Derivatives Foundation puzzles, solved step by step

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

Showing 61–70 of 100

061You and I each show heads or tails at the same time. You win Rs 3 if we both show heads, Rs 1 if we both show tails, and you lose Rs 2 if we show different faces. What mix should you play, and is the game worth playing?Games and logicCoreQuant tradingProp trading firms
Try it first
Two wins and two losses in the table. Is this game good for you?
Show the worked solution
Show heads 3/8 of the time, and do not play unless you are paid at least Rs 0.125 a round. If you show heads with probability p, your expected payoff is 5p - 2 when I show heads and 1 - 3p when I show tails. I will pick whichever is lower, so you choose p to make the lower line as high as possible, which is where they cross: p = 3/8, value - 1/8. Any other p lets me push you below that.
Why is the answer a mix rather than a single face?
Two children playing odds and evens learn fast that any pattern is punished: show heads every time and the other child shows tails every time. In a game where my best reply depends on what you do, any fixed choice is exploited, so you protect yourself by randomising in a ratio that leaves me with nothing to exploit. That ratio is found by making me indifferent between my two replies. If you show heads a fraction p of the time, my heads earns you 3p - 2(1 - p) = 5p - 2 and my tails earns you - 2p + (1 - p) = 1 - 3p. They are equal at p = 3/8.
Your expected payoff is 5p - 2 if I show heads and 1 - 3p if I show tails, and since I will always pick the lower line, the best you can do is the crossing at p = 3/8, where both lines give minus 1/8, so the game is worth minus Rs 0.125 to you per round.
How do you know minus 1/8 is the most you can guarantee?
Look at the lower of the two lines across all p. To the left of 3/8 the heads line is lower and rising; to the right the tails line is lower and falling, so the lower envelope peaks exactly at the crossing, and that peak is your guaranteed value. I have the same calculation from my side: if I show heads a fraction q of the time, you are indifferent when 3q - 2(1 - q) = - 2q + (1 - q), which again gives q = 3/8, and at that mix I hold you to - 1/8 whatever you do. Both sides landing on the same number is the minimax theorem, attributed to von Neumann, at work in a two by two table.
The relationship
$5p - 2 = 1 - 3p \;\Rightarrow\; p = \tfrac{3}{8}, \qquad V = 5\cdot\tfrac{3}{8} - 2 = -\tfrac{1}{8}$
p your probability of showing heads
5p - 2 your expected payoff when I show heads
1 - 3p your expected payoff when I show tails
V the value of the game to you per round
What it says in wordsEqualising your payoff across my two replies gives a three-eighths mix and a value of minus one eighth of a rupee per round.
What is the desk version of this question?
Quoting against a counterparty who sees your pattern. A market maker who always leans the same way after a fill is the child who always shows heads, and the counterparty who notices earns the difference, so randomised sizing and skew are the trading-floor form of the 3/8 mix. The limitation of the puzzle answer is that it assumes I play optimally; against an opponent who shows heads half the time out of habit, your best reply is pure heads, with an expected 0.5 x 3 - 0.5 x 2 = + Rs 0.50 a round, and the game becomes worth playing. Ask who you are playing before you quote the value.
Where candidates lose it
The common answer is that the game is fair or favourable, from summing the four cells. The sum of a payoff table says nothing when the opponent chooses the column. Set up the two lines and find where they cross.
The second loss is solving for the right p and then saying the game is fine because 3 and 1 are bigger than 2. State the value, minus 1/8, and say you need a fee of at least that to play.
What the interviewer asks next
What is my optimal mix, and what does it earn me?
Change the heads-heads payoff to Rs 4. Does the game become worth playing?
I am known to show heads 60% of the time regardless. What should you do now?
Why do both players end up with the same 3/8 here, and is that a coincidence?
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.
064A bag holds 9 fair coins and 1 coin with heads on both sides. You pull one out at random and flip it 5 times, getting 5 heads. What is the probability it is the two-headed coin, and what is the probability the next flip is heads?Conditional probability and BayesCoreQuant tradingHedge funds
Try it first
Five heads in a row from a coin that is two-headed one time in ten. How likely is it the two-headed one?
Show the worked solution
78.0% that it is the two-headed coin, and 89.0% that the next flip is heads. Prior odds are 1 to 9. The two-headed coin gives a head with certainty and a fair coin with chance 1/2, so each head multiplies the odds by 2; after five heads the odds are 32 to 9, which is 32/41. The next flip is heads with chance 32/41 x 1 + 9/41 x 1/2 = 73/82.
Why work in odds rather than probabilities?
A doctor who sees the same symptom five mornings running does not recompute the whole diagnosis each day; each new observation multiplies the odds of the condition by one fixed factor. In odds form, Bayes' rule is a multiplication: posterior odds equal prior odds times the likelihood ratio of each observation, and here every head has the same ratio of 1 to 1/2, which is 2. So the odds on the two-headed coin go 1:9, 2:9, 4:9, 8:9, 16:9, 32:9. Convert at the end: 32 divided by 32 plus 9 is 32/41. Doing it in probabilities means dividing by a different normaliser five times, which is where people slip.
Each head doubles the odds on the two-headed coin, lifting the probability from 10% to 78% after five heads, and the chance the next flip is heads, 89%, mixes a certain head from the two-headed coin with a coin flip from a fair one.
Why is the next flip not simply 78% heads?
Because the fair coin also produces heads. The next flip is heads if the coin is two-headed, with chance 32/41, or if the coin is fair and lands heads, with chance 9/41 times 1/2, and the two routes add to 73/82, about 89%. Candidates who answer 78% have confused the probability of the hypothesis with the probability of the outcome. The gap between the two is the fair coin's half chance of a head, weighted by the 22% chance you are holding a fair coin.
The relationship
$\frac{P(\text{two-headed}\mid 5H)}{P(\text{fair}\mid 5H)} = \frac{1}{9}\times\left(\frac{1}{1/2}\right)^5 = \frac{32}{9} \;\Rightarrow\; P = \frac{32}{41}, \qquad P(\text{next } H) = \frac{32}{41}\cdot 1 + \frac{9}{41}\cdot\frac{1}{2} = \frac{73}{82}$
1/9 the prior odds of drawing the two-headed coin
(1 / (1/2))^5 the likelihood ratio of five heads, two to the fifth
73/82 the chance of a sixth head, mixing both coins
What it says in wordsFive heads multiply the prior odds by thirty-two, giving thirty-two to nine, and the next flip mixes a sure head with a fair flip in those proportions.
How many heads would it take to be nearly sure, and what does a desk take from this?
Each head doubles the odds, so after 10 heads the odds are 1,024 to 9, about 99.1%, and after 5 you are only at 78%. Evidence that is merely consistent with a hypothesis moves you slowly when the alternative also produces it often, which is why five good months from a new trading strategy prove far less than people feel they do. The limitation is the prior: if the bag held 99 fair coins and one two-headed, five heads would leave you at 32 to 99, still under 25%, and no amount of looking at the flips alone tells you the composition of the bag.
Where candidates lose it
The fast wrong answer is 1 minus 1/32, about 97%, which is the chance a fair coin would not have done this. That number ignores that there are nine fair coins for every two-headed one. Start from the prior odds and double.
The second loss is giving 78% for the next flip. The next flip is a mixture: a certain head from the two-headed coin and a half chance from a fair one.
What the interviewer asks next
After how many heads does the probability it is the two-headed coin pass 99%?
The sixth flip is tails. What is the probability it is the two-headed coin now?
The bag has 99 fair coins and one two-headed coin. What are the two answers after five heads?
Why is the probability of the next head always between the fair coin's 1/2 and 1?
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.
066A company is worth a uniformly random amount between Rs 0 and Rs 100 crore to its owner, who knows the exact value. It is worth 1.5 times that amount to you. You make one take-it-or-leave-it offer, and the owner accepts if your offer exceeds the value. What should you offer?Games and logicHardHedge fundsMarket making
Try it first
The company is worth 50% more to you than to the owner. What do you bid?
Show the worked solution
Offer nothing. Every positive offer loses money on average. If you offer b and the owner accepts, you learn the value is below b, so its average is b/2, worth 1.5 x b/2 = 0.75b to you. You pay b, so each accepted deal loses 0.25b, and the deal is accepted a fraction b/100 of the time. Expected profit is minus 0.0025 b squared, negative for every b above zero. The 1.5 multiplier is not enough to overcome what acceptance tells you.
Why is the owner saying yes bad news for you?
A friend sells you their old scooter for any price you name, but only if your price beats what they privately think it is worth. If they take Rs 20,000 instantly, you have just learned the scooter is worth less than that to someone who knows it well. Acceptance is information: it tells you the true value lies below your bid, so the only companies you ever buy are the ones worth less than you paid, and the ones worth more walk away. Averaging over all possible values, as if you bought every company, is the mistake; you must average only over the values at which the owner says yes.
Expected profit from an offer b is minus a quarter of b times the acceptance chance b/100, a parabola that is zero at b = 0 and falls to minus 25 crore at b = 100, so no positive offer earns anything and the best move is not to bid; an offer of 60, for instance, buys a company worth 45 to you on average and loses 9 in expectation.
How do you set up the expected profit cleanly?
Condition on acceptance, then multiply by its probability. Given a bid b that is accepted, the value V is uniform on 0 to b, so E[V | accepted] = b/2, the company is worth 1.5 x b/2 = 0.75b to you, and the profit on an accepted deal is 0.75b - b = - 0.25b. Acceptance happens with probability b/100, so expected profit is - 0.25b x b/100 = - b squared over 400. At b = 50 that is minus 6.25 crore; at b = 100 it is minus 25 crore. The derivative is negative everywhere above zero, so the maximum is at b = 0.
The relationship
$E[\pi(b)] = \frac{b}{100}\left(1.5\cdot\frac{b}{2} - b\right) = -\frac{b^2}{400} < 0 \quad \text{for all } b > 0$
b your offer in Rs crore
b/100 the chance the owner accepts, since the value is uniform on 0 to 100
1.5 x b/2 what the company is worth to you on average once you know the value is below b
What it says in wordsThe expected profit from any offer is a negative multiple of the offer squared, so the best offer is zero.
At what multiplier does a bid start to make sense?
Replace 1.5 with a general m. Given acceptance, the company is worth m x b/2 to you against the b you pay, so the deal breaks even when m/2 = 1, that is m = 2. Unless the company is worth more than twice as much to you as to the owner, acceptance always costs you, and at exactly double every bid is a wash. That is the winner's curse in its purest form: the party with less information loses whenever the informed party decides whether to trade. The limitation is the uniform prior and the single offer; with a floor on the value, or a negotiation that reveals information, positive bids can be profitable.
Where candidates lose it
Candidates bid around 50 or 75 by averaging over the whole range of values, forgetting that they only buy when the owner accepts. The interviewer wants you to say that acceptance is information before you touch any arithmetic.
The second loss is getting zero and not being able to say what would change it. Give the general condition: the multiplier must exceed two for any positive bid to pay.
What the interviewer asks next
At what multiplier does a positive bid first break even, and what is the best bid at a multiplier of 3?
The value is uniform on Rs 50 to Rs 100 crore instead. Does a positive bid make sense now?
How is this the same problem as a market maker being hit only when they are wrong?
You get two offers rather than one, and the owner rejects the first. Does that change anything?
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.
068Your book is delta neutral with gamma of 2,000 shares per rupee on a stock trading at Rs 500. The stock jumps Rs 10. Roughly what is your P&L before you rehedge, and how many shares do you now need to trade?Option pricing intuitionCoreEquity derivativesVolatility trading
Try it first
Delta zero, gamma 2,000 shares per rupee, a Rs 10 jump. P&L?
Show the worked solution
About Rs 1,00,000 profit, and you need to sell about 20,000 shares. P&L from gamma is one half of gamma times the move squared: 0.5 x 2,000 x 10 squared = Rs 1,00,000. The delta picked up during the move is gamma times the move, 2,000 x 10 = 20,000 shares long, which you sell to get back to neutral. A Rs 10 fall would earn the same amount and leave you 20,000 shares short to buy back.
Why does a delta-neutral book make money on a move?
A cyclist at the bottom of a valley is on flat ground, but every metre up either slope gets steeper. Delta neutral means the P&L is flat at the current price only; gamma is how fast the slope changes, so as the stock moves the book acquires delta in the direction of the move and earns on it the whole way. With gamma of 2,000 shares per rupee, after the first rupee you are 2,000 shares long, after the fifth 10,000, after the tenth 20,000. The P&L is the area under that rising delta, a triangle with base 10 and height 20,000, which is 1,00,000.
A delta-neutral book with gamma of 2,000 shares per rupee earns one half of gamma times the move squared, Rs 1,00,000 on a Rs 10 move in either direction, and at the new price its slope is gamma times the move, 20,000 shares long, which is what must be sold to be flat again.
What is the arithmetic, and where does the one half come from?
Expand the book's value as a Taylor series in the stock price. The first-order term is delta times the move, zero here; the second-order term is one half of gamma times the move squared, 0.5 x 2,000 x 100 = Rs 1,00,000; and the new delta is the derivative of that, gamma times the move, 20,000 shares. The one half is the same one half as in the area of a triangle: delta started at zero and finished at 20,000, so on average it was 10,000 shares over the Rs 10 move. The limitation is that a jump also changes implied volatility and burns a day of theta, both ignored here.
The relationship
$\Delta P\&L \approx \Delta\cdot\delta S + \tfrac{1}{2}\Gamma\,(\delta S)^2 = 0 + \tfrac{1}{2}\times 2000\times 10^2 = 1{,}00{,}000, \qquad \Delta_{\text{new}} = \Gamma\,\delta S = 20{,}000$
delta the book's share-equivalent exposure, zero before the move
Gamma the change in delta per rupee of stock move, 2,000 shares
delta S the stock move, Rs 10
What it says in wordsThe profit is half of gamma times the move squared, and the delta to be hedged afterwards is gamma times the move.
What happens if you rehedge and the stock comes back?
You sell 20,000 shares at Rs 510. If the stock then falls back to Rs 500, the options give back their Rs 1,00,000 but the short stock earns 20,000 x Rs 10 = Rs 2,00,000, so you net Rs 1,00,000 from the round trip. That is what long gamma means in practice: each rehedge locks in half of gamma times the move squared, and a stock that moves a lot and comes back pays you twice. The cost is theta, the daily decay you pay for holding the options, and the trade only works if realised movement is larger than the implied volatility you paid for.
Where candidates lose it
Candidates say zero because the book is delta neutral, or they give gamma times the move squared without the one half and double the answer. Say the triangle: delta climbs from zero to 20,000, average 10,000, times Rs 10.
The second loss is confusing the two numbers. The P&L is in rupees and uses the move squared; the delta to trade is in shares and uses the move once.
What the interviewer asks next
The stock falls Rs 10 instead. What is the P&L and what do you trade?
You rehedge at Rs 510 and the stock returns to Rs 500. What have you made on the round trip?
What daily theta would make this book break even on a Rs 10 move per day?
The book is short gamma instead. Describe the same Rs 10 move.
069An index of ten equally weighted stocks has 15% implied volatility, and each of the ten stocks has 30% implied volatility. What average correlation is the market implying?Volatility and correlationHardVolatility tradingExotics trading
Try it first
Index vol is half the single-stock vol. Rough instinct for the implied correlation?
Show the worked solution
About 0.17, exactly one sixth. For n equally weighted stocks with the same volatility and the same pairwise correlation rho, index variance is stock variance times (1/n + (1 - 1/n) rho). Index variance over stock variance is 0.15 squared over 0.30 squared = 0.25, so 0.25 = 0.1 + 0.9 rho and rho = 0.15/0.9 = 1/6. The large-n shortcut of 0.25 overstates it because it ignores the 1/n term.
Why does the index have less volatility than its members?
Ten shopkeepers in one market each have noisy daily takings, but the market's total takings are steadier, because one shop's bad day is often another's good day. Only the part of the noise they share, the weather or a festival, survives the averaging. Index variance has two parts: the stocks' own noise, which averages away as 1/n, and the common movement, which survives in proportion to the correlation. With ten stocks at 30% and no correlation at all the index would still have 9.5% volatility, and that floor is why the implied correlation is below the naive 0.25.
For ten equally weighted stocks at 30% volatility, index volatility rises from 9.5% at zero correlation to 30% at a correlation of one, and a 15% index volatility is reached at an average correlation of one sixth, about 0.17.
How do you derive the formula on the spot?
Write the index as the average of n returns and expand its variance. There are n variance terms, each sigma squared over n squared, and n(n - 1) covariance terms, each rho sigma squared over n squared, so index variance is sigma squared times (1/n + (n - 1) rho / n). Set that equal to 0.15 squared with sigma = 0.30 and n = 10: 0.0225 = 0.09 x (0.1 + 0.9 rho), so 0.25 = 0.1 + 0.9 rho and rho = 1/6. The derivation takes four lines, and the interviewer wants to hear the count of covariance terms, n(n - 1), said out loud.
The relationship
$\sigma_I^2 = \sigma^2\left(\frac{1}{n} + \frac{n-1}{n}\rho\right) \;\Rightarrow\; \rho = \frac{\sigma_I^2/\sigma^2 - 1/n}{1 - 1/n} = \frac{0.25 - 0.1}{0.9} = \frac{1}{6}$
sigma_I index volatility, 15%
sigma single-stock volatility, 30% for every stock
n the number of equally weighted stocks, ten
rho the average pairwise correlation implied by the two volatilities
What it says in wordsIndex variance is stock variance times one over n plus the correlation times the rest, which solves to a correlation of one sixth.
What does a desk do with implied correlation?
It compares it with realised correlation and trades the gap. Selling index options and buying single-stock options is a short correlation position, called a dispersion trade, and it pays when the stocks move more independently than the 1/6 the market has priced in. The limitation of the puzzle is its symmetry: real indices have unequal weights and unequal volatilities, so the implied correlation is a weighted average that can differ from the simple formula, and the implied figure also moves with the skew of the index options, which the single number cannot show.
Where candidates lose it
The common wrong answers are 0.5 and 0.25: the first divides volatilities and the second divides variances but forgets the 1/n floor from the stocks' own noise. With only ten stocks the floor is a tenth of the variance, which is a big correction.
The second loss is knowing the formula but failing to say what trade it supports. Say dispersion, and say which side is short correlation.
What the interviewer asks next
Same numbers with 50 stocks instead of 10. What is the implied correlation?
What is the lowest possible index volatility for ten stocks at 30%, and at what correlation?
Implied correlation is 1/6 and realised correlation turns out to be 0.4. Which side of a dispersion trade made money?
The stocks have different weights and volatilities. How does the formula change?
070Make me a two-way price on the product of two dice rolls. I then show you that one of the dice is a 4. Requote.Market makingCoreOptiverChicago · 2026DRWChicago · 2025
Try it first
Fair value of the product of two dice, before anything is revealed?
Show the worked solution
First quote around 12.25: say 11.75 bid, 12.75 offer. After the 4 is shown, requote around 14: say 13.50 at 14.50. Two independent dice have an expected product of 3.5 x 3.5 = 12.25. Once one die is known to be a 4, the product is 4 times the other die, with expectation 4 x 3.5 = 14. The second market can be the same width or tighter, because only one die of uncertainty is left: the standard deviation of the outcome falls from about 8.9 to about 6.8.
Why does fair value come from multiplying the averages?
A canteen's daily takings are the number of customers times the average spend, and if the crowd size has nothing to do with how hungry people are, the average takings are the average crowd times the average spend. For two independent quantities the expectation of the product is the product of the expectations, so two dice thrown separately have an expected product of 3.5 x 3.5 = 12.25. You can check it on the grid: row i of the multiplication table averages 3.5 i, and the six row averages, 3.5 through 21, average 12.25. The product is skewed, with most of the 36 cells below the mean and a few large ones pulling it up.
Across the 36 equally likely products the mean is 3.5 x 3.5 = 12.25, so the opening market sits around 12.25; once one die is revealed as a 4 the outcome is 4 times a single die, with a mean of 14 and a standard deviation that falls from about 8.9 to about 6.8, so the requote moves up and can tighten.
How does the reveal change the quote, and why can the market tighten?
Replace the revealed die with its value and keep the other at its expectation. The product is now 4 times one unknown die, so fair value is 4 x 3.5 = 14 and the only uncertainty left is a single die, with a standard deviation of 4 times 1.71, about 6.8, against about 8.9 before. Less uncertainty means less risk per trade, so a market maker can quote the same width with more confidence or tighten it. What you must not do is anchor on the old 12.25: a quote that still straddles 12 after a 4 has been shown is a free trade for the other side, who lifts your offer and collects the difference.
The relationship
$E[XY] = E[X]\,E[Y] = 3.5 \times 3.5 = 12.25, \qquad E[XY \mid X = 4] = 4\,E[Y] = 14, \qquad \sigma_{XY\mid X=4} = 4\,\sigma_Y = 4\sqrt{35/12} \approx 6.83$
X, Y the two independent dice
E[XY | X = 4] the expected product once one die is known to be a 4
sigma_Y the standard deviation of one die, about 1.71
What it says in wordsBefore the reveal the fair product is twelve and a quarter; after seeing a four it is fourteen, with the uncertainty of one die rather than two.
What is the interviewer watching for in the requote?
Speed and direction first, then the width. A trader who says 14 inside a second, moves the market up without hesitation and gives a reason for the width is passing; one who recomputes from the grid, or leaves the old market up, is failing. The next step is usually a trade: if the interviewer lifts your 14.50 offer, you are short at above fair and should hold or edge the market up slightly rather than chase; if they hit your bid at 13.50, you are long below fair. The limitation is that this is a one-shot game with a known distribution; in a real market the reveal would itself be a signal about what else the counterparty knows.
Where candidates lose it
The first loss is the opening fair value: 18.5 from the midpoint of 1 and 36, or 15.17 from squaring one die. Say independent, say 3.5 times 3.5, and the grid check if asked.
The second loss is the requote. Candidates either freeze or adjust by a token amount. The product is now 4 times one die, fair value 14, and the market should move there immediately and may tighten.
What the interviewer asks next
Instead of showing a 4, I tell you the two dice are the same. Requote.
Now the revealed die is a 1. Where is your market, and how wide?
Make a market on the sum of the two dice, then on the sum given one is a 4.
I lift your 14.50 offer twice in a row. What do you do?
Asked at Optiver, Prop Trading, Chicago, 2026 (Wall Street Oasis): Market making game full simulation including fast mental math and quick ev/fair value calculation
Asked at DRW, Quantitative Trading, Chicago, 2025 (Wall Street Oasis): Market making and fermi estimation on random quantities

p	your probability of showing heads
5p - 2	your expected payoff when I show heads
1 - 3p	your expected payoff when I show tails
V	the value of the game to you per round

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

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

1/9	the prior odds of drawing the two-headed coin
(1 / (1/2))^5	the likelihood ratio of five heads, two to the fifth
73/82	the chance of a sixth head, mixing both coins

Try it first

Why is the answer a mix rather than a single face?

How do you know minus 1/8 is the most you can guarantee?

What is the desk version of this question?

Where candidates lose it

What the interviewer asks next

Try it first

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

Try it first

What does the likelihood actually say, in words?

How do you derive the peak in three lines?

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

Where candidates lose it

What the interviewer asks next

Try it first

Why work in odds rather than probabilities?

Why is the next flip not simply 78% heads?

How many heads would it take to be nearly sure, and what does a desk take from this?

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 is the owner saying yes bad news for you?

How do you set up the expected profit cleanly?

At what multiplier does a bid start to make sense?

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

Try it first

Why does a delta-neutral book make money on a move?

What is the arithmetic, and where does the one half come from?

What happens if you rehedge and the stock comes back?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the index have less volatility than its members?

How do you derive the formula on the spot?

What does a desk do with implied correlation?

Where candidates lose it

What the interviewer asks next

Try it first

Why does fair value come from multiplying the averages?

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

What is the interviewer watching for in the requote?

Where candidates lose it

What the interviewer asks next