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073

Case 073Market-making gamesCore

You are making a market on the number of sixes in twelve dice. After you quote, six of the dice are revealed and exactly one of them is a six. Reprice, and explain what happens to your width even though the fair value does not move.

1The situation

At Quillmere Trading's superday the interviewer puts twelve dice in a cup and asks you to make a two-way market on the number of sixes, each point worth Rs 100 a lot. You quote 1.5 bid, 2.5 offered. Nobody trades.

The interviewer tips six of the dice out onto the table. Exactly one of them shows a six. The other six dice stay in the cup. You are asked for a new market, and then why it should differ from the first if the fair value is the same.

2Your task

Compute the fair value and the spread of outcomes before and after the reveal, quote a new market, and explain how the width should respond to information that does not move the mean.

Quick check

One six among the six revealed dice: where does fair value go?

Worked solution

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

30-second answerThe answer to give first

Fair value stays at 2, but the standard deviation falls from 1.29 to 0.91, so the market should tighten from 1.5 / 2.5 to about 1.65 / 2.35. Before the reveal the count was binomial on twelve dice; after it, one six plus a binomial on six. The reveal landed exactly on its expectation, so the mean did not move, but half the dice are now certain and the variance halved. A market's width prices risk, not just the mean, so it narrows.

Step 1What was the market worth before anything was shown?

Twelve dice, each a six with probability one sixth: the count is binomial with mean 12/6 = 2 and variance 12 x 1/6 x 5/6 = 1.67, a standard deviation of 1.29. A 1.5 / 2.5 market is centred on fair value with half a point of edge each side, about 0.39 of a standard deviation, which is the usual shape for a first quote in a game where nobody yet knows more than you. The distribution is wide: there is a 11% chance of no sixes at all and a 13% chance of four or more, so a lot sold at 2.5 could lose Rs 350 if five sixes turn up.

Step 2What does the reveal change, and what does it not?

Think of a cricket chase where the first half of the innings goes exactly to the required rate: the projected total is unchanged, but the range of likely outcomes is much narrower than it was before a ball was bowled. Six dice shown with one six is exactly what was expected of them, so the fair value stays at 1 + 6/6 = 2; but the count is now 1 plus a binomial on six dice, with variance 6 x 5/36 = 0.83 and standard deviation 0.91, half the variance of before. The chance of an extreme outcome collapses: zero sixes is now impossible, and four or more has a 6% chance against 13%. Had the reveal shown no sixes, fair value would have dropped to 1; two sixes would have lifted it to 3. This reveal happened to move the mean by nothing and the risk by a lot.

Same fair value of 2, half the variance: the market should tighten, not moveBefore the reveal: 12 dice unknown11%027%130%220%39%43%51%6+mean 2, sd 1.29market: 1.5 bid, 2.5 offeredAfter: one six in the six shown, 6 dice unknown0%033%140%220%35%41%50%6+mean 2, sd 0.91market: 1.65 bid, 2.35 offeredNumber of sixes among the twelve dice. Vertical: probability
Before the reveal the number of sixes is spread around 2 with a standard deviation of 1.29, and after six dice show exactly one six it is still centred on 2 but with a standard deviation of 0.91, so the same 2 is a much safer number to quote around.
SixesBefore, 12 unknownAfter, 1 seen + 6 unknown
011.2%0.0%
126.9%33.5%
229.6%40.2%
319.7%20.1%
48.9%5.4%
52.8%0.8%
6 or more0.8%0.1%
The reveal leaves the mean at 2 while moving probability toward it: the chance of exactly 1, 2 or 3 sixes rises from 76% to 94%, and the chance of zero sixes falls from 11% to nothing.
Step 3Why should the width change when the mean does not?

A market maker's width pays for two things: the risk of holding the position until it settles, and the chance that the other side knows more. In this game nobody knows more, so the width is about risk alone, and risk has halved: the expected size of the miss from fair value falls from 0.99 points to 0.67. A width that scales with the standard deviation goes from 1.0 to about 0.71, so quote something like 1.65 bid, 2.35 offered, or 1.75 / 2.25 in the quarter-point steps of the game. Keeping the old 1.5 / 2.5 would be leaving half a point of unnecessary edge on the table, and in a game with other players someone will quote inside you and take the flow. The reverse holds too: if the interviewer had revealed nothing but added six more dice to the cup, fair value would rise and the market should widen with the extra variance.

Say the limitation. Scaling the width with the standard deviation is a rule of thumb; a real desk also prices inventory, the size the counterparty wants and how long the position lasts. The principle survives: price the mean from what you know and what you expect, and price the width from what you do not know.

Where candidates lose it

Candidates often move the fair value, up because a six appeared or down because five dice were not sixes. The revealed dice did exactly what was expected, so the mean is unchanged; the test is whether the candidate separates the mean from the uncertainty.

The second miss is re-quoting the same width. Information that halves the variance should halve the risk premium in the quote, and a candidate who does not tighten is showing that the width was a habit rather than a price.

What the interviewer asks next

  • The reveal shows three sixes among the six dice. Quote your new market.
  • The interviewer now offers to buy 20 lots at your new offer. Does the size change your quote?
  • Suppose the dice were loaded so that a six came up with unknown probability. How would a reveal with one six move your fair value then?
← Case 072A stock will join a benchmark tracked by Rs 20,000 crore of passive money at a 0.8% weight, and it trades Rs 60 crore a day. Estimate the passive demand in days of volume and the likely price pressure around the effective date.Case 074 →A book has a mean daily P&L of Rs 2 lakh with a daily standard deviation of Rs 10 lakh. Give 95% intervals for tomorrow's P&L, for the average daily P&L over the next 250 days, and for the year's total.

Company names and figures are illustrative.

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