Derivatives Foundation puzzles, solved step by step

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

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078Three calls on the same stock and expiry trade at: strike 90 for 14, strike 100 for 8, strike 110 for 1. Is there an arbitrage? Build it.Option payoffs and no-arbitrageHardMarket makingVolatility trading
Try it first
Which relationship between the three prices should you test first?
Show the worked solution
Yes. Buy the 90 call, sell two 100 calls, buy the 110 call, and you are paid 1 to own a payoff that is never below zero. The butterfly costs 14 - 16 + 1 = -1, so you receive 1 today. At expiry it pays nothing below 90, rises to 10 at 100 and falls to nothing above 110, never negative. Call prices must be convex in strike: the 100 call cannot exceed (14 + 1)/2 = 7.5, and it trades at 8.
Where does the 7.5 come from?
Picture three houses on one street at 90, 100 and 110 square metres, priced at 14, 8 and 1. The middle house should be worth no more than the average of its neighbours if each extra metre is worth less than the last; if it is priced above the average, you sell it and buy the two neighbours. A call's price falls as the strike rises, and it falls at a decreasing rate, so the price at the middle strike must sit on or below the straight line between its neighbours. The line from 14 at 90 to 1 at 110 passes through 7.5 at 100. The market says 8. That half point is the mispricing.
The butterfly of long one 90 call, short two 100 calls and long one 110 call pays nothing below 90, peaks at 10 at a stock price of 100 and pays nothing above 110, and because it costs minus 1 the whole payoff sits at least 1 above the price paid, while on the right the quoted 8 for the 100 call sits above the 7.5 chord between the 90 and 110 calls, which breaks convexity.
How do you prove the payoff is never negative?
Walk the stock price up. Below 90 nothing is in the money, payoff zero. Between 90 and 100 only the 90 call pays, so the payoff is the stock price minus 90, rising to 10. Between 100 and 110 the two short 100 calls start paying out, and the slope turns to 1 - 2 = -1, so the payoff falls from 10 back to zero at 110. Above 110 the long 110 call kicks in and the slope is 1 - 2 + 1 = 0: flat at zero. The payoff is zero or positive everywhere, you were paid 1 to hold it, so you have a riskless profit of at least 1 and up to 11.
The relationship
$C(100) \le \tfrac{1}{2}\,[C(90) + C(110)] = \tfrac{1}{2}(14 + 1) = 7.5 < 8$
C(K) the price of the call struck at K
7.5 the midpoint of the chord between the outer strikes
8 the quoted middle price, half a point too high
What it says in wordsWith equally spaced strikes, the middle call can never cost more than the average of the outer two, because the butterfly that tests it can only pay out zero or more.
Check the other bounds too, out loud, so the interviewer sees you are not pattern matching. A call spread can never be worth more than the gap between its strikes: the 90/100 spread costs 6 and the 100/110 spread costs 7, both under 10, so those pass. Prices fall as strike rises: 14, 8, 1, pass. Only convexity fails. The limitation: in a real screen you trade at bids and offers, not mids, and a half point of theoretical edge can vanish inside the spread. The structure is an arbitrage at these prices; whether you can execute it is a separate question.
Where candidates lose it
Candidates test the wrong bound. They check that each call is worth more than its intrinsic value, or that a spread is worth less than the strike gap, find both fine, and declare no arbitrage. Convexity is the bound that catches a dear middle strike, and it is the one most people forget.
The second loss is building the butterfly the wrong way round. If you sell the wings and buy the middle you pay 1 to hold a payoff that is zero or negative. Say which leg is long before you say the price.
What the interviewer asks next
Reprice the 100 call so there is no arbitrage. What is the highest price it can take?
If the strikes were 90, 100 and 120, how would you weight the legs?
Why does the same convexity rule hold for puts?
081Two assets show negative correlation in daily returns within each month, but positive correlation in monthly returns across a year. How is that possible? Build an example.Volatility and correlationHardSquarepoint CapitalMontreal · 2024
Try it first
What has to be true of the two return series for the sign to flip with the horizon?
Show the worked solution
Give both assets a shared monthly drift and opposing daily noise. Let A's daily return be m/21 + e + u and B's be m/21 - e + v, where m is a monthly drift common to both, e is noise that hits them in opposite directions, and u and v are their own noise. Inside a month m is fixed, so daily returns co-move through -e alone: correlation -0.50 with equal variances. Over a month e sums to a modest total while m is large, so monthly returns correlate at +0.66.
Why can one pair of series carry two correlations?
Two shops on the same street both do better in a festival month and worse in a quiet one, so their monthly takings rise and fall together. But on any given day, the customer who buys lunch at one is not buying it at the other, so their daily takings move against each other. Nothing is paradoxical: the two effects live at different speeds. Correlation is not a property of two assets, it is a property of two assets at a horizon, because different components of the return dominate at different horizons. The slow shared component barely moves within a day; the fast opposing component barely survives a month.
Two simulated paths with a shared drift of 12 points a month and opposing daily noise of 1 point zigzag against each other day by day yet climb and fall together month by month, giving an average within-month daily correlation of -0.54 and a correlation of 0.64 across the twelve monthly returns.
How do you put numbers on the construction?
Take a drift m with a standard deviation of 12 points a month, opposing noise e of 1 point a day, and own noise u and v of 1 point a day. Within a month m is a constant, so it drops out of the daily covariance: cov = -var(e) = -1, and each variance is var(e) + var(u) = 2, so the daily correlation is -1/2. Over 21 days the sums of e, u and v each have variance 21, and the monthly covariance is var(m) - 21 var(e) = 144 - 21 = 123 against a variance of 144 + 21 + 21 = 186. The monthly correlation is 123/186 = 0.66, positive, because the shared drift's variance grows with the square of the month while the noise variance only grows in proportion to it.
The relationship
$\rho_{\text{day}} = \frac{-\sigma_e^2}{\sigma_e^2 + \sigma_u^2} = -\tfrac{1}{2}, \qquad \rho_{\text{month}} = \frac{\sigma_m^2 - 21\sigma_e^2}{\sigma_m^2 + 21\sigma_e^2 + 21\sigma_u^2} = \frac{144 - 21}{186} = 0.66$
sigma_m the standard deviation of the shared monthly drift, 12 points
sigma_e the daily opposing noise, 1 point
sigma_u each asset's own daily noise, 1 point
What it says in wordsWithin a month only the opposing noise moves, so the correlation is negative; across months the shared drift dominates, so it turns positive.
Say where this shows up on a desk, because that is why the question is asked. Two stocks in one sector share the sector's slow repricing but trade against each other intraday as market makers hedge one with the other. A hedge ratio estimated from daily data will be wrong for a position held for months, and the reverse. The limitation of the example: a flat drift inside a month is a simplification, and in real data the slow component is itself noisy, so the sign flip is usually a fade from negative towards positive rather than a clean reversal.
Where candidates lose it
Candidates say it must be a small-sample fluke, which answers a question nobody asked and signals they have never looked at correlations across horizons. The interviewer wants a mechanism, with components at two speeds.
The second loss is building the example with lagged returns, where B follows A a month later. That gives positive cross-correlation at a lag, not positive contemporaneous monthly correlation, and it is not what was asked.
What the interviewer asks next
What hedge ratio would you use for a one-day holding period, and for a six-month one?
If the shared drift had a standard deviation of 5 points instead of 12, what happens to the monthly correlation?
How would you test for this in real data without assuming the structure?
Asked at Squarepoint Capital, Hedge Fund, Montreal, 2024 (Wall Street Oasis): correlation can be negative intra-month but positive across a year, how?
083A price starts at 100 and moves up or down 1 each day with equal chance for 10 days. What is the probability that it touches 104 at some point in those 10 days?Random walks and Markov chainsHardExotics tradingQuant trading
Try it first
Which quantity is easier to count, and how does it relate to touching?
Show the worked solution
232 of the 1,024 paths touch 104, about 22.7%. A path ending at or above 104 must have touched it: that needs at least 7 up days, 176 paths. A path that touches 104 and ends below it can be reflected in the 104 line after its first touch, and the reflection ends above 104, needing at least 8 up days: 56 paths. The pairing is one to one, so the count is 176 + 56 = 232.
Why is the reflection a fair way to count?
Imagine a walker who reaches a wall at some point and then wanders on. Flip every step after the first touch, up for down, and you get another walker who also touched the wall at the same moment and is now the same distance on the other side of it. Each flipped step is as likely as the original, so the two walks have equal probability. Reflection pairs every path that touches 104 and finishes below it with exactly one path that finishes above 104, so the hard count becomes an easy count of endpoints. Paths that finish at exactly 104 are their own partners and are counted once.
A path that climbs to 104 by day 4 and ends at 100 is mirrored in the 104 line after the first touch into a path that ends at 108, and counting by endpoint gives 176 paths ending at or above 104 plus 56 mirrors of paths ending above it, 232 of 1,024 paths, which is 22.7%.
How do the binomial counts fall out?
After 10 days of plus or minus 1, the position is 100 plus ups minus downs, which is 100 + 2 x ups - 10. Ending at or above 104 needs ups of at least 7: C(10,7) + C(10,8) + C(10,9) + C(10,10) = 120 + 45 + 10 + 1 = 176. Ending strictly above 104 means at or above 106, since the endpoint is always even, so ups of at least 8: 45 + 10 + 1 = 56. The probability of touching is (176 + 56)/1,024 = 232/1,024, about 22.7%, against only 176/1,024 = 17.2% for ending at or above 104. The gap is the 56 paths that visited the barrier and came back.
The relationship
$P(\max S \ge 4) = P(S_{10} \ge 4) + P(S_{10} > 4) = \frac{176 + 56}{1024} = \frac{232}{1024}$
S_10 the net move after ten days
P(S_10 >= 4) paths ending at or above the barrier, 176
P(S_10 > 4) paths ending strictly above, which are the reflections of touch-and-return paths, 56
What it says in wordsThe chance of touching a level equals the chance of ending at or past it plus the chance of ending strictly past it.
Say why a derivatives desk cares. A knock-out or one-touch option pays on exactly this event, and the reflection principle is how the closed-form barrier prices are derived in continuous time. The limitation is in the symmetry: reflection needs the up and down steps to be equally likely and the barrier to sit on the lattice. With a drift, or in continuous time with a non-zero rate, the reflected path is not equally likely and a correction factor appears, which is why barrier formulas carry that extra power of the barrier over the spot.
Where candidates lose it
The usual loss is answering the probability of ending at or above 104, 17.2%, and forgetting the paths that touched and came back. The interviewer asked about touching, and the difference is a quarter of the answer.
The second loss is trying to enumerate paths that stay below 104 by hand. There are 792 of them and no clean way to list them in the room. Reflection exists so you do not have to.
What the interviewer asks next
What is the probability the price touches 104 and ends at 100?
Change the odds to 55% up. Why does reflection stop working as stated?
How does this count turn into the price of a one-touch option?
084Orders join a queue at a limit price at random at 2 a minute and are filled at random at 3 a minute. What fraction of the time is the queue empty, and how many orders are in it on average?Random walks and Markov chainsHardDRWNew York · 2026
Try it first
What fraction of the time is the queue empty?
Show the worked solution
The queue is empty one third of the time and holds 2 orders on average. With arrivals at 2 a minute and fills at 3, the fill process is busy a fraction 2/3 of the time, so the queue is empty 1/3 of the time. Balancing flow across each arrow of the chain gives share(n + 1) = (2/3) share(n), a geometric ladder starting at 1/3, whose mean is (2/3)/(1/3) = 2.
Why does balancing flow across one arrow solve the whole chain?
Picture a ticket counter with one clerk. Over a long day, the number of times the queue grows from 3 people to 4 must equal the number of times it shrinks from 4 to 3, because you cannot go up through that boundary twice without coming back down through it once. In the long run the flow of arrivals across each boundary equals the flow of fills back across it, which pins the share of time in each state to the share in the state below. Arrivals push at 2 a minute times the share of time at n; fills pull at 3 a minute times the share at n + 1. Equal flows mean share(n + 1) = (2/3) share(n).
Arrivals at 2 a minute push the queue one state to the right and fills at 3 a minute pull it one state to the left, and balancing the two flows across every boundary makes the long-run share of each queue length two thirds of the one before, starting at one third for an empty queue and averaging 2 orders.
How do the shares add up to one, and what is the mean?
The shares form a geometric series: share(0), share(0) x 2/3, share(0) x 4/9 and so on. They must sum to 1, and a geometric series with ratio 2/3 sums to 3 times its first term, so share(0) = 1/3. The mean number in the queue is the sum of n times share(n), which for a geometric ladder is the ratio over one minus the ratio: (2/3)/(1/3) = 2. A queue fed at two thirds of its capacity holds two orders on average, and because the mean is rho/(1 - rho) it explodes as arrivals approach fills. At 2.7 arrivals a minute the average would be 9; at 3 it is unbounded.
The relationship
$\pi_{n+1} = \tfrac{\lambda}{\mu}\,\pi_n = \tfrac{2}{3}\pi_n, \qquad \pi_0 = 1 - \tfrac{\lambda}{\mu} = \tfrac{1}{3}, \qquad L = \frac{\lambda/\mu}{1 - \lambda/\mu} = 2$
lambda the arrival rate, 2 a minute
mu the fill rate, 3 a minute
pi_n the long-run share of time with n orders in the queue
L the mean number of orders in the queue
What it says in wordsEach queue length is two thirds as likely as the one below it, the empty state has share one third, and the average length is two.
Two sentences that show you can use it. Little's law says time in the system equals mean number over arrival rate, 2/2 = 1 minute, so an order joining this queue expects to wait a minute to be filled, which is what a market maker deciding whether to post at this price wants to know. The limitation: the chain assumes memoryless arrivals and fills and no cancellations, and real order books have cancellations that depend on queue position, so the geometric shape is a first model, not a description.
Where candidates lose it
Candidates write the balance equations for every state at once and try to solve a system, then run out of time. The cut across a single boundary is the whole method: flow up equals flow down, so each share is a fixed multiple of the one below.
The second loss is answering zero for the empty fraction because orders keep arriving. Fills outpace arrivals, so the queue drains to empty regularly; it is empty exactly the fraction of time the fill process is idle.
What the interviewer asks next
How long does an order expect to wait from joining to being filled?
What happens to the average queue if arrivals rise to 2.7 a minute?
Now orders can also be cancelled at 1 a minute each. How does the chain change?
Asked at DRW, Quantitative Research, New York, 2026 (Wall Street Oasis): one on birth death chains

088Rs 1 was invested in a broad equity index 30 years ago. If its log returns average 9% a year with 16% volatility, give a 95% range for what the Rs 1 is worth today, and explain why the range is lopsided.Distributions and statisticsHardOld Mission CapitalChicago · 2025

Try it first

Which quantity do you build the 95% range on first?

Show the worked solution

Roughly Rs 2.7 to Rs 82.9, around a median of Rs 14.9. Thirty years of log returns sum to a mean of 2.7 with a standard deviation of 0.16 x the square root of 30, about 0.876. The 95% band in logs is 2.7 plus or minus 1.72, from 0.98 to 4.42. Exponentiating gives Rs 2.7 to Rs 82.9. It is lopsided because e to the power stretches gains and compresses losses: 12.2 below the median, 68.0 above.

Why work in logs and not in rupees?

A savings balance grows by multiplying: up 10% then down 10% leaves you at 0.99, not 1. Multiplications are awkward to average, but their logarithms add, and sums of many independent pieces tend towards a bell curve. Log returns add across years, so their total over 30 years has a mean that grows with time and a spread that grows with the square root of time, and the 95% band is symmetric in logs. Here the mean is 30 x 0.09 = 2.7, the standard deviation is 0.16 x 5.477 = 0.876, and 1.96 standard deviations is 1.72. The log of the final value lies between 0.98 and 4.42 with 95% confidence.

Thirty years of log returns averaging 9% with 16% volatility give a final value of Rs 1 that peaks near Rs 6.9, has a median of Rs 14.9 and a mean of Rs 21.8, and lies between Rs 2.7 and Rs 82.9 with 95% confidence, a band that is symmetric in logs but runs 12.2 below the median and 68.0 above it in rupees.

Why does the rupee range lean so far to the right?

Converting back means raising e to each end. e to the 0.98 is 2.67 and e to the 4.42 is 82.9, around a median of e to the 2.7, 14.9. The same 1.72 step in logs is a factor of 5.6 either way, and dividing by 5.6 moves you 12.2 rupees while multiplying by 5.6 moves you 68.0. That is why the distribution of wealth has a long right tail: the typical outcome, the median, sits well below the average, Rs 21.8, which is pulled up by the rare decades that compound very well.

The relationship

\ln W_{30} \sim N\!\left(30\mu,\; 30\sigma^2\right) = N(2.7,\; 0.876^2), \qquad W_{30} \in \left[e^{2.7 - 1.96 \times 0.876},\; e^{2.7 + 1.96 \times 0.876}\right] = [2.67,\; 82.9]

W_30	the value after 30 years of Rs 1 invested at the start
mu	the average yearly log return, 9%
sigma	the yearly volatility of log returns, 16%
1.96	the number of standard deviations that covers 95% of a normal

What it says in wordsBuild a symmetric 95% band for the total log return, then raise e to each end to get a lopsided band in rupees.

Two cautions the interviewer will reward. First, the 9% and 16% here are assumptions for the exercise; confirm the index's actual history before quoting any real figure, and note that a 30-year average return is itself estimated with a wide error. Second, the band assumes yearly log returns are independent and normal. Real markets have fat tails and runs of bad years, which widen the lower end in particular. The model gets the shape right, a long right tail with the median below the mean, even if the exact ends are only approximate.

Where candidates lose it

The common loss is building the band in rupees with a symmetric plus or minus, which gives a lower bound below zero or a band centred on the wrong number. Wealth cannot go below zero, and any method that says it can has skipped the logs.

The second loss is scaling the volatility by 30 instead of the square root of 30. That gives a standard deviation of 4.8 in logs, a band from almost nothing to tens of thousands, and tells the interviewer you have not met the square-root-of-time rule.

What the interviewer asks next

What is the probability the Rs 1 is worth less than Rs 1 today?
Why is the mean of the final value higher than the median?
If volatility were 25% instead of 16%, what happens to the median and to the mean?

Asked at Old Mission Capital, Prop Trading, Chicago, 2025 (Wall Street Oasis): Confidence interval of portfolio value if you invested $1 in S&P 500 30 years ago

090An array of n distinct numbers in random order gets one left-to-right bubble pass: swap the first two if they are out of order, then the new second and third, and so on to the end. What is the probability the array is sorted after that single pass? Work n = 4, then general n.Probability and countingHardJump TradingChicago · 2018
Try it first
Where must the smallest number start for one pass to leave it at the front?
Show the worked solution
For n = 4 it is 8/24 = 1/3; in general it is 2^(n - 1)/n!. A pass moves any number left by at most one place, so the smallest must start first or second. If it starts first, the rest is the same problem on n - 1 numbers. If it starts second, it swaps to the front and whatever was first becomes the head of a fresh problem on n - 1 numbers. Two choices each time give 2^(n - 1) sortable orders out of n!.
What can one pass actually do to an array?
Picture a queue where the tallest person seen so far keeps stepping past whoever is behind them. The tall ones can travel a long way back; everyone else only gets stepped past, and each time that happens they move forward by one place. A single left-to-right pass carries the running maximum to the right and moves every other number left by at most one position. So a number that starts two or more places to the right of where it belongs cannot get home in one pass, and the array cannot come out sorted.
Running all six orders of 1, 2 and 3 through one bubble pass sorts the four that start 1 2 3, 1 3 2, 2 1 3 and 3 1 2 and leaves 2 3 1 and 3 2 1 unsorted, because in those two the 1 starts two places too far right, and the same count for four numbers is 8 of 24, which is 2^(n - 1) of n! in general.
How does that turn into a count of 2^(n - 1)?
Look at where the smallest number starts. It must be position 1 or 2. If it is first, the first comparison does nothing and the pass carries on over positions 2 to n, which is the same problem on n - 1 numbers. If it is second, the first comparison swaps it to the front and the number that was first now leads a pass over positions 2 to n, again the same problem on n - 1 numbers. Each step offers exactly two placements for the current smallest number, so the count of sortable orders doubles with each extra element: f(n) = 2 f(n - 1), f(1) = 1, which gives 2^(n - 1). For n = 4 that is 8 of 24, a third.
The relationship
$P(\text{sorted after one pass}) = \frac{2^{\,n-1}}{n!}, \qquad n=3: \tfrac{4}{6}, \quad n=4: \tfrac{8}{24} = \tfrac{1}{3}, \quad n=10: \approx 0.000141$
2^(n-1) the number of starting orders one pass sorts
n! the number of starting orders in all
What it says in wordsThe sortable orders double with each element while all orders multiply by n, so the chance collapses quickly.
Then verify on a case you can hold in your head, as the figure does for three numbers: 4 of 6 come out sorted. A brute-force check over every order up to seven numbers gives 1, 2, 4, 8, 16, 32 and 64 sortable orders, as the formula says. The equivalent condition is worth saying too: one pass sorts the array exactly when every number starts no more than one place to the right of its sorted position. The limitation is the assumption that all n! starting orders are equally likely; a nearly sorted array, which is what real data often is, comes out sorted far more often.
Where candidates lose it
The common loss is answering with the chance that the array was already sorted, 1/n!, or with the chance that the largest ends last, which is 1. One pass always puts the maximum at the end; the question is whether everything else is home too.
The second loss is trying to list the n = 4 cases one by one. With 24 orders you will miss some. Find the rule about how far left a number can move and the count follows in two lines.
What the interviewer asks next
What is the probability the array is sorted after two passes?
What is the expected number of swaps in one pass?
If the pass ran right to left instead, which number would be carried, and does the answer change?
Asked at Jump Trading, Quantitative Research, Chicago, 2018 (Wall Street Oasis): a math problem about the probability an array is sorted after swapping the first two if they're out of order
091I deal a shuffled deck face up one card at a time. At any moment you may say stop; if the next card is red you win, otherwise you lose. What strategy maximises your chance of winning, and what is that chance?Expected value and optimal stoppingHardJump TradingChicago · 2018
Try it first
You decide to wait until more reds than blacks remain, then stop. How does that compare with stopping straight away?
Show the worked solution
Every strategy wins with probability exactly 1/2, so there is nothing to optimise. The chance that the next card is red equals the share of red left in the deck, and that share is a fair game: whatever has been dealt, its expected value one card later is its value now. A neat way to see it: the card after you say stop has the same chance of being red as the last card in the deck, and the last card is red with probability 1/2 whatever you do.
Why can't waiting for a good moment help?
Imagine a jar of 26 red and 26 black sweets that a friend pulls out one at a time, and you bet on the colour of the next one. When more reds are left you feel you have edge, but the jar only gets red-heavy after blacks come out, and you do not control which comes out. The share of red remaining has the same expected value one card from now as it has now, so any rule for when to stop is a bet on a fair game and cannot beat 1/2. The fancy name is a martingale: a quantity whose best forecast of its next value is its current value.
Two simulated deals show the share of red among undealt cards wandering around 1/2, with deck A going red-heavy after 7 cards and deck B never going red-heavy and ending on a black card, and the rule of stopping when reds lead gets its chance in 26 deals out of 27 yet wins exactly 1/2, because the deals where it never comes are certain losses.
What is the one-line proof the interviewer wants?
Whatever rule you use, you say stop at some point and win if the next card is red. Swap that bet for a bet on the last card of the deck. Given everything dealt so far, the next card and the last card are both a random draw from the same set of undealt cards, so they are red with the same probability. Every stopping rule wins with the same probability as a bet on the bottom card of the deck, and the bottom card is red half the time. That holds for any strategy, so 1/2 is both the best and the worst you can do. A full check of every position from 26 red and 26 black confirms that the best achievable chance from r red and b black is exactly r/(r + b), the chance of stopping right there.
The relationship
$E\!\left[\frac{R_{k+1}}{N_{k+1}} \;\middle|\; \frac{R_k}{N_k}\right] = \frac{R_k}{N_k} \;\Rightarrow\; P(\text{win}) = \frac{R_0}{N_0} = \frac{26}{52} = \frac{1}{2}$
R_k red cards still undealt after k cards
N_k all cards still undealt after k cards
R_k / N_k the chance the next card is red if you stop now
What it says in wordsYour winning chance at any moment has the same expected value tomorrow as today, so no rule for when to stop can raise its starting value of a half.
Put a number on why the waiting rule fools people. Starting from an even deck, the first black card dealt leaves 26 red in 51, so reds lead immediately half the time, and over the whole deal the moment arrives in 26 deals out of 27. But it typically arrives at a share only just above 1/2, and in the 1 deal in 27 where it never arrives, the last card is black for certain. The edge in the good deals is paid for exactly by the sure loss in the bad ones. The limitation of the result is the payoff: if you were paid more for winning late, or could bet different amounts, the game would no longer be fair and timing could matter.
Where candidates lose it
The common loss is proposing the wait-for-reds-to-lead rule and claiming a small edge, usually around 51%. The interviewer then asks what happens if reds never lead, and the edge disappears. Count both branches before you claim anything.
The second loss is starting a dynamic programme over 27 by 27 states in the room. It works, but it takes far too long. The last-card argument settles it in one sentence and is what the question is testing.
What the interviewer asks next
Now you win Rs 2 if the next card is red and lose Rs 1 if black, and you must stop at some point. Does timing matter?
With 3 red and 1 black, what is your chance, and can any strategy change it?
What changes if you may skip a card without it being revealed?
Asked at Jump Trading, Research, Chicago, 2018 (Wall Street Oasis): I'm dealing a deck of poker, you can stop me anytime. If the next card is red, you win.
097I pick a whole number from 1 to 100 and know it; you do not. You quote 45 at 55 and I buy at 55. You requote 58 at 68 and I buy again at 68. What have you learned, and where should your next quote be?Market makingHardJane StreetNew York · 2025
Try it first
After the first lift at 55, where is the middle of what the number could still be?
Show the worked solution
The number is between 69 and 100, so centre the next quote on 84.5, say 80 at 89. Someone who knows the number buys only when your offer is too low. The lift at 55 put it in 56 to 100, midpoint 78, and your requote at 58 to 68 was far too low, so it was lifted too. Now it is in 69 to 100. You are short two at 55 and 68, worth about -46 against a fair value of 84.5, and the next quote should sit on what is left, not drift up from the last one.
What does a trade tell you when the other side knows the answer?
If a friend who has already peeked at the exam paper offers to bet you that the pass mark is above 55, you do not need to ask why. A counterparty who knows the number only trades when your price is wrong in their favour, so every lift of your offer is proof that the number is above it, and your fair value should jump to the middle of what is left. After the lift at 55 the number lies in 56 to 100, centred on 78. The requote at 58 to 68 treated the lift as noise and moved only 13 points. A quote centred near 78, say 73 at 83, would have used the information.
The first lift at 55 leaves 56 to 100 with a midpoint of 78, the requote of 58 at 68 sits 15 below that midpoint and is lifted again, which leaves 69 to 100 with a midpoint of 84.5, so the next quote of 80 at 89 is centred on what can still be true.
Where should the next quote sit, and what have the trades cost?
After the second lift the number is in 69 to 100: 32 values, mean 84.5. Centre the next quote on the middle of the remaining range, 84.5, so that whichever side is hit, the range halves, which is the fastest way to stop losing to someone who knows more than you. A quote of 80 at 89 does that. The damage so far: you sold at 55 and 68 something now worth about 84.5, an expected loss of 29.5 + 16.5 = 46. Most of it came from the timid requote; a quote centred on 78 would have been lifted only if the number was above 83.
The relationship
$E[X \mid X > 55] = \frac{56 + 100}{2} = 78, \qquad E[X \mid X > 68] = \frac{69 + 100}{2} = 84.5$
X the number I picked, uniform on 1 to 100 before any trade
X > 55 what my buying at 55 reveals
84.5 the fair value after both lifts, where the next quote should be centred
What it says in wordsEach lift cuts the range to everything above your offer, and fair value moves to the middle of what is left.
Then say how the answer changes with the counterparty, because that is the judgement being marked. If I might be guessing rather than knowing, a lift is weaker evidence and you should move less, perhaps halfway towards 78. If I know the number exactly, any tight quote inside the range loses whenever it trades, and the only quote that cannot lose is one that spans the whole range, 69 bid, 100 offered, which is no market at all. Real desks live between the two: they widen against flow they think is informed and tighten against flow they think is not.
Where candidates lose it
The common loss is moving the quote up a few points after each lift, as the 58 at 68 requote did. It treats an informed trade as a random one and pays for the lesson again on the next trade.
The second loss is the opposite overreaction: refusing to quote, or quoting 1 at 100. The interviewer wants a market that uses the information and still trades, centred on the conditional mean of what is left.
What the interviewer asks next
Instead I sell to you at 45 on the first quote. What is your next quote?
Suppose I know the number only half the time and guess otherwise. How far should the first lift move you?
How many lifts or hits does it take, at most, to find the number if you always quote around the middle of the range?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): It was a probability theory based quant trading style market making questions which were intense
099Daily returns come from a normal with 1% volatility on 90% of days and a normal with 3% volatility on the other 10%, both with mean zero. What are the overall volatility and kurtosis, and what does this do to out-of-the-money option prices?Distributions and statisticsHardTwo SigmaNew York · 2025
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What is the overall daily volatility?
Show the worked solution
Volatility 1.34% a day and kurtosis 8.33, against 3 for a normal, which makes far out-of-the-money options worth much more than a single-volatility model says. Variance is 0.9 x 1 + 0.1 x 9 = 1.8. The fourth moment is 3 x (0.9 x 1 + 0.1 x 81) = 27, so kurtosis is 27/1.8^2 = 8.33. With the same volatility, a 4% move is 6.4 times as likely as the normal says, and a one-day put 4% out of the money is worth about 23 times as much.
How do the moments of a mixture combine?
Picture a road with quiet days and occasional storms. Most days the traffic varies a little; on storm days it varies a lot. Averaged over a year, the variability you see is dominated by the storm days, out of proportion to how rare they are. For a mixture with mean zero, each even moment is the weighted average of the components' moments: variances add up by weight, and fourth moments, which grow with the fourth power of volatility, are dominated by the rare wild days. Variance: 0.9 x 1 + 0.1 x 9 = 1.8, a volatility of 1.34%. Fourth moment: each normal's is 3 sigma^4, so 3 x (0.9 x 1 + 0.1 x 81) = 27. The 3% days are a tenth of the time but supply 90% of the fourth moment.
Against a normal with the same 1.34% daily volatility, the mixture of 1% and 3% days is taller in the middle and lies above the normal in the tails beyond about 3.7%, so its kurtosis is 8.33 instead of 3 and a move beyond 4% either way is 6.4 times as likely.
What does the fat tail do to option prices?
Kurtosis is the fourth moment over the square of the variance: 27/3.24 = 8.33. A single normal fitted to the same data has the same volatility, 1.34%, but kurtosis 3, so it puts far too little weight on big moves. An option far out of the money pays only on a big move, so its value is driven by the tail, and a mixture that keeps the volatility but fattens the tail makes it worth many times what the matched normal says. A one-day put 4% out of the money is worth 0.0127% of spot under the mixture against 0.00055% under the normal, about 23 times as much. To match the mixture's price, a single-volatility model needs 1.91% for that put but only 1.20% at the money: the smile.
The relationship
$\sigma^2 = 0.9(1)^2 + 0.1(3)^2 = 1.8, \qquad \kappa = \frac{3\,[\,0.9(1)^4 + 0.1(3)^4\,]}{(1.8)^2} = \frac{27}{3.24} \approx 8.33$
sigma^2 the variance of the daily return, in square percentage points
kappa the kurtosis, 3 for any single normal
3 sigma^4 the fourth moment of a normal with mean zero
What it says in wordsVariances average by weight, fourth moments average by weight, and the ratio comes out near 8.3 because the rare 3% days dominate the fourth power.
Say what it means on a desk. This mixture is the simplest model of regime-switching volatility, and it reproduces two things a single normal cannot: a peak that is too tall and tails that are too fat, the shape seen in daily returns of most liquid assets. It is also why at-the-money options can look rich and far wings cheap if you price both off one historical volatility. The limitation: the mixture here is symmetric and draws each day independently. Real markets cluster their wild days and fall harder than they rise, which adds skew to the smile, not just curvature.
Where candidates lose it
The common loss is averaging the volatilities, 0.9 x 1 + 0.1 x 3 = 1.2%, and then getting the kurtosis wrong as a result. Volatilities never average in a mixture; variances do.
The second loss is computing the moments correctly and stopping, without connecting kurtosis to option prices. The question asks what the fat tail does to the wings, and the answer is a smile: more implied volatility the further out of the money you go.
What the interviewer asks next
What is the kurtosis if the wild days are 5% volatility but only 4% of days?
Which is more mispriced by a single-volatility model, an at-the-money option or a far out-of-the-money one, and in which direction?
How would you add skew to this model?
Asked at Two Sigma, Quantitative Research, New York, 2025 (Wall Street Oasis): They asked a couple questions involving Mixture Gaussians (e.g., probability density and moments).

C(K)	the price of the call struck at K
7.5	the midpoint of the chord between the outer strikes
8	the quoted middle price, half a point too high

sigma_m	the standard deviation of the shared monthly drift, 12 points
sigma_e	the daily opposing noise, 1 point
sigma_u	each asset's own daily noise, 1 point

S_10	the net move after ten days
P(S_10 >= 4)	paths ending at or above the barrier, 176
P(S_10 > 4)	paths ending strictly above, which are the reflections of touch-and-return paths, 56

lambda	the arrival rate, 2 a minute
mu	the fill rate, 3 a minute
pi_n	the long-run share of time with n orders in the queue
L	the mean number of orders in the queue

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