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088

Case 088Market-making gamesCore

The interviewer asks for a market on the maximum of four fair dice, then lifts your offer twice in a row at 5.4. What is the contract worth, how wide should the market be given its skewed distribution, and how do you requote?

OptiverSan Francisco · 2026

1The situation

The Varunika Markets interviewer rolls four fair dice behind a screen, looks at them, and asks you for a two-way market on a contract that settles at the highest of the four. You may trade one contract at a time, and the interviewer may buy, sell or pass.

You open at 4.9 bid, 5.4 offer. The interviewer buys at 5.4. You requote, and the interviewer buys again at your offer. You are now short two contracts.

2Your task

Price the contract, explain how its skewed distribution should shape your width, and decide how to requote after being lifted twice.

Quick check

What is the expected maximum of four fair dice?

Worked solution

Try it on paper, then open one step at a time.

30-second answerThe answer to give first

Fair value is 5.24, but half the probability sits on 6, so the risk is lopsided: an offer can lose at most the gap to 6, a bid can lose almost 4. Open at 4.9 bid, 5.4 offer, the bid further from fair than the offer. Two lifts from someone who has seen the dice say the max is probably 6: if half your counterparties are informed, the chance of 6 rises to 91% and fair value to 5.85. Requote 5.7 at 6.0.

Step 1What is the contract worth?

The trick is to count the complement. The maximum is at most k only if all four dice are at most k, which has probability (k/6) to the fourth. So P(max = k) is (k/6)^4 minus ((k-1)/6)^4, which puts 51.8% on 6, 28.5% on 5 and only 6.2% on 3 or below. Weighting gives an expected value of 5.245, with a standard deviation of 0.95. Note that the median and the most likely value are both 6, while the mean is pulled down to 5.24 by the long tail of low outcomes.

The relationship
P(max⁡=k)=(k6)4−(k−16)4E[max⁡]=∑k=16k P(max⁡=k)=67971296≈5.24P(\max = k) = \Big(\frac{k}{6}\Big)^4 - \Big(\frac{k-1}{6}\Big)^4 \qquad E[\max] = \sum_{k=1}^{6} k\,P(\max = k) = \frac{6797}{1296} \approx 5.24
ka possible value of the maximum, 1 to 6
(k/6)^4chance all four dice are at most k
12966 to the fourth, the number of equally likely rolls
What it says in wordsThe chance the highest die equals k is the chance all four are at most k minus the chance all four are at most k minus 1; weighting by k gives an average of about 5.24.
Max of four dice: heavily stacked on 5 and 6, with a long left tail20%40%0.1%11.2%25.0%313.5%428.5%551.8%6fair 5.244.95.4left tail: buying at 4.9 can lose up to 3.9selling at 5.4 can lose at most 0.6Maximum of four fair dice
The maximum of four dice lands on 6 about 52% of the time and on 5 about 28%, so its fair value of 5.24 sits close to the cap of 6 with a long tail below, and the opening market of 4.9 at 5.4 leans to that shape.
Step 2How should the skew shape the width?

Think of the two sides separately. Selling at 5.4 can lose at most 0.6, because the contract cannot settle above 6; buying at 4.9 can lose up to 3.9 if all four dice come up 1. Against a counterparty who has seen the dice, the side with the bigger worst case needs the bigger cushion, so the bid sits 0.34 below fair and the offer only 0.16 above. A symmetric market of 5.0 at 5.5 would look tidy and give a seller who saw three low dice a far better trade than a buyer who saw a six. The width of 0.5 is about half a standard deviation, tight enough that an uninformed player will trade.

Step 3What do two lifts at 5.4 tell you?

Someone who has seen the dice buys at 5.4 only if the max is 6, since 5 or less settles below 5.4. Model the interviewer as informed half the time and trading at random otherwise. A lift then happens 75% of the time when the max is 6 and 25% of the time when it is not, so each lift triples the odds on a six. The prior odds are 0.518 to 0.482. After one lift the chance of a six is 76% and fair value is 5.63; after two it is 91% and 5.85, using 4.43 as the average maximum when it is below 6.

Two lifts at 5.4 say the max is probably 6OpeningP(6) = 52%4.95.4fair 5.24After lift 1P(6) = 76%5.55.8fair 5.63After lift 2P(6) = 91%5.76.0fair 5.856 is the cap: an offer at 6.0 cannot lose4.65.05.45.86.2
Each lift at 5.4 triples the odds that the hidden maximum is 6, moving Varunika's fair value from 5.24 to 5.63 and then 5.85, so the market moves up to 5.5 at 5.8 and then 5.7 at 6.0.

So requote 5.5 at 5.8 after the first lift and 5.7 at 6.0 after the second. An offer at 6.0 cannot lose, because the contract can never settle above it, which makes it the natural place to stand when the evidence points to a six. You are short two at 5.4 and marked at about 5.85, down about 0.9: the price of having learned. The limitation is the model: the 50% informed share is a guess, and an interviewer who lifts to test your nerve would make the update smaller. Say the assumption out loud; the interviewer is grading how you update, not the decimal.

Where candidates lose it

The common error is pricing the max at 6 because 'some die will be a six', or at 4.5 from intuition. The complement trick gives 5.24 in a few seconds and the interviewer expects to see it.

The second is treating the two lifts as inventory to lean against and lowering the offer to get flat. The counterparty can see the dice; one-way buying at 5.4 is the clearest signal the game offers, and the market should move up, not down.

What the interviewer asks next

  • What is the expected minimum of four dice, and how would the width change?
  • The interviewer now sells to your bid of 5.7. What do you conclude?
  • How would your opening market change for the maximum of ten dice?
  • What price would you pay to see one of the four dice before quoting?

Asked at Optiver, Quant Research Interview, San Francisco, 2026 (Wall Street Oasis): They do ask one round of market making game-like question.

← Case 087Aviratam holds two positions of Rs 50 crore each, with annual volatilities of 20% and 25%. What are the portfolio volatility and one-day 99% VaR at the modelled correlation of -0.3, and at a crisis correlation of +0.8?Case 089 →Parvanta asks you to design the train, validation and test split for eight years of daily data with 20-day forward-return labels. How many days must be purged and embargoed around each boundary, and how many walk-forward folds with one-year test windows remain?

Company names and figures are illustrative.

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