Quant puzzles, solved step by step

91/216 of a rupee, about 42 paise. A fair price for a ticket paying Rs 1 is the probability of winning. The fastest route is the complement: the chance of no six on three dice is 5/6 x 5/6 x 5/6 = 125/216, so the chance of at least one six is 1 - 125/216 = 91/216, or 0.421. Adding 1/6 three times gives 50 paise and overcounts.

Why is the price just a probability?

If a raffle pays Rs 100 and you win one time in four, playing many times earns you Rs 25 a ticket on average, so Rs 25 is the break-even price. A ticket paying Rs 1 on some event is worth exactly the probability of that event, because that is its average payout. Trading firms phrase probability questions as prices on purpose: it makes you answer in the units a desk uses, and it sets up the next question, which is where you would quote a bid and an offer.

Why is at least one a signal to use the complement?

At least one six covers exactly one six, exactly two, or three, and each needs its own count. The opposite event, no six at all, is a single clean case: every die avoids six, and independent dice multiply. So the complement takes one line. Adding 1/6 + 1/6 + 1/6 fails because the three events overlap: a roll of 6, 6, 2 is counted once for the first die and again for the second. With ten dice the same mistake would give a probability above 1.

What does a trader add after the number?

A fair value is the centre of a market, not the market itself. A market maker quotes a bid below 42 paise and an offer above it, and the width depends on how confident they are in the number and how much risk one ticket adds to their book. Here the fair value is exact, so a tight market such as 40 bid, 44 offer is defensible. Saying that sentence turns a probability answer into a trading answer, which is what the question format is inviting.

Centre it at 50 and quote around 46 at 54. The fair value is exactly 50. The standard deviation is √(100 x 0.5 x 0.5) = 5, so settlement lands between 45 and 55 about 73% of the time. A market 4 either side of fair earns 4 per lot on any trade, loses on a single sale at 54 only 18% of the time, and leaves room to move the quote if the other side seems to know something.

Where does the centre come from, and what sets the width?

A shopkeeper selling mangoes by the dozen knows the fair price; the margin he adds depends on how much the price of the next crate can swing and on whether the buyer knows something he does not. The centre of your market is the expected value, and the width is a choice about risk and information, scaled by how much the outcome can move. Here the expected value is 100 x 0.5 = 50, and nobody can know more than you about fresh flips of a fair coin, so the width is about risk alone.

How do you justify 46 at 54 rather than 49 at 51?

Use the standard deviation as the ruler. The count has variance 100 x 0.5 x 0.5 = 25, so a standard deviation of 5. A quote 4 either side of fair earns 4 on each lot traded, against a settlement that typically moves 5, so every trade has an edge worth a large fraction of its risk. If someone buys at 54, you lose only if the count finishes at 55 or more, about 18% of the time. A tight 49 at 51 earns 1 per lot and a sale at 51 loses whenever the count reaches 52, about 38% of the time. Tighter wins more trades and earns less on each; in an interview game, start around one standard deviation wide and tighten as you learn.

Then say how you would react to trades, because that is the follow-up. If the interviewer lifts your 54 again and again, either they are testing your nerve or they know something, perhaps that the coin is not fair or that some flips are already done. Repeated one-way trading is information: move your market toward it and cut your size, rather than defending 50. The limitation of the simple answer is exactly that it assumes nobody knows more than you.