1) is close to half of ln(4n) plus half the Euler constant, which for n = 100 gives 3.28 against the exact 3.284. Doubling the number of noodles adds only about 0.35 to the expected number of loops, so a bowl of a thousand noodles still gives only about four loops. The intuition is that early ties almost never close a loop: they build a few very long strands, and the loops appear at the end when there are only two or three strands left. The limitation is that this is an expectation only; the distribution is skewed, and ending with a single loop is the most likely outcome.
Where candidates lose it
Most candidates try to track the configuration of strands, which explodes. The interviewer wants the one observation that each tie closes a loop with chance 1/(2n - 1) regardless of history, and then linearity of expectation.
The second loss is guessing a large number. Say out loud that the early ties almost never close a loop, so the answer is a slowly growing sum, and that the harmonic-style sum of 100 terms is around 3, not 50.
What the interviewer asks next
What is the probability that all 100 noodles end up in a single loop?
Approximately how many noodles would you need for the expected number of loops to reach 5?
What is the variance of the number of loops?
Now you tie ends only across different strands when you can. How many loops then?
Asked at D.E. Shaw, Research, New York, 2026 (Wall Street Oasis): What is the expected number of loops from tying 100 noodles' ends together randomly
062I am going to roll a die six times. Make me a market on the number of different faces that show up.Market makingHardOptiverSan Francisco · 2026
Try it first
Before quoting: where is the fair value of the number of distinct faces?
Show the worked solution
Fair value is about 3.99, so quote something like 3.9 bid, 4.1 offer. A given face is absent from all six rolls with probability (5/6) to the sixth, about 0.335, so it appears at least once with probability 0.665. The number of distinct faces is the sum of six such indicators, and linearity of expectation gives 6 x 0.665 = 3.99, without listing a single case.
Why does linearity let you skip the cases?
Six friends each toss a letter into one of six boxes at random, and you want the expected number of boxes that end up non-empty. Counting the ways the boxes can fill is a mess; asking each box whether it got anything is easy. The number of distinct faces is one plus one plus one over the six faces, each one counting if that face appeared, and the expectation of a sum is the sum of the expectations even though the six events overlap. Each face is missed on every roll with chance (5/6) to the sixth, so the expected count is 6 times one minus that, about 3.99.
Each of the six faces appears at least once with probability 0.665, so the expected number of distinct faces is 6 x 0.665 = 3.99, and the exact distribution peaks at four faces with standard deviation 0.78, which is what a market of 3.9 at 4.1 is priced around.
How wide should the market be, and how do you defend it?
Width comes from how uncertain the outcome is and how much the other side may know. The exact distribution puts about 50% of the mass on four faces, 23% on three and 23% on five, with a standard deviation of 0.78, so a market 0.2 wide around 3.99 is tight relative to the noise but still symmetric around fair. If the interviewer lifts your offer at 4.1, you are short at a price above fair and should keep the quote where it is; if they lift it twice, ask yourself whether they know something, such as that the die is loaded, and move the market up rather than argue with the flow.
The relationship
$E[D] = \sum_{i=1}^{6} P(\text{face } i \text{ appears}) = 6\left(1 - \left(\tfrac{5}{6}\right)^6\right) \approx 6 \times 0.6651 = 3.991$
D the number of distinct faces in six rolls
(5/6)^6 the chance a particular face is missed on all six rolls
6 the number of faces, each contributing one indicator
What it says in wordsThe expected number of distinct faces is six times the chance that any one face appears at least once.
What is the check, and what changes with more rolls?
Check the ends. With one roll there is exactly one distinct face and the formula gives 6(1 - 5/6) = 1. With many rolls the missed chance collapses and the expectation approaches 6. At six rolls you are at 3.99, meaning two faces are typically missing, which surprises people who expect six rolls to nearly cover six faces. The limitation of the quote is that it prices only the mean; a counterparty who wants to bet on exactly six distinct faces needs a different market, and that chance is only 6!/6^6, about 1.5%.
Where candidates lose it
Candidates start enumerating outcomes, or quote 6 because there are six rolls. The interviewer is looking for the indicator trick: one event per face, sum the probabilities, no cases.
The second loss is a market with no reasoning behind its width. Say the standard deviation, say the market is symmetric around fair, and say what you would do when the other side trades with you twice in the same direction.
What the interviewer asks next
What is the probability that all six faces appear in six rolls?
Make a market on the number of distinct faces in twelve rolls.
The interviewer hits your bid three times in a row. What do you do with the quote?
What is the variance of the number of distinct faces, and does it matter for the quote?
Asked at Optiver, Quant Research Interview, San Francisco, 2026 (Wall Street Oasis): They do ask one round of market making game-like question
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.
065A contract pays, in rupees, the amount by which a fair die roll exceeds 4, and nothing otherwise. What is it worth? What about the matching put, which pays the amount by which the roll falls short of 4?Option payoffs and no-arbitrageWarm upBelvedere TradingChicago · 2021
Try it first
The die call with strike 4. What is it worth?
Show the worked solution
The call is worth Rs 0.50 and the put is worth Rs 1.00. The call pays 1 on a 5 and 2 on a 6, so its average payoff is 3/6. The put pays 3 on a 1, 2 on a 2 and 1 on a 3, so its average is 6/6. The put is worth more because the strike of 4 sits above the die's mean of 3.5, and call minus put equals 3.5 minus 4, which is minus 0.5.
Why is a die option priced by averaging the payoffs?
A school raffle with six equally likely tickets where ticket 5 pays Rs 1 and ticket 6 pays Rs 2 is worth exactly the average prize, Rs 0.50, because nothing else is uncertain and nobody can hedge a die roll. With equally likely outcomes and no hedge available, the fair price of a payoff is its expected value, so you list the payoff on each face and average. For the call with strike 4: 0, 0, 0, 0, 1, 2, which averages 0.5. For the put: 3, 2, 1, 0, 0, 0, which averages 1.0. Each payoff is floored at zero, which is what makes it an option rather than a forward.
Face by face, the strike-4 call pays 0, 0, 0, 0, 1 and 2 for an average of 0.5, the put pays 3, 2, 1, 0, 0 and 0 for an average of 1.0, and the difference of minus 0.5 equals the expected roll of 3.5 minus the strike of 4.
What is the parity check, and why does it work on a die?
Add the call and subtract the put on every face. Call minus put on any single face equals the roll minus 4 exactly, because whichever side is in the money pays the gap and the other pays nothing, so the average of call minus put is the average roll minus the strike: 3.5 - 4 = - 0.5. That is put-call parity with no interest and no dividends, and it gives a one-line check: once you have the call at 0.5, the put must be 0.5 + 0.5 = 1.0. On a real option the same identity holds with the forward in place of the expected roll.
The relationship
$C = \frac{1+2}{6} = \tfrac{1}{2}, \qquad P = \frac{3+2+1}{6} = 1, \qquad C - P = E[\text{roll}] - K = 3.5 - 4 = -\tfrac{1}{2}$
C, P the values of the die call and die put with strike 4
E[roll] the expected face of a fair die, 3.5
K the strike, 4
What it says in wordsThe call is worth half a rupee, the put one rupee, and their difference equals the expected roll minus the strike, which is parity.
What does the interviewer ask next, and where does the analogy stop?
The next question is usually a different strike, or a market. Move the strike to 3 and the call pays 1, 2 and 3 on the top three faces, worth 1.0, while the put pays 2 and 1, worth 0.5, so the two swap values because the strike is now below the mean. Where the analogy stops is hedging: a real option is priced not by the expected payoff under your view but by the cost of replicating it with the underlying, which shifts the probabilities to the risk-neutral ones; on a die there is nothing to trade against, so the expectation under the real probabilities is the price.
Where candidates lose it
The fast wrong answer is to compute 3.5 minus 4 and say the call is worth minus 0.5, or to count two paying faces and say 1/3. A call never pays a negative amount: list the payoffs face by face and average.
The second loss is pricing the put from scratch and getting it right while missing the parity relation. Say call minus put equals 3.5 minus 4 and the interviewer hears that you know what parity is.
What the interviewer asks next
Price the call and the put with strike 3.
What is the value of a contract that pays the square of the roll minus 10, floored at zero?
Make a two-way market on the strike-4 call.
Why does put-call parity on a real stock use the forward rather than the expected price?
Asked at Belvedere Trading, Generalist, Chicago, 2021 (Wall Street Oasis): Pricing an option contract on a game involving rolling a die.
067A staircase has 10 steps and you climb either one or two steps at a time. How many different ways are there to reach the top?Probability and countingWarm upTower Research CapitalNew York · 2012
Try it first
Ten steps, singles or doubles. How many routes?
Show the worked solution
89 ways. Think about the last move. Either it was a single step from step 9 or a double step from step 8, and those two cases cannot overlap, so ways(10) = ways(9) + ways(8). With ways(1) = 1 and ways(2) = 2, the sequence runs 1, 2, 3, 5, 8, 13, 21, 34, 55, 89. It is the Fibonacci rule, shifted by one place.
Why count by the last move rather than the first?
Ask how many ways there are to arrive at a railway junction and you count the lines coming in, not the stations people set out from. Every route to step n arrives from exactly one of two places, step n - 1 by a single or step n - 2 by a double, so the routes to n are the routes to those two places added together. The first move works just as well, but the last move makes the recursion read naturally from the top down, and it is the habit that generalises to harder counting problems where you condition on the final event.
Each count is the sum of the two before it, because the last move is a single step from n - 1 or a double from n - 2, and from ways(1) = 1 and ways(2) = 2 the sequence reaches 89 at ten steps.
How do you check 89 a second way?
Count by how many doubles you use. With k doubles and 10 - 2k singles you make 10 - k moves in total, and the number of orderings is 10 - k choose k, so the total is the sum over k from 0 to 5 of C(10 - k, k). That is 1 + 9 + 28 + 35 + 15 + 1 = 89, the same answer by a route that does not use the recursion at all. Two methods agreeing is the thing to say out loud; it also hands you the next question, since the terms tell you that four doubles and two singles is the most common shape of route.
The relationship
$w(n) = w(n-1) + w(n-2), \quad w(1) = 1,\ w(2) = 2 \;\Rightarrow\; w(10) = 89 = \sum_{k=0}^{5}\binom{10-k}{k}$
w(n) the number of ways to climb n steps in singles and doubles
k the number of double steps used in a route
C(10 - k, k) the ways to place k doubles among 10 - k moves
What it says in wordsThe count follows the Fibonacci rule and equals the sum over the number of doubles of the ways to arrange them.
Where does this pattern appear in trading, and where does it stop?
In anything built from steps of two sizes: the number of ways a price can move up to a level in ticks of one and two, or the number of paths in a recombining tree. The recursion is also the warm-up for dynamic programming, where the value of a position is built from the values of the positions it can reach, which is how an American option is priced on a lattice. The limitation is that Fibonacci only appears when every move is a one or a two; allow a three-step jump and the rule becomes a sum of the previous three terms, with 274 ways for ten steps.
Where candidates lose it
The common wrong answer is 2 to the 10, from imagining a free choice at every step. A double skips a step, so the choices are not independent. Set up the recursion by the last move and the structure appears.
The second loss is starting the sequence at the wrong place. Ways(1) is 1 and ways(2) is 2, so ten steps give 89 and not 55 or 144.
What the interviewer asks next
Now you may also take three steps at a time. How many ways for 10 steps?
How many of the 89 routes use exactly three double steps?
What is the probability a random route uses no doubles at all?
How is this recursion related to pricing an option on a binomial tree?
Asked at Tower Research Capital, Intern Interview -, New York, 2012 (Wall Street Oasis): How many ways can you jump up stairs if you can only jump either 1 or 2 steps? Answer: Fibonacci sequence.

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

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

a	the nearest round number, here 1,000 or 100
b, c	the small distances from the round number, here 2 and 3, or 1
bc	the small product that carries the sign most people get wrong

rho_XY, rho_YZ	the two given correlations, 0.2 and 0.5
rho_XZ	the correlation being bounded
the square root term	the half-width of the feasible band, which shrinks as the given correlations strengthen

Try it first

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

Try it first

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

Why is the round number always the first move?

How do you handle percentages that are not round?

What does the interviewer listen for beyond the answer?

Where candidates lose it

What the interviewer asks next

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Why can the third correlation not be anything it likes?

How do you get the band without writing out eigenvalues?

Where does this show up on a desk?

Where candidates lose it

What the interviewer asks next

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Why is the answer odds, not a probability?

How do you check the stake with expected value?

What would a trader add after the number?

Where candidates lose it

What the interviewer asks next

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Why does every tie either close a loop or shorten the list by one strand?

Where does 1/(2n - 1) come from?

What is the approximation, and why is the answer so small?

Where candidates lose it

What the interviewer asks next

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Why does linearity let you skip the cases?

How wide should the market be, and how do you defend it?

What is the check, and what changes with more rolls?

Where candidates lose it

What the interviewer asks next

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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 is a die option priced by averaging the payoffs?

What is the parity check, and why does it work on a die?

What does the interviewer ask next, and where does the analogy stop?

Where candidates lose it

What the interviewer asks next

Try it first

Why count by the last move rather than the first?

How do you check 89 a second way?

Where does this pattern appear in trading, and where does it stop?

Where candidates lose it

What the interviewer asks next