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.

Rs 2. Without information the bet is worth 0.5 x 10 - 6 = -1, so you decline and your value is 0. With the colour known, you play on red and make 4, and skip black and make 0, which averages 2. Information is worth the improvement in your best decision: 2 - 0 = 2. If it would not change what you do, it is worth nothing.

How do you value a piece of information?

Suppose a weather forecast costs money and you are deciding whether to carry an umbrella. If you would carry it anyway, the forecast is worthless to you; it is valuable only if some answer would change what you do. The value of information is the expected value of your best decision with it, minus the expected value of your best decision without it. Work out both decision trees separately and subtract. Never value information by the size of the payout it relates to.

Why is it not worth Rs 4 or Rs 5?

Rs 4 is what you make when the card is red, but it is red only half the time. Rs 5 is half the payout, which ignores the Rs 6 you pay to play. The information saves you from the losing half of the bet and lets you keep the winning half, and that is worth half of Rs 4, which is Rs 2. Pay more than Rs 2 and you would do better declining the offer of information and declining the bet.

When is information worth the most?

Vary the price of the bet. At a price of 5 you are exactly indifferent blind, and the information is worth 2.50, its maximum; at a price of 0 you would always play, and it is worth 0. Information is valuable when you are close to indifferent and the decision could go either way. The formula also has the shape of an option payoff: knowing first lets you exercise only when it pays, which is why traders talk about paying for optionality and paying for information in the same breath.

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.

At least one six in four rolls is more likely, 51.8% against 49.1%, so take that side. The chance of no six in four rolls is (5/6)^4 = 625/1296, and the chance of no double six in 24 rolls is (35/36)^24. At even money the six earns about 3.5 paise per rupee staked; the double six loses about 1.7 paise and needs 25 rolls to become favourable.

Why does the obvious 2/3 come out the same for both bets?

The old gamblers' shortcut, sometimes called the rule of proportion, says that if one try succeeds with chance p, then n tries succeed with chance n times p. Four rolls at 1/6 gives 4/6, and 24 rolls at 1/36 gives 24/36, the same 2/3. This puzzle is the one the Chevalier de Mere is said to have brought to Pascal, because his winnings disagreed with the shortcut. Adding chances only works for events that cannot happen together, and two sixes in four rolls can. The shortcut counts a hand with two sixes twice, and that is why it breaks down completely after seven rolls, where it would claim 7/6.

A family checking the weather forecast makes the same slip: three days at a 40% chance of rain do not make a 120% chance of a wet weekend. The right question is the chance that it stays dry on all three days, 0.6 cubed, about 22%, so rain on at least one day is about 78%.

How do you compute each chance in your head?

Go through the complement. At least one success is one minus the chance that every try fails, and independent failures multiply. No six in four rolls is (5/6)^4 = 625/1296, which is just under one half, so the six wins 671/1296, or 0.5177. For the double six, use the approximation that (1 - 1/36)^24 is close to e to the power -24/36, which is e^(-2/3), about 0.513; the exact figure is 0.5086, so the double six comes up only 0.4914 of the time.

What edge does each side carry at even money?

At even money you win a rupee with probability p and lose a rupee otherwise, so the expected gain per rupee is p - (1 - p) = 2p - 1. The six gives 2 x 0.5177 - 1 = +0.0355, about 3.5 paise per rupee, while the double six gives -0.0172. That is a thin edge: over 100 bets of Rs 1 the expected profit is about Rs 3.55 with a standard deviation near Rs 10, so it takes thousands of bets before the edge shows reliably. The break-even for the double six is the smallest n with (35/36)^n below one half, which is 25 rolls, giving 0.5055. The limitation: the dice must be fair and independent; any bias in the dice swamps an edge this size.

Stake 10% of the bankroll; that grows wealth by about 0.97% a bet, and twice Kelly grows it by only about 0.07%. Kelly is edge over odds: (2 x 0.4 - 0.6)/2 = 0.1. The growth rate is 0.4 ln(1.2) + 0.6 ln(0.9). At 20% the losses compound away almost the whole edge, and above about 20.4% the bankroll shrinks in the long run despite a positive expected value.

Why not bet as much as possible on a good bet?

Think of a shopkeeper with a profitable weekly sale who puts the entire shop's stock on it every week. The average week is good, but one bad week ends the business. With repeated bets, wealth multiplies, so what matters is the average of the log of each outcome, not the average outcome, and a big loss costs more in log terms than an equal gain earns. This bet has a clear edge: each rupee staked returns 0.4 x 2 - 0.6 = Rs 0.20 on average. The question is how much of that edge survives compounding at each stake size.

How do you get the Kelly fraction and the growth rate?

Stake a fraction f. A win multiplies wealth by 1 + 2f, a loss by 1 - f, so the growth per bet is g(f) = 0.4 ln(1 + 2f) + 0.6 ln(1 - f). Set the derivative to zero: 0.8/(1 + 2f) = 0.6/(1 - f), giving f = 0.1. The Kelly stake is the edge divided by the odds, (bp - q)/b = 0.2/2 = 10%. Plugging in, g = 0.4 x 0.1823 - 0.6 x 0.1054, about 0.97% a bet, so the typical path doubles its wealth roughly every 71 bets.

Why is overbetting so much worse than underbetting?

Near the peak the growth curve is close to a parabola, so the cost of a sizing error grows with its square. Half Kelly gives up only about a quarter of the growth, while twice Kelly gives up nearly all of it, and three times Kelly shrinks wealth at about 2.6% a bet. Real edges are estimated, not known, so a trader who thinks the win rate is 40% but faces 35% is already overbetting at the full 10%. That asymmetry is why desks size at a fraction of Kelly.

Decline it: same colour comes up 25 times in 51, so the bet loses about Rs 1.96 per Rs 100. Whatever the first card is, 25 of the remaining 51 share its colour and 26 do not. The bet is fair only if a win pays Rs 26 for every Rs 25 risked, that is Rs 104 on a win against Rs 100 on a loss.

Why is same colour less likely than different colour?

Imagine a classroom with 26 girls and 26 boys. Pick one child, then pick a second: the second is slightly more likely to be of the other sex, because the first pick removed one of their own. Drawing without replacement makes the second card lean away from the first card's colour, since that colour is now short by one. The first card is irrelevant to the answer, whichever colour it is. Only what is left in the deck matters: 25 matching cards and 26 non-matching.

What is the bet worth, and what makes it fair?

Expected value is the win times its chance less the loss times its chance: 100 x 25/51 - 100 x 26/51 = -100/51, about Rs -1.96. A bet is fair when the payout ratio equals the odds against winning, here 26 to 25, so the win must pay Rs 104 for each Rs 100 at risk. Quoting it as odds rather than a probability is how a trader would answer: you would take the bet at 26 to 25 or better, and decline anything worse.

Why would an interviewer offer such a small edge?

To see whether you notice it at all, and then whether you size your response to it. An edge of about 2% per bet is small for one play but decisive over many, so the right answer is to decline at even money and quote the price at which you would play. It also tests whether you separate the first card, which is free, from the conditional chance of the second. With replacement, or with an infinite deck, the bet would be exactly fair; the whole edge comes from the deck being finite, and it shrinks as more decks are shuffled together.

Centre on today's level, about 20,000, with a 90% range of roughly 16,800 to 23,600. Four months is a third of a year, so one standard deviation is 18% x sqrt(1/3), about 10.4%. In log terms the 90% band is 1.645 of those either side, which gives 16,767 and 23,601. The median sits a little below 20,000 and the upside tail is longer than the downside.

Why is a single number the wrong answer?

Ask a cab driver how long the airport run takes and a good one says forty minutes, maybe an hour in traffic. The range is the useful part, because you plan your flight around it. A trading interviewer asking where an index closes wants a distribution, because a market maker quotes against the spread of outcomes, not against a guess. Your central value should not be a view on the economy either: with no edge, the best central estimate of a traded index is roughly its forward, which for four months is close to today's 20,000 once financing and dividends roughly offset.

The width comes from the implied volatility the market already quotes. Volatility grows with the square root of time, because independent daily moves add their variances, not their standard deviations. Four months is a third of a year, so one standard deviation is 18% x sqrt(1/3) = 10.4%, about 2,078 index points.

Why is the range lopsided, and where does the median sit?

A fall of 10% and a rise of 10% are not mirror images in log space. Normal log returns make the upside tail longer: the 90% band stretches 3,601 points up but only 3,233 points down. The same convexity pushes the median below the mean: if the mean is 20,000, the median is 20,000 x exp(-sigma squared T / 2), about 19,892. That gap of about 108 points is small here, but it grows with volatility and time, and a candidate who names it shows they know the difference between the most central outcome and the average one.

What would you add before quoting a market on it?

Two honest caveats. Implied volatility is a price, not a forecast: it tends to sit above the volatility that is later realised, because option sellers charge for bearing crash risk, so the band built from it is usually a little wide. Against that, real index returns have fatter tails than the lognormal, so the 5% tails are more likely to hold a larger move than the curve suggests. Say both, then give your market: a tight two-way price around 20,000 if asked for the level, and the 90% band as the range you would sell outside of.

About 9.4 million a day, and I would open at 8 bid, 11 offered, in millions. Assume about 700,000 births a year, 2.5 years in nappies, six changes a day and 90% disposable: 9.45 million. Multiplying the low and high ends of each input gives 5.5 to 15.0 million, so a quote of 8 at 11 is tight enough to trade and still honest about the uncertainty.

How do you build the estimate so the interviewer can follow it?

Chain it through things you can reason about. Children in nappies are roughly births a year times the years each child spends in them. Assume about 700,000 births a year, a round number worth checking against the latest official statistics, and 2.5 years in nappies: about 1.75 million children. Each child uses about six a day on average, more as a newborn and fewer as a toddler, and assume 90% of families use disposables: 1.75 million x 6 x 0.9 = 9.45 million a day. Say each assumption out loud and give it a range as you go.

Where should the width of the market come from?

From the ranges, multiplied. A shopkeeper who is unsure of both price and quantity is more unsure of revenue than of either. Put a low and a high on every input and multiply the lows together and the highs together: here 5.5 million to 15.0 million, about a factor of 1.6 either side of the centre. Centre the quote near the middle on a multiplicative scale, the geometric mean of the ends, 9.1 million, which sits close to the central estimate.

Then choose the width you will actually trade. A market as wide as the whole range, 5.5 at 15, is useless: nobody trades against it and it tells the interviewer you have no view. Quote tighter, 8 at 11, and move it as they trade: if they keep buying at 11, raise both sides, because their trades carry information. Say the scope questions too: does the count include adult incontinence products, and a school-age child in night-time pants? Those can move the answer more than any of the four inputs.

Set the ask at 106, and by the same logic the bid at 104. If the stock is worth 110, a buy arrives with probability 0.2 + 0.8 x 0.5 = 0.6; if it is worth 100, with probability 0.4. Given a buy, Bayes puts the chance of 110 at 0.6, so the stock is worth 106 to the market maker selling it. The spread of 2 is the price of trading against informed flow.

Why can't the market maker just quote the expected value of 105?

A second-hand car dealer who pays the average price for every car will find that the owners of good cars go elsewhere and the owners of bad ones queue up. Who chooses to trade with you is information. A market maker does not care what the stock is worth on average; it cares what the stock is worth given that someone has just chosen to buy from it. At an ask of 105, noise buyers are harmless, a loss of 5 when the stock is worth 110 and a gain of 5 when it is worth 100. Informed buyers only appear in the 110 world, and they cost 0.5 per arriving trader on average, so 105 is a losing quote.

What sets the width of the spread?

The share of informed traders and the size of what they know. With a share alpha informed, a buy is alpha + (1 - alpha)/2 likely in the high world and (1 - alpha)/2 in the low world, and the ask works out to 105 + 5 alpha. The spread is a fee for adverse selection: it is zero when nobody is informed and widens to the full 100 to 110 range when everyone is. The table runs the formula for a few shares. Order processing and inventory costs add to this in real markets, but the information component is what makes spreads jump around earnings and news.

What happens after the first trade?

The market maker updates. After one buy, the chance of 110 is 0.6, and if a second buy arrives the same Bayes step lifts it to 0.692, so the next ask is about 106.92. Each order moves the quotes towards the true value, which is how prices come to reflect what the informed traders know. The limitation is that the model has one share size, no inventory risk and no competition between market makers; real desks also skew quotes to manage position, which this puzzle leaves out.