Derivatives Foundation puzzles, solved step by step

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

Showing 51–60 of 100

051You have the same 60% coin as before and even money on every flip, but now you bet half your capital every time. After 100 bets, what does your typical outcome look like, even though every single bet has positive expected value?Betting and sizingHardRisk managementHedge funds
Try it first
Each bet has a 10% edge on the money at risk. Where does betting half your stack 100 times leave the typical player?
Show the worked solution
The typical player ends with about 3% of the starting capital, while the average outcome is about 13,781 times the start. With 60 wins and 40 losses the stack is multiplied by 1.5 to the 60 and 0.5 to the 40, which is 0.033. Betting half the stack is 2.5 times the Kelly fraction of 20%, and anything beyond twice Kelly turns a positive-edge game into a losing one for the person actually playing it.
Why can every bet be favourable and the typical result still be a loss?
Think of a shopkeeper who doubles the stock every time a line sells and halves it every time it does not. A good line sells six times in ten, yet after a hundred cycles the shelf is nearly bare, because growth is a chain of multiplications and the order of operations does not rescue you. Expected value is an average across many imaginary players, but your capital follows one path, and a path is a product of factors, not a sum. Betting half the stack makes each win a factor of 1.5 and each loss a factor of 0.5. Sixty wins and forty losses, the most likely split, multiply to 0.033. The arithmetic mean of 13,781 is carried by the rare paths with 80 or 90 wins, which almost nobody lives on.
On a shared log scale, the average path compounds at 1.1 per bet to about 13,781 times the start after 100 bets, while the typical path of 60 wins and 40 losses shrinks by 3.4% per bet to about 3% of the start, because capital multiplies and the half-stack bet makes the loss factor too harsh.
What is the right quantity to look at instead of the mean?
Look at the expected growth rate of the logarithm of capital, because the log turns products into sums and the law of large numbers then works. Each bet changes log capital by 0.6 x log(1.5) + 0.4 x log(0.5), which is -0.034, a negative number, so the typical player loses about 3.4% of capital per bet on average. Over 100 bets that is a factor of exp(-3.4), the 0.033 you saw above. The chance of finishing below the start is 76%, so the mean is a number you will almost never meet.
The relationship
$g(f) = p\,\ln(1+f) + (1-p)\,\ln(1-f) \qquad g(0.5) = 0.6\ln 1.5 + 0.4\ln 0.5 \approx -0.034 \qquad f^* = 2p-1 = 0.2$
g(f) the expected growth of log capital per bet when you stake fraction f
p the chance of winning each even-money bet, here 0.6
f* the Kelly fraction that maximises g, here 20% of capital
What it says in wordsStaking half the stack makes the expected log growth negative, while the Kelly stake of a fifth makes it positive.
Where does the sizing go wrong, and what would fix it?
The Kelly fraction for an even-money bet is 2p minus 1, here 20% of capital, and it maximises the growth rate at +0.0201 per bet, which turns 100 bets into a typical multiple of about 7.5. Growth is zero at exactly twice Kelly, 40% here, and negative beyond it, so a half-stack bettor is on the wrong side of the hill even though the coin is in their favour. The limitation is that the Kelly rule assumes you know p exactly and will play many times; with an uncertain edge, desks size below Kelly on purpose, often at half of it.
Where candidates lose it
Most candidates say the player ends up rich, because they multiply the 10% edge through 100 bets and quote the mean. The interviewer wants to hear the word median, or typical, and the observation that capital compounds multiplicatively, so wins and losses of equal size do not cancel.
The second loss is stopping at the picture without the growth rate. Say the log growth per bet, show it is negative at a half stake and positive at a fifth, and name twice Kelly as the point where the game turns against you.
What the interviewer asks next
What fraction maximises the typical growth, and what is the growth rate there?
At what stake does the typical player break even despite the edge?
If you only get to play 5 times rather than 100, does the answer change?
Your estimate of the 60% is itself noisy. Does that push the stake up or down?
052A bus makes three stops. At each stop half the people on board get off, and then the number on board grows by a third. The bus reaches the end of the route with 16 people. How many were on board at the start?Mental maths and estimationCoreProp trading firms
Try it first
Before working it: each stop multiplies the number on board by what?
Show the worked solution
54 people. Each stop halves the load and then adds a third of what is left, so the load is multiplied by 1/2 x 4/3 = 2/3 at every stop. Undoing three stops means dividing 16 by 2/3 three times: 16 to 24 to 36 to 54. The forward check works: 54 to 27 to 36, then 18 to 24, then 12 to 16.
Why work backwards rather than guess a start?
If someone tells you a shirt was marked down by half, then marked up by a third, and now costs Rs 16, you do not guess the original tag; you undo the two moves in reverse order. Each stop is a pair of multiplications, and multiplications are undone by dividing in the opposite order, so the end number walks back to the start without any trial and error. The growth of a third comes last at each stop, so it is undone first: divide by 4/3, which is multiply by 3/4. Then undo the halving by doubling. From 16: times 3/4 is 12, times 2 is 24. Twice more gives 36 and then 54.
Starting from 16 at the end of the route and undoing each stop, first multiplying by 3/4 then by 2, the count climbs 16 to 24 to 36 to 54, and running the three stops forward from 54 returns exactly 16, because each stop multiplies the load by 2/3.
What is the one-line version an interviewer wants to hear?
Collapse each stop to a single factor. Losing half is x 1/2 and gaining a third is x 4/3, so a stop is x 2/3, and three stops are x 8/27. The start is 16 x 27/8 = 54. Saying that in one breath shows you saw the structure rather than the arithmetic, which is what the question is for. The limitation is that it only works because every stop has the same two moves; a route where the fractions change needs the step-by-step walk back.
The relationship
$N_0 = N_3 \times \left(\tfrac{1}{2}\times\tfrac{4}{3}\right)^{-3} = 16 \times \left(\tfrac{3}{2}\right)^3 = 16 \times \tfrac{27}{8} = 54$
N_0 the number on board at the start
N_3 the number at the end of the route, 16
(3/2)^3 undoing the per-stop factor of 2/3 three times
What it says in wordsThe start is the end count divided by the per-stop factor of two thirds, three times over.
Where do the whole-number checks help you?
The bus carries people, so every intermediate count must be a whole number, and that is a free check. If the backwards walk ever produces a fraction, you have undone the moves in the wrong order. Undoing the halving first from 16 gives 32, then times 3/4 gives 24, which happens to be whole here, but on the next stop it would give 48 then 36, and the order mistake would have cost you nothing visible until the end, where 72 x 3/4 = 54 again by luck. Order matters in general even when the numbers hide it, so say the order out loud.
Where candidates lose it
The common slip is treating the two moves at each stop as a net loss of a sixth, since a half minus a third is a sixth. The third is taken on the smaller number, so the stop is a factor of 2/3, not 5/6.
The second slip is undoing the moves in the wrong order. Growth came last, so it is undone first. Say the per-stop factor, then walk back 16, 24, 36, 54, and finish with the forward check.
What the interviewer asks next
If instead a third get off and then the number on board doubles at each stop, what is the per-stop factor?
How many stops would it take for a bus starting with 54 to get below 5 people?
The bus starts with 54 and the pattern continues. After how many stops is the number no longer a whole number?
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
055A token sits on one corner of a square. Every second it moves to one of the two neighbouring corners, chosen by a fair coin. What is the expected number of seconds until it first reaches the opposite corner?Random walks and Markov chainsCoreQuant trading
Try it first
The opposite corner is two steps away. Expected time to get there?
Show the worked solution
4 seconds. By symmetry, the two corners next to the start are the same state, call it adjacent. From the start you always move to adjacent in one step. From adjacent, half the time you reach the target and half the time you return to the start. So E(start) = 1 + E(adjacent) and E(adjacent) = 1 + E(start)/2, giving E(adjacent) = 3 and E(start) = 4.
Why collapse four corners into three states?
If you are lost in a town with a river on one side, what matters is how far you are from the river, not which street you are on. The square is the same: standing at either corner next to the start, the token's future looks identical, one coin flip from the target and one from the start. Grouping corners by their distance from the target turns a four-state chain into a three-state line, and a line is solved with one equation per unknown. You have two unknowns, the expected time from the start and from an adjacent corner, because the target itself takes zero time.
Grouping the two corners next to the start into one state gives a three-state chain in which the start always moves to adjacent, and adjacent finishes or returns to the start with equal chance, so E(adjacent) = 3 and E(start) = 4 seconds.
How do you set up and solve the equations in the room?
Each equation says the same sentence: one step, plus the average of what is left from where you land. From the start, every step lands on an adjacent corner, so E(start) = 1 + E(adjacent); from an adjacent corner, half the steps finish and half return, so E(adjacent) = 1 + (1/2) x 0 + (1/2) x E(start). Substitute the first into the second: E(adjacent) = 1 + (1/2)(1 + E(adjacent)), so E(adjacent)/2 = 3/2, E(adjacent) = 3, and E(start) = 4. Saying the sentence before the symbols is what keeps the equations honest.
The relationship
$E_0 = 1 + E_1, \qquad E_1 = 1 + \tfrac{1}{2}E_0 \;\Rightarrow\; E_1 = 3,\; E_0 = 4$
E_0 the expected steps to the target from the starting corner
E_1 the expected steps from either corner next to the start
1/2 the chance a step from an adjacent corner lands on the target
What it says in wordsFrom the start you always move one step closer; from there a coin flip either finishes or sends you back, and the two equations give four steps on average.
What is the check, and what does the general case look like?
Check it with the geometric picture. After the first step you are adjacent, and from there each attempt either finishes in one step or costs two steps, back to the start and out again, before you are adjacent once more. Each attempt succeeds with chance 1/2, so E(adjacent) = (1/2) x 1 + (1/2) x (2 + E(adjacent)), which again gives 3, and the first step makes it 4. On a cube the same method with four distance classes gives 10 steps to the opposite vertex; on a general graph the method is the same, but the number of distance classes grows and the arithmetic stops being mental.
Where candidates lose it
The fast wrong answer is 2, the length of the shortest path. The interviewer is checking whether you see that the token can bounce back, and whether you reach for the first-step equations rather than trying to sum a series.
The second loss is writing four equations, one per corner. Say the symmetry out loud, collapse to three states, and the whole thing is two lines.
What the interviewer asks next
What is the expected time to return to the starting corner for the first time?
Same walk on the eight corners of a cube. Expected time to the opposite vertex?
What is the probability the token reaches the opposite corner within 4 steps?
The coin is biased: it moves clockwise with chance 0.7. Does the expected time change?
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?
059On a stock that pays no dividend, the three-month 100-strike call trades at 6.50 and the six-month 100-strike call trades at 6.10. Is there an arbitrage, and how would you capture it?Option payoffs and no-arbitrageCoreMarket makingVolatility trading
Try it first
Two calls, same strike, the longer one cheaper. What do you do?
Show the worked solution
Yes. Sell the three-month call at 6.50, buy the six-month call at 6.10, and lock in 0.40 today with no risk. On a non-dividend stock, a call with more time to expiry is worth at least as much as the same strike with less, because at three months the six-month call is still worth at least its intrinsic value, which is exactly what the expiring call pays out. The position never owes money, so the 0.40 is a pure profit.
Why must the longer call be worth at least the shorter one?
A ticket that lets you buy a train seat any time in the next six months is worth at least as much as one that expires in three; you can always do with the long ticket exactly what you would have done with the short one, and then keep it. At the three-month expiry the short call pays max(S - 100, 0), and the six-month call at that moment is worth at least max(S - 100, 0) too, because a call on a non-dividend stock is never worth less than its intrinsic value and is usually worth more. So holding the long and being short the short can never leave you with a negative balance at three months, and the 0.40 you took in today is yours.
At-the-money call value rises with time to expiry, so the six-month call must be worth at least the 6.50 the three-month call trades at; the quoted 6.10 sits below that floor, and selling the three-month while buying the six-month locks in 0.40 with a spread that can only pay more.
What happens at three months in each case?
Walk through both branches. If the stock is above 100, the short call is exercised against you, you deliver stock for 100, and you still hold a six-month call worth more than S - 100, so you exercise or sell it and finish ahead; if the stock is at or below 100, the short call expires worthless and you are left holding a live six-month call for free. Either way you keep the 0.40 from today and own something worth zero or more. The limitation is the non-dividend assumption: with a large dividend before six months, early exercise of an American three-month call could matter, and the comparison needs the dividend's present value subtracted.
The relationship
$C(K, T_2) \ge C(K, T_1) \quad \text{for } T_2 > T_1 \text{ on a non-dividend stock} \qquad 6.10 < 6.50 \Rightarrow \text{sell } T_1,\ \text{buy } T_2,\ \text{keep } 0.40$
C(K, T) the price of a call with strike K and time to expiry T
T_1, T_2 three and six months
0.40 the cash locked in today, the smallest profit the trade can make
What it says in wordsA longer-dated call at the same strike can never be worth less than a shorter one, so a longer call quoted cheaper is sold against the shorter one for a riskless profit.
Why would a market maker see this quote and what does it say about the vol surface?
The three-month price of 6.50 corresponds to an implied volatility of about 33% with zero rates, and at that volatility a six-month call would be worth around 9.18. A quote of 6.10 for the six-month implies a term structure of volatility so inverted that no volatility at all could justify it, which is why this is an arbitrage and not merely a view on calendar spreads. Real quotes rarely break the bound outright; what you see instead is a long-dated call a few ticks above the floor on an illiquid name, and the question is then whether the bid-ask spread swallows the edge before you can trade both legs.
Where candidates lose it
Candidates treat it as a volatility question and start talking about the term structure. Implied volatility cannot push a longer call below a shorter one at the same strike, so the answer is a bound, not a view.
The second loss is getting the direction wrong under pressure. Sell what is dear, the three-month at 6.50; buy what is cheap, the six-month at 6.10. Then walk the two branches at the first expiry out loud.
What the interviewer asks next
Does the same bound hold for puts on a non-dividend stock?
The stock pays a large dividend in month four. Can the three-month call now be worth more than the six-month?
What if the two calls had different strikes, say 100 and 105?
Both quotes are mid prices and the bid-ask is 0.30 on each. Is the arbitrage still there?
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

g(f)	the expected growth of log capital per bet when you stake fraction f
p	the chance of winning each even-money bet, here 0.6
f*	the Kelly fraction that maximises g, here 20% of capital

N_0	the number on board at the start
N_3	the number at the end of the route, 16
(3/2)^3	undoing the per-stop factor of 2/3 three times

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

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