Derivatives Foundation puzzles, solved step by step

Puzzles: 100
Traced to a firm: 66
Topics: 12
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076You roll a fair die until you have seen every even number (2, 4 and 6) at least once. Given that the roll which completes the set is a 2, what is the probability that the first roll was a 1, and why is the answer not 1/5?Probability and countingCoreSquarepoint CapitalLondon · 2026

Try it first

Before you work it: given the game ends on a 2, which first rolls are more likely than the others?

Show the worked solution

The probability is 1/6, not 1/5. By symmetry the game ends on a 2 one third of the time. A first roll of 1 happens one sixth of the time and leaves all three evens unseen, so 2 is last with probability 1/3. The joint probability is 1/18, and 1/18 divided by 1/3 is 1/6. The naive 1/5 treats the five possible first rolls as equally likely given the ending, and they are not.

Why does the ending change the odds of the start?

Think of a race where you learn only who finished last. If you are told that runner C came last, the start line was still fair, but some starting arrangements make C last more often than others, and you should update towards those. The die works the same way. A first roll of 4 leaves only two evens unseen, so 2 comes last half the time; a first roll of 1 leaves three evens unseen, so 2 comes last only one third of the time. The ending is twice as consistent with a first-roll 4 as with a first-roll 1, and a first-roll 2 is ruled out entirely, because a 2 that has already appeared cannot complete the set.

An odd first roll has probability 1/2 and leaves a 1/3 chance that the 2 arrives last, a first roll of 2 leaves no chance, and a first roll of 4 or 6 has probability 1/3 and leaves a 1/2 chance, so the joint weights are 1/6, 0 and 1/6, and given the game ends on a 2 the first roll was a 1 with probability 1/6 and a 4 with probability 1/4.

How do you set up the Bayes calculation in thirty seconds?

Group the first roll into three cases rather than six, because 1, 3 and 5 are interchangeable and so are 4 and 6. Odd first roll: probability 1/2, and then 2 ends the game with probability 1/3, giving a joint weight of 1/6. First roll 2: joint weight 0. First roll 4 or 6: probability 1/3, then 2 ends the game with probability 1/2, joint weight 1/6. The weights add to 1/3, which is the unconditional chance of ending on a 2, as symmetry says they must. Given the ending, the first roll was odd with probability 1/2 and was 4 or 6 with probability 1/2, so each odd face carries 1/6 and each of 4 and 6 carries 1/4.

The relationship

P(\text{first}=1 \mid \text{end on } 2) = \frac{\tfrac{1}{6}\cdot\tfrac{1}{3}}{\tfrac{1}{3}} = \frac{1}{6}

1/6	the chance the first roll is a 1
1/3 in the numerator	the chance 2 is the last even to appear when all three are still unseen
1/3 in the denominator	the unconditional chance the game ends on a 2, by symmetry across the three evens

What it says in wordsMultiply the chance of the start by the chance of the ending given that start, then divide by the chance of the ending.

The sanity check the interviewer wants to hear: 3 x 1/6 + 2 x 1/4 = 1, so the posterior weights over the five possible first rolls add up. Then say the general point. An odd roll is a wasted roll that tells you nothing about which even finishes last, which is why its posterior weight is simply its prior, 1/6, unchanged. The information in the ending all goes into shifting weight from the 2, which is now impossible, onto 4 and 6.

Where candidates lose it

The fast wrong answer is 1/5: the first roll cannot be 2, five faces remain, so each gets a fifth. It fails because the ending is not equally likely after each of those five starts. Candidates who say 1/5 have forgotten that conditioning reweights, it does not just delete.

The second loss is doing the Bayes sum face by face and running out of time. Group the odd faces together and the 4 and 6 together, use symmetry for the denominator, and the whole thing is three lines.

What the interviewer asks next

Given the game ends on a 6, what is the probability the first roll was a 2?
What is the expected number of rolls to see all three evens?
Now condition on the game ending on roll 5 exactly. Does the first-roll distribution change again?

Asked at Squarepoint Capital, Quant Research Intern Interview, London, 2026 (Wall Street Oasis): why is the probability of seeing a 1 on our first roll, given that we end on a 2, not 1/5

077A fund charges 2% of assets a year plus 20% of gains and expects a gross return of 10%. If it cuts the management fee to 1%, what performance fee keeps expected revenue unchanged, and why is the performance fee really a call option the investors have written?Option payoffs and no-arbitrageCoreTwo SigmaNew York · 2026
Try it first
At the expected return of 10%, what performance fee replaces the lost 1% of management fee?
Show the worked solution
A 30% performance fee holds revenue at 4% of assets at the mean, but the swap is not neutral once returns vary. Two and twenty earns 2 + 0.2 x 10 = 4; one and thirty earns 1 + 0.3 x 10 = 4. The performance fee pays 20% of max(return, 0), which is a call on the fund's return struck at zero. Investors have written it, and its value rises with volatility, so moving fee from fixed to performance raises what the manager expects to collect.
How do you make the two schedules equal at the mean?
A shopkeeper who swaps a fixed monthly rent for a share of sales asks one question first: at my usual sales, what share leaves the landlord no worse off? Do the same here, in percent of assets. Two and twenty collects 2 fixed plus 20% of a 10% gain, 4% of assets; cutting the fixed fee to 1 leaves 3 points to be earned from a 10% return, which needs a 30% performance fee. That is the arithmetic the interviewer wants first, and it is one line: 1 + 10x = 4, so x = 0.3.
Two and twenty is a flat 2% that kinks upward at a zero return with slope 0.2, one and thirty is a flat 1% that kinks upward with slope 0.3, and the two cross at a 10% return where both pay 4%, but with returns spread around 10% with 20 points of volatility the expected fee is 4.79% under two and twenty and 5.19% under one and thirty.
Why is the performance fee an option, and who has written it?
The manager receives 20% of the gain when the fund is up and nothing when it is down. That is the payoff of a call on the fund's return with a strike of zero: convexA payoff that bends upward, so the average of the payoff over a spread of outcomes is higher than the payoff at the average outcome. in the return, floored at nothing. Investors are on the other side; they have granted the manager that call and are paid for it only through the management fee they do not have to pay. Because a call is worth more when the underlying is more volatile, the performance fee is worth more than its value at the mean return, and the more of the fee you move into performance, the more the schedule is worth for the same expected return.
The relationship
$E[\max(R,0)] = \mu\,\Phi\!\left(\tfrac{\mu}{\sigma}\right) + \sigma\,\phi\!\left(\tfrac{\mu}{\sigma}\right) = 0.10\,\Phi(0.5) + 0.20\,\phi(0.5) \approx 0.1396$
mu the expected return, 10%
sigma the volatility of the yearly return, taken as 20 points
Phi, phi the normal distribution and density functions
What it says in wordsWith returns spread normally around 10% with 20 points of volatility, the average floored gain is about 14.0%, not 10%, because losses are cut off at zero but gains are not.
Put numbers on it. With that spread the expected performance fee under two and twenty is 0.2 x 14.0% = 2.79%, so the manager expects 4.79% of assets, not 4%. Under one and thirty it is 1 + 0.3 x 14.0% = 5.19%. The swap that looked neutral at the mean adds about 0.40% of assets a year in expected revenue. Say the limitation too: real schedules carry hurdles and high-water marks, which raise the strike and cut the option's value, and the fee is charged on the net of the management fee, which shaves a little off both sides.
Where candidates lose it
Candidates get 30% and stop, as if the question were arithmetic. The interviewer is listening for the word option. Without it the answer is a shopkeeper's answer, not a derivatives answer.
The second loss is saying the fee is an option and then claiming volatility makes it worth less because the fund might lose money. The fund's loss is the investor's, not the manager's; the manager's payoff is floored at zero, which is exactly why volatility helps the manager.
What the interviewer asks next
Add a hurdle of 5%. Does the neutral performance fee rise or fall?
A high-water mark means losses must be recovered before fees resume. Which Greek of the option does that change most?
If the fund's volatility doubles, roughly how much does the 20% performance fee gain in expected value?
Asked at Two Sigma, Equity Capital Markets, New York, 2026 (Wall Street Oasis): the 2/20 rule, and if one part of this equation changed, how would the other variable make up for it
079What are the last two digits of 4^3000?Mental maths and estimationCoreBelvedere TradingNew york · 2021
Try it first
What is the shape of the method, before any arithmetic?
Show the worked solution
76. Multiplying by 4 and keeping only the last two digits gives 04, 16, 64, 56, 24, 96, 84, 36, 44, 76, and then 76 x 4 = 304 returns to 04. The cycle has length 10 starting at 4^1, and 3000 is a multiple of 10, so 4^3000 ends like 4^10, in 76. Check: 76 x 76 = 5,776, so 76 reproduces itself under squaring, which is what a power of 4^10 must do.
Why do only the last two digits matter at each step?
When you work out what time it will be 3,000 hours from now, you do not count the hours; you note that the clock face has 24 positions and ask where 3,000 lands on it. Last two digits are a clock with 100 positions. The last two digits of a product are fixed by the last two digits of the factors alone, so the sequence of 4^n mod 100 can only visit 100 states and has to fall into a cycle. Walk it: 04, 16, 64, 56, 24, 96, 84, 36, 44, 76, and 76 x 4 = 304 brings you back to 04. Ten steps, then it repeats.
The last two digits of 4^n for n from 1 to 10 are 04, 16, 64, 56, 24, 96, 84, 36, 44 and 76, and 76 x 4 = 304 returns the ring to 04, so because 3000 is 300 full turns of ten, 4^3000 lands on the same box as 4^10, which is 76.
How do you reduce 3000 without miscounting the start of the cycle?
The cycle begins at 4^1 = 04, not at 4^0 = 01, because 01 is never revisited: once a power of 4 is a multiple of 4 it stays one, and 01 is not. So the positions n = 1, 11, 21 and so on share last digits 04, and the positions n = 10, 20, 30 and so on share 76. 3000 is a multiple of 10, so it sits in the same slot as 10, and 4^10 ends in 76. Candidates who start counting from n = 0 land one step off and say 44.
The relationship
$4^{3000} = (4^{10})^{300}, \qquad 4^{10} \equiv 76 \pmod{100}, \qquad 76^2 = 5776 \equiv 76$
mod 100 keep only the last two digits
76^2 = 5776 76 squared ends in 76, so every power of 76 ends in 76
What it says in wordsAny power of 4^10 ends in 76 because 76 reproduces itself whenever it is multiplied by itself.
A second route, worth one sentence: split 100 into 4 and 25. Any 4^n with n at least 1 is 0 mod 4, and 4^10 = 1,048,576 is 1 mod 25, so 4^3000 is 1 mod 25. The number under 100 that is 0 mod 4 and 1 mod 25 is 76. Two methods agreeing is the finish an interviewer wants. The limitation: the cycle trick is only this short because 4 is small; for a base like 7 the cycle mod 100 is 4 steps, for 3 it is 20, and you should check rather than assume.
Where candidates lose it
The fast loss is counting the cycle from the wrong end. People list ten residues, say the cycle is 10, then divide 3000 by 10 and read off the first entry, 04, or count from 4^0 and get 44. Decide whether your list starts at 4^1 and say so.
The slower loss is trying to use Euler's theorem with 4 and 100 not coprime, which it does not allow. Either walk the cycle or split 100 into 4 and 25.
What the interviewer asks next
What are the last two digits of 7^3000?
What is the last digit of 3^3000, and how long is that cycle?
Why does 76 reproduce itself under multiplication by any power of 4 above 4^9?
Asked at Belvedere Trading, Equities, New york, 2021 (Wall Street Oasis): Some basic number theory (4^3000 modulo 100), basic probability calculations, and combinatorics puzzles
080Make me a market on the number of nappies used in the UK each day.Market makingCoreDRWLondon · 2025
Try it first
Which of these is the interviewer actually marking?
Show the worked solution
About 10 million a day, so quote 8 million bid, 12 million offered. Take a population of roughly 67 million, births around 1.1% a year, so about 737,000 babies, children in nappies for about two and a half years, giving 1.84 million children, at five or six nappies a day each: 10.1 million. Running every input at its low gives 5.2 million and at its high 17.6 million, so the quote sits inside that range with room to be wrong.
How do you turn a guess into a market?
If a friend asks you to bet on how many samosas a canteen sells a day, you would not name one number. You would count the seats, guess the sittings, guess how many order a samosa, and then say a range you would bet either side of. A market is that range with a price on each end. Decompose the quantity into inputs you can bound, carry a low and a high through each step, and let the spread of the outputs set the width of your quote. Population times birth rate gives babies a year; times years in nappies gives children in nappies; times nappies per child per day gives the answer. The population and birth rate are the kind of round figures you carry in your head; the interviewer accepts any sensible anchor and marks the structure.
Carrying a low and a high through population, birth rate, years in nappies and nappies per day gives 1.3 to 2.5 million children and 5.2 to 17.6 million nappies a day, and the market of 8 million bid, 12 million offered sits around the point estimate of 10.1 million inside that range rather than at its edges.
How wide should the market be?
Not as wide as the full range. Every input at its low or every input at its high is unlikely, because the errors are mostly independent and partly cancel. A quote about one third as wide as the all-low to all-high range, centred on the point estimate, is tight enough to be a real market and wide enough that you expect to be inside it. Here the range runs from 5.2 to 17.6 million, so 8 at 12 is a working quote. Say the limitation: the tree ignores adult incontinence products and children over three, both of which push the true figure up, so if anything you would skew the market higher rather than lower.
The relationship
$67\text{m} \times 1.1\% \times 2.5 \times 5.5 \approx 10.1\text{ million a day}$
67m x 1.1% births a year, about 740,000
2.5 years a child spends in nappies
5.5 nappies per child per day, more for newborns and fewer for toddlers
What it says in wordsBabies a year times years in nappies gives children in nappies, and times nappies a day each gives the daily total.
Then expect the interviewer to trade. If they lift your offer at 12 million, ask yourself what they know: perhaps the adult market, perhaps a higher nappies per day. Move your quote up and tighten, do not freeze it. If they hit your bid, do the reverse. The exercise is not the answer; it is whether you update your market when someone trades against you, which is what market making is.
Where candidates lose it
Candidates give a point estimate and stop, or give a market so wide it is meaningless, like 1 million at 100 million. Both are refusals to make a market. The interviewer wants a two-sided quote you will honour and a reason for its width.
The second loss is getting a trade and not reacting. Whoever lifts your offer is telling you something; a quote that does not move after a trade is the first thing a desk trains out of you.
What the interviewer asks next
I lift your offer at 12 million. Where is your next quote?
Now make me a market on the number of nappies sold in India each day. Which inputs change most?
Would you quote tighter or wider if I told you I was a nappy manufacturer?
Asked at DRW, Trading, London, 2025 (Wall Street Oasis): Make me a market on the amount of diapers used in the UK daily

086100 passengers board a 100-seat plane in order. The first has lost his boarding pass and sits in a seat chosen at random. Every later passenger takes their own seat if it is free, and otherwise a random free seat. What is the probability that the last passenger gets their own seat?Probability and countingCoreBelvedere TradingChicago · 2022

Try it first

Before any algebra: which seats can the last passenger possibly end up in?

Show the worked solution

One half. By the time the last passenger boards, seats 2 to 99 are always taken, because each owner either sat in theirs or found it taken. The free seat is either seat 1 or seat 100. Every passenger who chooses at random, the first and each one displaced after him, faces seat 1 and seat 100 with exactly the same chance, so the game is equally likely to settle on either. The answer is 1/2 for any plane with two or more seats.

Why do only two seats matter?

Think of a cloakroom where one guest hangs his coat on a random hook. Each later guest uses their own hook if it is free and a random free hook if not. Every hook except two has an owner who will turn up and either use it or find it used. The last passenger can only end up in seat 1 or seat 100, because every seat in between has an owner who boards earlier and leaves it occupied. So the whole question is which of those two seats is still empty when the last passenger walks down the aisle.

In one example passenger 1 picks among 100 free seats and takes seat 23, passenger 23 picks among 78 and takes seat 61, and passenger 61 picks among 40 and takes seat 1, after which everyone sits in their own seat, and at every one of those choices seat 1 and seat 100 carried exactly the same chance, which is why the answer is 1/2.

Why are the two endings equally likely?

Follow the chain of displaced people. Passenger 1 picks at random. If he takes seat 1, nobody is ever displaced and the last passenger gets seat 100. If he takes seat 100, the last passenger is shut out. If he takes some other seat k, everyone up to k sits normally and passenger k inherits the problem: a random choice among the free seats, which still include both seat 1 and seat 100. Every random chooser in the chain picks seat 1 and seat 100 with equal probability, and the chain stops the moment either is taken, so the two endings carry the same total weight, 1/2 each. The other seats only pass the problem along; they never decide it.

The relationship

f(n) = \frac{1}{n}\cdot 1 + \frac{1}{n}\cdot 0 + \frac{1}{n}\sum_{k=2}^{n-1} f(n-k+1), \qquad f(2) = \tfrac{1}{2} \;\Rightarrow\; f(n) = \tfrac{1}{2}

f(n)	the chance the last of n passengers gets their own seat
1/n x 1	passenger 1 takes seat 1: the last passenger is safe
1/n x 0	passenger 1 takes the last seat
f(n - k + 1)	passenger 1 takes seat k, and passenger k faces the same problem with fewer seats

What it says in wordsIf the answer is a half for every smaller plane, the recursion gives a half for this one too, and with two seats it is plainly a half.

Check it on the smallest case aloud: with two seats, passenger 1 takes his own or the other with equal chance, so 1/2. Running the recursion for every plane from 2 to 100 seats returns exactly 1/2 each time. A useful extension the interviewer often asks next: passenger j, for j from 2 to n, gets their own seat with probability (n - j + 1)/(n - j + 2). Passenger 2 is almost always fine; passenger 99 is fine two times in three. The limitation is in the rules: if displaced passengers preferred seats near the front, the symmetry between seat 1 and seat 100 breaks and so does the half.

Where candidates lose it

The common loss is reaching for 1/100, on the idea that the last passenger is one of a hundred equally unlucky people. That treats the last seat as a random seat. It is not: by the end only two seats can be free, so the answer has to be large.

The second loss is starting the recursion and drowning in cases. Say the two-seat argument first, then use the recursion only as a check. The interviewer is listening for the symmetry between seat 1 and seat 100.

What the interviewer asks next

What is the chance that passenger 50 gets their own seat?
What is the expected number of passengers who end up out of their own seat?
Now the first two passengers have both lost their passes. Does the last passenger's chance change?

Asked at Belvedere Trading, Equity Capital Markets, Chicago, 2022 (Wall Street Oasis): Drunk passenger on a plane, what's the probability the Nth passenger gets his assigned seat

089Four doors hide one prize. You pick a door. The host, who knows where the prize is, opens one of the other three that is empty and offers you a switch to either of the two remaining closed doors. Should you switch, and what are your chances each way?Conditional probability and BayesCoreSquarepoint CapitalLondon · 2026
Try it first
After the host opens an empty door, what is the chance your original door hides the prize?
Show the worked solution
Switch: either other closed door wins with probability 3/8, against 1/4 for staying. Your door had a 1/4 chance and the host's action cannot change it, because he can always open an empty door among the other three. The remaining 3/4 sat on three doors and now sits on two, so each carries 3/8. Switching does not make you a favourite, it only beats staying, 37.5% against 25%.
Why does your own door keep its 1/4?
A friend who knows the answers to a quiz looks at three questions you did not attempt and crosses out one wrong option. She would have done that whether your own answer was right or wrong, so her crossing out says nothing about your answer. Information only moves probability onto or off your door if the action could have depended on what is behind it, and the host can open an empty door among the other three every time. So your door stays at 1/4, exactly where it started.
Before the host acts each of the four doors carries 1/4, and after he opens empty door 4 your door 1 still carries 1/4 while the 3/4 that sat on doors 2, 3 and 4 now sits on doors 2 and 3, 3/8 each, so switching wins 37.5% of the time against 25% for staying.
How does Bayes give 3/8 for each of the other doors?
Say the host opens door 4. If the prize is behind your door 1, he chooses among doors 2, 3 and 4 at random, so he opens door 4 with chance 1/3. If it is behind door 2, he must choose between 3 and 4, chance 1/2; the same for door 3. If it is behind door 4, he never opens it. Multiply each by the prior of 1/4: 1/12, 1/8, 1/8 and 0, which add to 1/3. Divide through and the posterior is 1/4 for door 1, 3/8 for door 2, 3/8 for door 3 and nothing for door 4. The host is more likely to open door 4 when the prize sits behind door 2 or 3 than when it sits behind yours, and that difference is the information.
The relationship
$P(\text{door 2} \mid \text{opens 4}) = \frac{\tfrac{1}{4}\cdot\tfrac{1}{2}}{\tfrac{1}{4}\cdot\tfrac{1}{3} + \tfrac{1}{4}\cdot\tfrac{1}{2} + \tfrac{1}{4}\cdot\tfrac{1}{2} + 0} = \frac{1/8}{1/3} = \frac{3}{8}$
1/4 the prior on each door
1/2 the chance the host opens door 4 when the prize is behind door 2
1/3 the chance he opens door 4 when the prize is behind your door
What it says in wordsEach closed door you did not pick ends up with three eighths, half again as much as your own quarter.
Then generalise, which is what the question tests. With n doors and one opened, staying wins 1/n and each other closed door wins (n - 1)/(n(n - 2)). With three doors that is 2/3, the familiar answer; with 100 doors it is 99/9,800, barely above the 1/100 for staying. The edge from switching shrinks as the host reveals a smaller share of the doors. The limitation is the host's rule: if he opened a door at random and it merely happened to be empty, he would carry no information about the other doors and switching would gain nothing.
Where candidates lose it
The intuitive loss is 1/3 each: three closed doors, so a third apiece. It treats the host's choice as if he had picked blindly, which he did not. The interviewer asks this version precisely to see whether you say why the intuitive answer is wrong rather than recite 2/3.
The second loss is reciting the three-door logic and saying switching wins 3/4. Switching moves you to one of two doors, not to both, so you collect only half of the 3/4.
What the interviewer asks next
Now the host opens two empty doors and offers you the last closed one. What are the chances?
If the host opens a door at random and it happens to be empty, should you switch?
With 100 doors and one opened, by how much does switching beat staying?
Asked at Squarepoint Capital, Quant Research Intern Interview, London, 2026 (Wall Street Oasis): I was asked why the 'intuitive answer' was not true rather than just what the correct answer was, related to the Monty Hall problem
092Screen quotes: the 95 call is 7.00 bid, 7.40 offered, and the 100 call is 4.10 bid, 4.50 offered. You buy the 95 call and sell the 100 call. What do you pay, what is the most you can make and lose, and where do you break even at expiry?Option payoffs and no-arbitrageCoreWolverine Trading, Chicago, ILUSA · 2019
Try it first
What does the call spread cost you to put on?
Show the worked solution
You pay 3.30; the most you can make is 1.70, the most you can lose is 3.30, and you break even at 98.30. You buy the 95 call at the offer, 7.40, and sell the 100 call at the bid, 4.10. At expiry the spread is worth nothing below 95, the stock minus 95 between the strikes, and 5 above 100. So the profit runs from minus 3.30 to 5 - 3.30 = 1.70, and is zero when the stock is 95 + 3.30 = 98.30.
Which side of each quote do you deal on?
At a currency counter at the airport there are two rates on the board: the one they buy at and the one they sell at, and you always get the worse one for you. Option screens are the same. If you want to trade now, you buy at the offer and sell at the bid, so a two-leg trade pays the spread on both legs. Buy the 95 call at 7.40, sell the 100 call at 4.10, net debit 3.30. At the mid prices, 7.20 and 4.30, the same trade would cost 2.90. The 0.40 difference is the cost of crossing two bid-offer spreads of 0.40 each, half of each.
Bought across the bid-offer for 3.30, the 95/100 call spread loses 3.30 below 95, gains one for one between the strikes, makes 1.70 above 100 and breaks even at 98.30, while the same spread at mid prices would cost 2.90 and break even at 97.90.
How do you read off the maximum profit, loss and break-even?
Walk the stock price up. Below 95 both calls expire worthless, so you lose the 3.30 you paid. Between 95 and 100 only the long call pays, one for one, so the profit rises from minus 3.30. Above 100 the short call pays out as fast as the long call pays in, so the value is capped at the strike gap of 5. A call spread can never be worth more than the gap between its strikes, so the most you can make is 5 minus what you paid, 1.70, and you break even where the stock has risen 3.30 above the lower strike, at 98.30.
The relationship
$\text{cost} = 7.40 - 4.10 = 3.30, \qquad \text{max profit} = (100 - 95) - 3.30 = 1.70, \qquad S^{*} = 95 + 3.30 = 98.30$
7.40 the offer on the 95 call, what you pay to buy it
4.10 the bid on the 100 call, what you receive to sell it
S* the break-even stock price at expiry
What it says in wordsPay the offer, receive the bid, and the spread's payoff is boxed between losing the premium and making the strike gap less the premium.
Two checks to say aloud. First, the price passes the no-arbitrage bounds: a 95/100 call spread must cost between 0 and 5, and 3.30 does. Second, the risk-reward: you risk 3.30 to make 1.70, which only makes sense if you think the stock finishes above 98.30 more than 3.30/5 = 66% of the time. The limitation: this is the payoff at expiry. Before expiry the spread's value moves with volatility and time, and you could unwind it, but again only by selling the 95 at its bid and buying the 100 at its offer.
Where candidates lose it
The common loss is using mid prices and answering 2.90, as if the screen would trade with you at the middle. The interviewer gave you two-sided quotes precisely to see whether you know which side you hit.
The second loss is getting the maximum profit wrong by forgetting the cap, and saying it is unlimited because you own a call. You also sold one, and above 100 the two cancel; the most a 5-wide spread can ever be worth is 5.
What the interviewer asks next
What would you pay if you could work both orders at mid?
At what price would you buy the 95 call and sell the 100 call so that you risk exactly as much as you can make?
Build the same view with puts. What are the costs on this screen if the 95 put is 1.80 at 2.10 and the 100 put is 3.70 at 4.00?
Asked at Wolverine Trading, Prop Trading, Chicago, IL, USA, 2019 (Wall Street Oasis): pricing options given an ask and a bid price for options with different strikes if you were to short one and long another
093A stock at 100 will be 120 with probability 70% or 90 with probability 30% in one period, and interest rates are zero. Price a call struck at 100. Why does the 70% not appear in your answer?Option pricing intuitionCoreQuant trading
Try it first
What is the call worth?
Show the worked solution
The call is worth 20/3, about 6.67, and the 70% does not matter because the call can be copied with stock and cash. Hold 2/3 of a share and borrow 60: at 120 the copy is worth 80 - 60 = 20, at 90 it is worth 60 - 60 = 0, matching the call in both states. The copy costs 66.67 - 60 = 6.67, so the call must too. The real-world odds are already in the stock price, which the copy uses.
How do you copy the call?
If a shop sells a gift box of two items for more than the items cost separately, you buy the items and skip the box; the box's price is pinned by what goes in it. Options work the same way. Find a mix of stock and cash that pays exactly what the call pays in every state, and the call must cost what the mix costs, whatever anyone believes about the odds. The call pays 20 or 0, a swing of 20, while the stock swings from 120 to 90, a swing of 30. So hold 20/30 = 2/3 of a share. At 120 that is 80, which is 60 too much, and at 90 it is 60, also 60 too much: borrow 60 today and repay it in either state.
Two thirds of a share financed with 60 of borrowing pays 20 when the stock goes to 120 and 0 when it goes to 90, exactly like the call, so the call costs what that portfolio costs, 2/3 x 100 - 60 = 6.67, and the real-world 70% chance of an up move, which would give an average payoff of 14, never enters.
Where did the 70% go?
It is in the stock price. A stock that goes up 70% of the time to 120 and is still priced at 100 today is one the market demands a return on, because it is risky. The copy buys the stock at that price, so it inherits whatever the market thinks of the odds and the risk. The option is priced relative to the stock, not relative to anyone's forecast, so the real-world probability cancels out of the answer. What does appear is a different probability, q, the one that makes the stock earn the risk-free rate: 100 = q x 120 + (1 - q) x 90, so q = 1/3. Discounting the call's payoff at q gives 1/3 x 20 = 6.67, the same answer by another route.
The relationship
$\Delta = \frac{20 - 0}{120 - 90} = \frac{2}{3}, \qquad C = \Delta S - B = \tfrac{2}{3}(100) - 60 = 6.67, \qquad q = \frac{100 - 90}{120 - 90} = \tfrac{1}{3}$
Delta shares held in the copy, the call's swing over the stock's swing
B cash borrowed, 60, so the copy pays nothing in the down state
q the risk-neutral probability of the up move, not a forecast
What it says in wordsHold two thirds of a share, borrow sixty, and you have built the call for 6.67; the risk-neutral probability of a third gives the same number.
Then show the arbitrage, since that is what makes the answer binding. If the call traded at 8, sell it and buy the copy for 6.67: you pocket 1.33 today and the two positions cancel in both states. If it traded at 5, do the reverse. The limitation is the one-step world: real prices take many values, so the copy has to be rebalanced as the stock moves, which is where the Black-Scholes model comes from, and where trading costs and jumps make the copy imperfect.
Where candidates lose it
The common loss is answering 14, the call's expected payoff under the stated odds. It is the natural first move and the one the question is built to catch. Expected payoff under real-world odds is not a price unless everyone is indifferent to risk.
The second loss is getting q = 1/3 and then calling it the true chance of an up move. It is not a forecast. Say what it is: the probability that makes the stock's expected return equal the risk-free rate.
What the interviewer asks next
Price the put struck at 100 in the same tree, and check put-call parity.
If interest rates were 5% for the period, what is q and what is the call worth?
The stock's up probability rises to 90% but its price stays at 100. What happens to the call price, and why?
094A broad equity index has returned an average of 11% a year over the past 30 years, with annual volatility of 16%. Give a 95% confidence interval for its true expected annual return.Distributions and statisticsCoreOld Mission CapitalChicago · 2025
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How wide is the 95% interval for the true expected return, either side of 11%?
Show the worked solution
About 5.3% to 16.7%. The standard error of a 30-year average is the yearly volatility over the square root of 30: 16/5.48 = 2.92 points. A 95% interval is 1.96 standard errors either side, about 5.7 points, so 11% plus or minus 5.7%. Thirty years of data leave the true expected return anywhere from modest to spectacular, and narrowing it to plus or minus one point would take nearly a thousand years.
Why is the interval so wide after 30 years?
Weigh yourself on a scale that is off by up to two kilos each time. One reading tells you little; the average of four readings is better, but only twice as good, not four times, because errors cancel in proportion to the square root of the count. The uncertainty in an average shrinks with the square root of the number of observations, so 30 noisy years cut a 16-point yearly spread only to about 2.9 points. That is the standard error. Multiply by 1.96 for 95% and you get roughly 5.7 points either side of 11%.
The 95% interval for the true expected return narrows only with the square root of the years of data, so with an 11% average and 16% volatility it runs from about -3% to 25% after 5 years, from 5.3% to 16.7% after 30 years, and would need about 983 years to shrink to plus or minus one point.
What does the interviewer do with the answer?
Usually they push on what it means. A 30-year history cannot tell a 6% market from a 16% market with any confidence, which is why long-run return assumptions are judgements, not measurements. To get the interval down to plus or minus 2 points needs (1.96 x 16/2)^2 = 246 years, and to plus or minus 1 point, 983 years. Volatility, by contrast, is estimated far better from the same data, because daily or monthly returns give thousands of observations of spread, while there is only one 30-year path for the mean. Sampling more often does not help the mean: the average depends only on the first and last levels.
The relationship
$\bar{r} \pm 1.96\,\frac{\sigma}{\sqrt{n}} = 11\% \pm 1.96 \times \frac{16\%}{\sqrt{30}} = 11\% \pm 5.73\% = [5.27\%,\; 16.73\%]$
r bar the sample average yearly return, 11%
sigma the volatility of one year's return, 16%
n the number of years, 30
1.96 the multiplier for a 95% normal interval
What it says in wordsThe average of 30 noisy years is itself noisy, with an error of the volatility over the square root of 30, so the interval is about 5.7 points either side.
Say the assumptions, because the interviewer will. The interval treats the 30 yearly returns as independent draws from one unchanging distribution, which real markets are not; regime shifts make the true uncertainty wider. The 11% and 16% are the inputs given for this exercise, so confirm any real index's figures before using them. And an arithmetic average of yearly returns is not the compound growth rate; with 16% volatility the compound rate is roughly 1.3 points lower, which is a separate question worth flagging rather than answering.
Where candidates lose it
The common loss is using 16% as the error, giving 11% plus or minus 31 points, which confuses the spread of one year with the uncertainty of an average. The opposite loss is a tiny interval from forgetting that only 30 independent years exist.
The second loss is reaching for daily data to shrink the interval. More frequent sampling sharpens the volatility estimate, not the mean: the total return over 30 years depends only on where the index started and where it ended.
What the interviewer asks next
How many years of data would you need to tell a 6% expected return from an 8% one?
Why does more frequent data help estimate volatility but not the expected return?
If returns were autocorrelated, would the interval be wider or narrower?
Asked at Old Mission Capital, Prop Trading, Chicago, 2025 (Wall Street Oasis): Confidence interval on S&P 500 return past 30 years
095Three people stand in a line facing forward, wearing hats drawn from 3 red and 2 blue. The back person sees the two hats ahead, the middle person sees only the front hat, and the front person sees none. The back says "I don't know my colour", then the middle says "I don't know my colour". What colour is the front person's hat?Games and logicCoreWolverine Trading, Chicago, ILUSA · 2019
Try it first
What does the back person's "I don't know" rule out?
Show the worked solution
Red. The back person would know his hat if he saw two blues, since only two exist; his silence rules that out. The middle person now knows the two front hats are not both blue. If he saw blue on the front person, he would know his own was red, and he would say so. His silence means the front hat is not blue. So the front person, who sees nothing, deduces red from the two silences alone.
What does a silence tell everyone else?
If a friend who can see the scoreboard says she cannot tell who is winning, you learn that the scores are close. Her not knowing is information, because you know what she would have said if they were not. Each "I don't know" rules out every world in which that person would have known, and everyone behind and in front can use it. The back person would know only in one world: two blue hats ahead, which forces his own to be red. So his silence removes that world, and the middle and front people both hear it.
Of the four hat pairs the back person might see on the middle and front people, his silence crosses out blue and blue, the middle person's silence then crosses out a red middle with a blue front, and both rows that survive have a red hat on the front person, which is why the front person knows the answer without seeing anything.
How does the middle person's silence finish it?
The middle person now knows that he and the front person are not both blue. He looks at the front hat. If it is blue, he cannot be blue too, so he is red, and he would say so. He does not. The middle person's silence can only mean he sees a red hat in front, because a blue one would have told him his own colour. The front person runs the same reasoning, does not need to see anything, and says red. The full check enumerates the seven possible hat triples, removes the ones where the back person would know, then the ones where the middle person would, and every survivor has red in front.
The relationship
$\{\text{back silent}\} \Rightarrow \neg(M = B \wedge F = B), \qquad \{\text{middle silent}\} \Rightarrow \neg(F = B) \;\Rightarrow\; F = R$
M, F the middle and front hats
B, R blue and red
the arrow what each silence lets everyone conclude
What it says in wordsThe first silence removes the case of two blues ahead, and the second removes a blue in front, which leaves only red.
Then say what it rests on, because that is the interview point. The puzzle needs common knowledge: everyone knows the hat counts, everyone reasons perfectly, and everyone knows the others do too. If the middle person might simply be slow, his silence carries no information and the front person learns nothing. On a trading floor the same logic runs in the other direction: a counterparty who could have traded and chose not to has told you something, and reading those non-events is part of the job.
Where candidates lose it
The common loss is saying the front person cannot know anything because they see nothing. That ignores that the two silences are data. The question is built to see whether you treat a non-answer as information.
The second loss is running the logic from the front. Start with the person who has the most information, the back, ask what would have let him know, and strike that case. Then move forward one person at a time.
What the interviewer asks next
Suppose the back person says "I know". What can the other two conclude?
With 2 red and 3 blue hats, does the same chain of silences tell the front person anything?
If only the back person speaks and says "I don't know", what can the middle person conclude about his own hat when he sees red in front?
Asked at Wolverine Trading, Prop Trading, Chicago, IL, USA, 2019 (Wall Street Oasis): a brain teaser about the hat problem where 3 people go into a room with I think 3 red and 2 blue hats

mu	the expected return, 10%
sigma	the volatility of the yearly return, taken as 20 points
Phi, phi	the normal distribution and density functions

mod 100	keep only the last two digits
76^2 = 5776	76 squared ends in 76, so every power of 76 ends in 76

67m x 1.1%	births a year, about 740,000
2.5	years a child spends in nappies
5.5	nappies per child per day, more for newborns and fewer for toddlers

1/4	the prior on each door
1/2	the chance the host opens door 4 when the prize is behind door 2
1/3	the chance he opens door 4 when the prize is behind your door

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Why does the ending change the odds of the start?

How do you set up the Bayes calculation in thirty seconds?

Where candidates lose it

What the interviewer asks next

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How do you make the two schedules equal at the mean?

Why is the performance fee an option, and who has written it?

Where candidates lose it

What the interviewer asks next

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Why do only the last two digits matter at each step?

How do you reduce 3000 without miscounting the start of the cycle?

Where candidates lose it

What the interviewer asks next

Try it first

How do you turn a guess into a market?

How wide should the market be?

Where candidates lose it

What the interviewer asks next

Try it first

Why do only two seats matter?

Why are the two endings equally likely?

Where candidates lose it

What the interviewer asks next

Try it first

Why does your own door keep its 1/4?

How does Bayes give 3/8 for each of the other doors?

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What the interviewer asks next

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Which side of each quote do you deal on?

How do you read off the maximum profit, loss and break-even?

Where candidates lose it

What the interviewer asks next

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How do you copy the call?

Where did the 70% go?

Where candidates lose it

What the interviewer asks next

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Why is the interval so wide after 30 years?

What does the interviewer do with the answer?

Where candidates lose it

What the interviewer asks next

Try it first

What does a silence tell everyone else?

How does the middle person's silence finish it?

Where candidates lose it

What the interviewer asks next