Derivatives Foundation puzzles, solved step by step

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

Showing 1–4 of 4 · filtered from 100Clear filters

009Three players each hold one hidden card from a standard deck, ace counting 1 up to king counting 13. You can see only your own card, a 10. You must make a two-way market on the total of all three cards. The player on your left lifts your offer, you requote, and he lifts again. What do you do now?Market makingHardOptiverChicago · 2025

Try it first

Before the second lift: what is a fair value for the total when all you know is your own 10?

Show the worked solution

Raise the market and widen it; do not sell a third time near the old level. With only your 10 known, the fair total is 24 and the two unknowns give a standard deviation of about 5.3. A player who sees his own card and lifts your offer is telling you his card is high: if it is 8 or more the fair total is 27.5; after a second lift, 11 or more, it is 29. You are short 2 at an average near 27 against a value near 29. Quote something like 28 at 32.

What does a lift tell you that the cards do not?

If you are selling a second-hand bike and the first viewer pays your asking price without haggling, you have probably priced it low; if he immediately asks to buy a second one, you certainly have. A counterparty who can see something you cannot and keeps buying is telling you the value is higher than your offer, and every trade you do with him before repricing is a trade you will regret. This is adverse selection, and a market-making game exists to see whether you notice it in time.

With your 10 the prior fair total is 24; one lift suggests the buyer's card is 8 or more and moves the fair value to 27.5; a second lift suggests 11 or more and moves it to 29, so the quote climbs from 23 at 25 to 28 at 32 and widens from 2 to 4 while you sit short 2.

The relationship

E[\text{total}] = 10 + E[L] + 7,\qquad E[L \mid L \ge 8] = 10.5,\quad E[L \mid L \ge 11] = 12 \;\Rightarrow\; 24 \to 27.5 \to 29

L	the card held by the player who keeps lifting
7	the mean of the third player's card, still unknown and uniform on 1 to 13
L >= 8, L >= 11	the rough information in a first and a second lift: his card is above what your offer implied

What it says in wordsEach lift raises your estimate of the lifter's card, and the fair total moves by exactly that amount.

How much do you move, and how much do you widen?

Move at least as far as the information says and widen because your uncertainty about his behaviour has grown. After one lift a quote of 26 at 29 sits around the new estimate of 27.5; after two lifts 28 at 32 sits around 29 and is twice as wide, because the next lift would mean his card is 12 or 13 and the total near 30 or more. The width is not a penalty on him; it is the price of your own blindness. A standard deviation of 5.3 on the total from the two unknown cards is the natural scale for the width before any lifts.

What about the position you already have?

You are short 2 at an average near 27 and the value is near 29, so you are losing about 4 on paper. Do not try to earn it back by selling more at a worse price, and do not flip to buying from the other player at any cost; skew your quote up so the next trade is more likely to reduce the short than add to it. Say your position out loud when the interviewer asks; the game checks whether you can hold the fair value, the quote and the inventory in your head at once. The limitation to state: the thresholds 8 and 11 are a rough model of his behaviour, and a player who bluffs changes the inference.

Where candidates lose it

The common failure is to keep quoting 23 at 25 after the first lift, and to sell a third unit at 25 after the second, because the cards have not changed. Your information has changed. Two lifts from a player who sees his own card are worth more than the deck statistics.

The second loss is overreacting: moving the quote to 35 at 40 after one lift. He may hold a 9. Move to what the evidence supports, widen for what it does not, and keep your position in mind.

What the interviewer asks next

Now the third player hits your bid at 28. How do you reprice?
What if the lifter can see your card as well as his own?
Make a market on the product of the three cards instead of the sum. What changes about the width?

Asked at Optiver, Quantitative Research, Chicago, 2025 (Wall Street Oasis): The next round was a poker style market making game as well as a separate behavioural interview

024I roll a fair die, then flip as many fair coins as the die shows. Make me a market on the number of heads.Market makingHardOptiverAustin · 2025
Try it first
Before the variance: what is the fair value, the centre of your market?
Show the worked solution
Centre on 1.75 and quote something like 1.6 bid, 1.9 offered. The die averages 3.5 coins and each coin gives half a head, so the mean is 1.75. The spread comes from two sources: the coin flips, E[N]/4 = 0.875, and the uncertain number of coins, Var(N)/4 = 0.729, a variance of 1.604 and a standard deviation of 1.27. The fair value is exact, so the quote can be tight; the spread tells you how hard to size it.
How do you get the centre?
Suppose a shop's daily customers vary and each spends Rs 200 on average. Average takings are average customers times Rs 200, whatever the day-to-day mix. When a random number of random things are added up, the mean is the expected count times the expected size of each, so here 3.5 coins times half a head = 1.75. That is the law of total expectation in one line. Note that 1.75 is not a possible outcome, and the single most likely outcome is one head, with a chance of 0.312; a market is centred on the mean because that is where neither side has an edge.
The number of heads has mean 1.75 and most of its mass on one and two heads, and its variance of 1.604 splits into 0.875 from the coin flips and 0.729 from not knowing how many coins the die will give, so the standard deviation is 1.27.
Why is the spread bigger than the coins alone suggest?
Because you are uncertain about two things at once: how many coins, and how they land. The variance of a random sum is the average of the inner variance plus the variance of the inner mean: E[N] x 1/4 from the coins, plus Var(N) x 1/4 from the die, 0.875 + 0.729 = 1.604. If you knew the die would show 3.5 coins and ignored its own wobble, you would quote a standard deviation of 0.94 instead of 1.27 and size too large. The die's contribution is almost half the total, which is the step most candidates leave out.
The relationship
$E[H] = \frac{E[N]}{2} = 1.75,\qquad \text{Var}(H) = \frac{E[N]}{4} + \frac{\text{Var}(N)}{4} = \frac{3.5}{4} + \frac{35/12}{4} = 1.604$
H the number of heads
N the number of coins, the die roll, with mean 3.5 and variance 35/12
1/4 the variance of one fair coin, and the square of its half-head mean
What it says in wordsThe mean is half the expected number of coins, and the variance adds the coin noise to the noise in the coin count.
How tight should the market be, and what if the other side saw the die?
Width pays you for two risks: not knowing the fair value, and trading with someone who knows more. Here the fair value is exact and nobody has seen anything, so a tight quote around 1.75, such as 1.6 at 1.9, is right, and the standard deviation of 1.27 governs how many contracts you take, not where you centre. Change one fact and the answer changes: if the counterparty has seen the die, a buyer is telling you the die was high. A die of 6 implies 3 heads on average, and a die of 1 only 0.5, so widen sharply or ask to see the die before quoting. The limitation is that real games price in that information risk from the first quote.
Where candidates lose it
The common loss is centring the market on two, the most likely outcome of the coins, or on 1.5 from a guessed three coins. A market is centred on the expected value, and the expected value takes the die's average into account exactly.
The second loss is computing the spread as if the number of coins were fixed at 3.5. That leaves out the variance of the die, almost half the total, and makes you size the position as if it were safer than it is.
What the interviewer asks next
I buy 5 from you at your offer. Where is your market now, and does it matter whether I saw the die?
What is the probability of zero heads?
Now the die decides the number of coins, and each head pays the die's value. What is the expected payout?
Asked at Optiver, Quantitative Research, Austin, 2025 (Wall Street Oasis): Technical (Simulated EV Poker like game, with cards, coins and dice; Market Making and Taking)
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
097I pick a whole number from 1 to 100 and know it; you do not. You quote 45 at 55 and I buy at 55. You requote 58 at 68 and I buy again at 68. What have you learned, and where should your next quote be?Market makingHardJane StreetNew York · 2025
Try it first
After the first lift at 55, where is the middle of what the number could still be?
Show the worked solution
The number is between 69 and 100, so centre the next quote on 84.5, say 80 at 89. Someone who knows the number buys only when your offer is too low. The lift at 55 put it in 56 to 100, midpoint 78, and your requote at 58 to 68 was far too low, so it was lifted too. Now it is in 69 to 100. You are short two at 55 and 68, worth about -46 against a fair value of 84.5, and the next quote should sit on what is left, not drift up from the last one.
What does a trade tell you when the other side knows the answer?
If a friend who has already peeked at the exam paper offers to bet you that the pass mark is above 55, you do not need to ask why. A counterparty who knows the number only trades when your price is wrong in their favour, so every lift of your offer is proof that the number is above it, and your fair value should jump to the middle of what is left. After the lift at 55 the number lies in 56 to 100, centred on 78. The requote at 58 to 68 treated the lift as noise and moved only 13 points. A quote centred near 78, say 73 at 83, would have used the information.
The first lift at 55 leaves 56 to 100 with a midpoint of 78, the requote of 58 at 68 sits 15 below that midpoint and is lifted again, which leaves 69 to 100 with a midpoint of 84.5, so the next quote of 80 at 89 is centred on what can still be true.
Where should the next quote sit, and what have the trades cost?
After the second lift the number is in 69 to 100: 32 values, mean 84.5. Centre the next quote on the middle of the remaining range, 84.5, so that whichever side is hit, the range halves, which is the fastest way to stop losing to someone who knows more than you. A quote of 80 at 89 does that. The damage so far: you sold at 55 and 68 something now worth about 84.5, an expected loss of 29.5 + 16.5 = 46. Most of it came from the timid requote; a quote centred on 78 would have been lifted only if the number was above 83.
The relationship
$E[X \mid X > 55] = \frac{56 + 100}{2} = 78, \qquad E[X \mid X > 68] = \frac{69 + 100}{2} = 84.5$
X the number I picked, uniform on 1 to 100 before any trade
X > 55 what my buying at 55 reveals
84.5 the fair value after both lifts, where the next quote should be centred
What it says in wordsEach lift cuts the range to everything above your offer, and fair value moves to the middle of what is left.
Then say how the answer changes with the counterparty, because that is the judgement being marked. If I might be guessing rather than knowing, a lift is weaker evidence and you should move less, perhaps halfway towards 78. If I know the number exactly, any tight quote inside the range loses whenever it trades, and the only quote that cannot lose is one that spans the whole range, 69 bid, 100 offered, which is no market at all. Real desks live between the two: they widen against flow they think is informed and tighten against flow they think is not.
Where candidates lose it
The common loss is moving the quote up a few points after each lift, as the 58 at 68 requote did. It treats an informed trade as a random one and pays for the lesson again on the next trade.
The second loss is the opposite overreaction: refusing to quote, or quoting 1 at 100. The interviewer wants a market that uses the information and still trades, centred on the conditional mean of what is left.
What the interviewer asks next
Instead I sell to you at 45 on the first quote. What is your next quote?
Suppose I know the number only half the time and guess otherwise. How far should the first lift move you?
How many lifts or hits does it take, at most, to find the number if you always quote around the middle of the range?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): It was a probability theory based quant trading style market making questions which were intense

H	the number of heads
N	the number of coins, the die roll, with mean 3.5 and variance 35/12
1/4	the variance of one fair coin, and the square of its half-head mean

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

Try it first

What does a lift tell you that the cards do not?

How much do you move, and how much do you widen?

What about the position you already have?

Where candidates lose it

What the interviewer asks next

Try it first

How do you get the centre?

Why is the spread bigger than the coins alone suggest?

How tight should the market be, and what if the other side saw the die?

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 a trade tell you when the other side knows the answer?

Where should the next quote sit, and what have the trades cost?

Where candidates lose it

What the interviewer asks next