Derivatives Foundation puzzles, solved step by step

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

Showing 31–40 of 52 · filtered from 100Clear filters

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?
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?
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?
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.
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
071You walk on a grid from (0,0) to (6,4), each step one unit right or one unit up. The point (3,2) is blocked. How many routes avoid it?Probability and countingCoreSusquehanna International GroupLondon · 2026
Try it first
Before the block: how many routes from (0,0) to (6,4) with right and up steps only?
Show the worked solution
110 routes. Without the block there are C(10,4) = 210 routes, one for each way of placing 4 ups among 10 moves. A route through (3,2) is a route from (0,0) to (3,2), C(5,2) = 10 ways, followed by a route from (3,2) to (6,4), another C(5,2) = 10 ways, so 100 routes pass through the block. Subtract: 210 - 100 = 110.
Why is a lattice route a choice of positions rather than a sequence of decisions?
Think of a delivery driver in a city laid out as a grid who only ever drives east or north. Whatever order the turns come in, the trip is six blocks east and four blocks north, and the only freedom is which of the ten blocks are the north ones. A monotone route is fully described by choosing which 4 of its 10 moves go up, so the number of routes is 10 choose 4, which is 210, and no decision tree is needed. The same logic prices any question that asks how many ways a count can reach a level in fixed-size steps.
Of the 210 monotone routes from (0,0) to (6,4), every route through (3,2) is one of 10 routes into the block followed by one of 10 routes out of it, so 100 routes pass through it and 110 avoid it.
Why does multiplying the two legs count each bad route exactly once?
Because a monotone route visits a given point at most once; it can never come back. A route through (3,2) splits uniquely into the part before the block and the part after, so the number of such routes is the product of the two leg counts, 10 x 10 = 100, with no double counting to correct. Each leg is three rights and two ups, so C(5,2) = 10. If there were two blocked points, the same idea works but needs inclusion and exclusion: subtract the routes through each, then add back the routes through both.
The relationship
$N = \binom{10}{4} - \binom{5}{2}\binom{5}{2} = 210 - 10\times 10 = 110$
C(10,4) all routes: 10 moves, choose which 4 go up
C(5,2) C(5,2) routes into the block times routes out of it
N the routes that never touch the blocked point
What it says in wordsCount every route, subtract the ones that pass through the blocked point, which are the product of the two legs.
How do you check 110 another way?
Fill the grid with counts. Each point's count is the sum of the counts to its left and below, with the blocked point set to zero, and the corner comes out at 110. That dynamic-programming check takes a minute on paper and catches arithmetic slips in the binomials. It also answers the probability version the interviewer sometimes asks: if each step is right or up with equal chance, the chance a random walk reaches (6,4) at all is not 1, because it can overshoot, so the probability of avoiding the block among routes that do arrive is 110/210, about 52%, which is a different question from the probability for a free walk.
Where candidates lose it
Candidates try to count the avoiding routes directly and get lost in cases. The move is to count the complement: all routes less the routes through the block, with the block routes as a product of two binomials.
The second loss is a wrong binomial, often C(10,6) confused with something else or C(5,2) miscounted as 20. Say the legs out loud: three rights and two ups, 5 choose 2, is 10.
What the interviewer asks next
Now both (3,2) and (2,3) are blocked. How many routes avoid both?
Each step is right or up with probability one half. What is the probability a random walk from (0,0) passes through (3,2) before leaving the grid?
How many routes from (0,0) to (6,4) pass through (3,2) or (4,1)?
What is the general formula for routes from (0,0) to (m,n) avoiding a single point (a,b)?
Asked at Susquehanna International Group, Quantitative Research, London, 2026 (Wall Street Oasis): Probability about crossing from (0,0) to (6,4). Some point in the middle cannot pass through
073You roll a die and are paid the face value in rupees, but you may reject the first roll and roll once more, taking whatever the second roll shows. When should you reroll, and what is the game worth? Now the reroll costs Rs 1. What changes?Expected value and optimal stoppingCoreWolverine TradingChicago · 2016Old Mission CapitalNew York · 2018
Try it first
Free reroll. What is the game worth?
Show the worked solution
Reroll any 1, 2 or 3; keep a 4, 5 or 6; the game is worth 4.25. With a Rs 1 fee, keep a 3 as well and the value falls to 3.83. A fresh roll is worth 3.5, so you reroll only faces below it. The value is (4 + 5 + 6)/6 + (3/6) x 3.5 = 4.25. With the fee, a fresh roll nets 2.5, so a 3 is now worth keeping, and the value is (3 + 4 + 5 + 6)/6 + (2/6) x 2.5 = 23/6.
Why is the threshold the value of a fresh roll?
You are offered a mango from a basket; you can keep the one in your hand or swap it blind for another. You swap only if the one you hold is worse than the average mango. The reroll replaces a known face with the average of an unknown one, so you take it exactly when the face you hold is below that average, which is 3.5 for a fair die. Faces 1, 2 and 3 are below, so reroll them; 4, 5 and 6 are above, so keep them. There is no face equal to 3.5, so there is no tie to argue about.
With a free reroll you keep 4, 5 or 6 and reroll anything lower, for a value of 4.25; with a Rs 1 fee a fresh roll nets only 2.5, so a 3 is kept too and the value drops to 3.83.
How does the fee change the decision and the value?
The fee lowers what a fresh roll is worth, from 3.5 to 2.5, and the threshold moves with it. A 3 was worth rerolling for free, since 3 is below 3.5, but with the fee a 3 beats the 2.5 a reroll now nets, so you keep it, and only a 1 or a 2 is rerolled. The value becomes (3 + 4 + 5 + 6)/6 + (2/6) x 2.5 = 3 + 0.833 = 3.83. Check the alternative: keeping only 4 and above with the fee gives 2.5 + (3/6) x 2.5 = 3.75, which is worse, so the threshold really does move. The fee costs you 0.42 of value in total, less than the Rs 1 charge because you pay it only a third of the time.
The relationship
$V = \sum_{f \ge t}\frac{f}{6} + \frac{\#\{f < t\}}{6}\,(3.5 - c) \qquad c = 0:\ t = 4,\ V = \tfrac{17}{4} \qquad c = 1:\ t = 3,\ V = \tfrac{23}{6}$
t the smallest face you keep, one above the value of a fresh roll
c the cost of a reroll, zero or one rupee
3.5 - c what a reroll is worth net of its cost
What it says in wordsYou keep any face worth more than a reroll, and the value of the game is the kept faces plus the chance of rerolling times the net value of a fresh roll.
What is the general pattern a desk is looking for?
Backward induction. Value the last decision first, then use that value as the threshold for the decision before it; with two rerolls the second-stage value of 4.25 becomes the bar for the first roll, so you keep only a 5 or a 6 and the game is worth (5 + 6)/6 + (4/6) x 4.25 = 4.67. This is how an American option is priced on a tree, with exercise now compared against the continuation value. The limitation is that the puzzle has a known distribution; real stopping problems have to estimate the continuation value, and a wrong estimate moves the threshold.
Where candidates lose it
Candidates say the game is worth 3.5 because a die averages 3.5. The reroll is an option, exercised only when it helps, and options are worth something. Say the threshold, then the value.
With the fee, the loss is keeping the same threshold and only subtracting the cost. The threshold moves: a 3 is now kept. Compute both ways if you are unsure and pick the higher.
What the interviewer asks next
You get two rerolls instead of one, both free. Threshold and value?
The reroll costs Rs 2. Does the threshold move again, and what is the game worth?
How much would you pay for the right to one free reroll?
How does this relate to the exercise decision on an American option?
Asked at Wolverine Trading, Prop Trading, Chicago, 2016 (Wall Street Oasis): If you had to roll a dice and then roll another, what would be the value I would need in order to roll another dice?
Asked at Old Mission Capital, Finance, New York, 2018 (Wall Street Oasis): What is the expected value of rolling a fair dice? What if you can re-roll? What if the re-roll cost 1 dollar
074What is the expected number of fair coin flips needed to see two heads in a row? And how many to see a head followed by a tail?Random walks and Markov chainsCoreSquarepoint CapitalLondon · 2025
Try it first
HH and HT each have probability 1/4 on any two flips. Do they take the same expected time to appear?
Show the worked solution
6 flips for HH and 4 flips for HT. Track one state: whether the last flip was a head. For HH, a tail after a head sends you back to the start, so E(start) = 1 + E(one H)/2 + E(start)/2 and E(one H) = 1 + E(start)/2, giving 6. For HT, a head after a head leaves you still holding a head, so E(one H) = 1 + E(one H)/2 = 2 and E(start) = 2 + 2 = 4.
Why do two patterns with the same probability take different times?
Two queues at a counter: in one, a mistake sends you to the back; in the other, a mistake keeps your place. Both queues move at the same speed, but one is a much longer wait. Waiting for a pattern is a race with restarts, and what matters is how much progress a failure destroys: a tail after a head destroys everything for HH, while a head after a head destroys nothing for HT. That asymmetry, not the probability of the pattern, sets the expected time. It is also why HT and TH take 4 while HH and TT take 6.
Waiting for HH takes 6 flips because a tail after a head sends you back to the start, while waiting for HT takes 4 because a second head after a head still counts as a first head, so no progress is lost.
How do you set up and solve the equations in the room?
Two states, two equations, each saying one flip plus the average of what remains. For HH: from the start, a head takes you to one H and a tail keeps you at the start, so E0 = 1 + E1/2 + E0/2; from one H, a head finishes and a tail returns you to the start, so E1 = 1 + E0/2; substitute to get E1 = 4 and E0 = 6. For HT the second equation changes to E1 = 1 + E1/2, since a head keeps you at one H, giving E1 = 2, and the first equation gives E0 = 2 + E1 = 4. The first equation is the same in both problems; only the failure branch differs.
The relationship
$\text{HH: } E_0 = 1 + \tfrac{1}{2}E_1 + \tfrac{1}{2}E_0,\ E_1 = 1 + \tfrac{1}{2}E_0 \Rightarrow E_0 = 6 \qquad \text{HT: } E_1 = 1 + \tfrac{1}{2}E_1 \Rightarrow E_1 = 2,\ E_0 = 4$
E_0 expected flips remaining from the start, no useful progress
E_1 expected flips remaining once the last flip was a head
1/2 the chance of a head or a tail on each flip
What it says in wordsThe pattern whose failures throw away progress takes six flips on average, and the pattern whose failures keep progress takes four.
What is the quick check, and where does this matter beyond coins?
There is a general rule: the expected time to a pattern is the sum of 2 to the k over every length k at which the pattern overlaps itself. HH overlaps itself at lengths 1 and 2, so 2 + 4 = 6; HT overlaps only at its full length 2, so 4. The same arithmetic prices a bet on which of two patterns appears first, and it shows up on a desk whenever a signal needs a run of confirmations: a rule that resets on any contradicting tick waits far longer than one that keeps partial progress. The limitation is the fair coin; with a biased coin the overlap rule still holds but the powers of two become products of the relevant probabilities.
Where candidates lose it
The common answer is that HH and HT take the same time because each has probability 1/4. The interviewer is testing whether you see that waiting time depends on what a failure costs, not on the pattern's probability.
The second loss is setting up the HT chain with a return to the start after a second head. A head after a head is still one head; the state does not change.
What the interviewer asks next
What is the expected number of flips to see three heads in a row?
Which appears first on average, HHT or HTH, and why are their waiting times different?
The coin lands heads with probability 0.6. Expected time to HH?
You flip until you see HT and I flip until I see HH. What is the chance you finish first?
Asked at Squarepoint Capital, Quantitative Research, London, 2025 (Wall Street Oasis): a few siimple questions on statistical problems e.g. # of throws expected to get 2 heads in a row
075Five cards are dealt face down from a standard 52-card deck, with ace counting 1 and king 13. Make me a market on their total. Two of the cards are then turned face up: a king and a 3. Requote.Market makingCoreOptiverChicago · 2025
Try it first
Five cards, ace 1 to king 13. Where is fair value for the total?
Show the worked solution
Open around 35, say 33 bid at 37 offer. After the king and the 3, requote around 36.9, say 35.5 at 38.5. Each card averages 7, so five average 35. Once a 13 and a 3 are known, 16 points are fixed and three cards remain from a 50-card deck whose average is now (364 - 16)/50 = 6.96, so the total is 16 + 3 x 6.96 = 36.88. Three unknown cards carry less spread than five, so the market can tighten.
Why is the opening fair value simply five times seven?
Five friends each pick a sweet from a jar without looking; the expected total weight is five times the average sweet, even though each pick changes what is left for the next. Linearity of expectation holds whether or not the draws are independent, so the expected total of five cards is 5 x 7 = 35 and the dealing-without-replacement detail changes only the spread, not the centre. The standard deviation of the total is about 8.0, slightly below the independent-draw figure because the finite deck pulls the cards apart, and a market four wide around 35 is a reasonable opening quote.
Before any reveal the five cards average 7 each for a total of 35; once a king and a 3 are turned up, 16 points are fixed and the three remaining cards average 6.96 from the 50-card remainder, so the centre moves to 36.9 and the market can narrow because only three cards are still uncertain.
What does the reveal change, and what do people get wrong?
Two things move. The two revealed cards swap their expectation of 7 each for their actual values of 13 and 3, which lifts the total by 2, and the remaining deck has lost a high card and a low card, so its average drops from 7 to 6.96, which trims 0.12 off the three unknown cards. The careful centre is 36.88; the quick answer of 16 + 21 = 37 is off by only 0.12, and in the room 37 with a note that the deck is slightly poorer is a fine answer. What is not fine is leaving the market at 35, or widening it when the uncertainty has fallen from five cards to three.
The relationship
$E[\text{total}] = 5 \times 7 = 35, \qquad E[\text{total} \mid K, 3] = 16 + 3 \times \frac{364 - 16}{50} = 16 + 3 \times 6.96 = 36.88$
7 the average of a card, ace 1 to king 13
364 the total points in the deck, 4 x (1 + 2 + ... + 13)
50 the cards left once two are shown
What it says in wordsBefore the reveal the five cards are worth thirty-five; after it, the two known cards add sixteen and the three unknown ones average slightly under seven each.
How should the width change, and what trade do you expect next?
Width tracks the remaining uncertainty. The standard deviation of the total falls from about 8.0 to about 6.2 once only three cards are unknown, so a market that was 4 wide can go to 3 wide without taking more risk per trade. The next step is usually a trade: if the interviewer lifts your 38.5 offer, you are short at above fair and hold; if they hit 35.5 you are long below fair. The limitation is that the puzzle assumes a fair deck and honest reveals; in the real version of this game the counterparty may have seen a card you have not, and a run of trades in one direction is the tell.
Where candidates lose it
The opening loss is overthinking the dealing without replacement and quoting something other than 35 as the centre. Linearity handles it: five cards times seven.
The requote loss is anchoring on the old centre or, less often, forgetting that the revealed cards change the remaining deck. Replace the two cards with their values, adjust the remaining average down slightly, and tighten the market.
What the interviewer asks next
Instead of a king and a 3, the two revealed cards are both kings. Requote.
What is the standard deviation of the five-card total, and why is it below the independent-draw figure?
I lift your offer twice after the reveal. What does that tell you and what do you do?
Make a market on the highest of the five cards rather than the total.
Asked at Optiver, Future focus Interview, Chicago, 2025 (Wall Street Oasis): This interview was a standard market making game with cards and CPUs quoting prices.

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

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

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 must the longer call be worth at least the shorter one?

What happens at three months in each case?

Why would a market maker see this quote and what does it say about the vol surface?

Where candidates lose it

What the interviewer asks next

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

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 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 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

Try it first

Why is a lattice route a choice of positions rather than a sequence of decisions?

Why does multiplying the two legs count each bad route exactly once?

How do you check 110 another way?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the threshold the value of a fresh roll?

How does the fee change the decision and the value?

What is the general pattern a desk is looking for?

Where candidates lose it

What the interviewer asks next

Try it first

Why do two patterns with the same probability take different times?

How do you set up and solve the equations in the room?

What is the quick check, and where does this matter beyond coins?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the opening fair value simply five times seven?

What does the reveal change, and what do people get wrong?

How should the width change, and what trade do you expect next?

Where candidates lose it

What the interviewer asks next