Quant puzzles, solved step by step

10/12, about 0.833. A uniform prior on the win rate, updated with 9 wins and 1 loss, gives a Beta(10, 2) posterior. The chance of winning tomorrow is the posterior mean, (9 + 1)/(10 + 2). The answer is below 0.9 because ten days cannot rule out a lower true win rate, and averaging over that uncertainty pulls the estimate towards one half.

Why is 0.9 too confident?

A new restaurant with nine five-star reviews out of ten looks excellent, but you would not bet that the next diner rates it five stars with 90% certainty. Ten reviews is a small sample, and a restaurant that truly earns five stars 70% of the time could easily post nine out of ten. The right forecast averages over every win rate the evidence still allows, and with only ten days that includes plenty of rates below 0.9. The average of that spread of possibilities is what you should quote for tomorrow.

How does the Bayesian update give 10/12?

Start with every win rate p between 0 and 1 equally likely. The chance of the observed record is proportional to p^9 (1 - p), so the posterior is proportional to that, which is the Beta(10, 2) distribution. The chance of a win tomorrow is the average of p over the posterior, and the mean of a Beta(a, b) is a/(a + b), here 10/12. The shortcut is Laplace's rule of succession: add one imaginary win and one imaginary loss to the record, then divide.

When does the pull towards one half stop mattering?

When the record is long. At 90 wins out of 100 the rule gives 91/102, about 0.892, almost exactly the raw 0.9, because a hundred days of data swamp the one imaginary win and loss. The limitation is the prior itself. A uniform prior says a 99% win rate was as plausible as a 50% one before you saw any data, which no trader believes about a new strategy. With a sceptical prior centred near one half, ten days would pull the estimate even further below 0.9. The posterior also tells you more than one number: the chance of winning both of the next two days is (10 x 11)/(12 x 13), about 0.705, not 0.833 squared.

About 4 years at a Sharpe of 1, and about 16 years at 0.5. The t-statistic for a mean return is the Sharpe ratio times the square root of the number of years, so reaching t = 2 needs (2 / SR)^2 years. Halving the Sharpe ratio quadruples the evidence you need, and sampling daily instead of yearly does not shorten it.

Why does the t-statistic come out as Sharpe times root years?

A t-test on a mean divides the average return by its standard error, which is the volatility over the square root of the number of observations. With annual observations that ratio is (mean / volatility) x sqrt(years), and mean over volatility is exactly the annual Sharpe ratio. So a Sharpe of 1 gives t = sqrt(years): 2 after 4 years. A Sharpe of 0.5 gives t = 0.5 x sqrt(years): 2 only after 16 years. With the textbook 1.96 in place of 2 the numbers are 3.8 and 15.4 years; the round figures are what you say in the room.

Can you shortcut it with daily data?

This is the follow-up that separates candidates. Sampling daily gives about 252 times as many observations a year, but the daily Sharpe ratio is smaller by the square root of 252, because daily mean scales with time and daily volatility with its square root. The two effects cancel exactly, so the t-statistic depends on calendar time, not on how finely you slice it. Think of estimating a river's average level: measuring every minute instead of every day does not help if the river's slow swings are the uncertainty.

What makes the real requirement even longer?

Three things, each worth one sentence. Returns are not independent from year to year, and positive autocorrelation inflates the true standard error. If you tested twenty strategies and kept the best, a t of 2 is easy to get by luck, so the bar has to rise with the number of ideas tried. And the Sharpe ratio itself drifts as markets change, so a sixteen-year record may be measuring two different strategies. A desk that says a Sharpe 0.5 strategy is proven after three years is reading noise.

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.

1/sqrt(2), about 0.71. The covariance of X with X + Y is Var X plus Cov(X, Y), and the second term is zero, so it is sigma squared. The variance of X + Y is 2 sigma squared, because independent variances add. Dividing sigma squared by sigma times sqrt(2) sigma leaves 1/sqrt(2). Squared, that is 0.5: X explains half of the sum's variance.

Why isn't the answer one half?

Picture two people each tossing a coin for a rupee, and a pot holding their combined winnings. One player's result explains exactly half of the pot's variability, and the other half comes from the other player. Half is the share of variance explained, R squared, and correlation is its square root, so the correlation is 1/sqrt(2), not 1/2. This is the most common slip on the question, and it comes from mixing up the two measures.

How does it change if the variances differ?

Let Var Y be k times Var X. The covariance is still Var X, and Var(X + Y) becomes (1 + k) Var X. The correlation is 1/sqrt(1 + k): the noisier Y is, the less the sum tracks X. At k = 1 you get 0.707; at k = 4, 0.447; at k = 0.25, 0.894. This is exactly the signal-plus-noise model: if a price move is a true signal plus independent noise of equal size, the best-case correlation between your signal and the move is about 0.71.

Where does this show up on a desk?

Any time one piece is part of a total. A stock's return is market return plus its own specific return; if the two had equal variance, the stock would correlate 0.71 with the market. The same arithmetic tells you the ceiling on a predictor: if half of tomorrow's move is unpredictable noise, no model can correlate more than 0.71 with it. The limitation is independence; if X and Y are correlated, add 2 Cov(X, Y) to the variance of the sum and Cov(X, Y) to the covariance.

89 ways. Split by the last move: you arrive at step 10 with a single from step 9 or a double from step 8, so ways(10) = ways(9) + ways(8). With 1 way to reach step 1 and 2 ways to reach step 2, the counts run 1, 2, 3, 5, 8, 13, 21, 34, 55, 89: the Fibonacci numbers, with 89 at the top.

How do you count without listing every route?

Imagine a friend at the top of the stairs asks how you got there. There is one thing you can say for certain about your last move: it was either a single from step 9 or a double from step 8, and never both. Every route to step 10 is a route to step 9 followed by a single, or a route to step 8 followed by a double, so the count at step 10 is the sum of the counts at steps 9 and 8. The same holds at every step, which turns the puzzle into a running sum.

Can you check 89 a second way?

Count by how many double steps you take. With k doubles, you make 10 - 2k singles, so 10 - k moves in all, and you only have to choose which k of those moves are the doubles. Summing the binomial counts over k = 0 to 5 gives 1 + 9 + 28 + 35 + 15 + 1 = 89, the same answer by a completely different road. Saying a second check aloud is worth more than the answer itself in a first round, because it shows you do not trust a pattern you have not tested.

What does the interviewer usually ask next?

Two things. First, allow steps of one, two or three: the same last-move argument gives w(n) = w(n - 1) + w(n - 2) + w(n - 3), and the count for ten steps becomes 274. Second, the coding version. A recursive function that calls itself for n - 1 and n - 2 recomputes the same steps again and again: for 30 steps it makes 1,664,079 calls to return 1,346,269. Storing each step's count once, or just keeping the last two numbers in a loop, does the job in 30 additions. The counts grow by about 1.618, the golden ratio, per step, which is the limitation of any approach that lists routes rather than counting them.

No die is best: red beats green, green beats blue and blue beats red, each with probability 5/9. So who picks first matters a great deal. Let me choose, then take the die that beats mine and win 5/9 of the time. If you are forced to pick first, every choice loses 5/9 of the time against an opponent who knows the cycle.

How do you work out who beats whom?

Each matchup has only nine equally likely pairs of faces, so write the 3 by 3 grid and count. Red against green: red's 2 beats only the 1, while its 6 and 7 each beat the 1 and the 5, for 1 + 2 + 2 = 5 wins out of 9. Do the same for the other two pairs and every matchup comes out 5 to 4: red over green, green over blue, blue over red. It is rock, paper, scissors built out of dice, and in rock, paper, scissors nobody asks which hand shape is best.

Why does green lose to red when green has the higher average?

The averages are red 5, green 6 and blue 5. Winning is about how often, not by how much. Green's 12 wins every time it shows, but it shows only a third of the time, and green's other two faces, 1 and 5, lose to both of red's high faces. A higher mean and a higher chance of winning are different things, and the gap between them is the whole puzzle. Change the rules so the winner collects the difference between the two numbers, and green's expected margin against red is 6 - 5 = +1: now you want green against red, and blue against red is a dead heat at 0.

Where does a trader meet the same thing?

Head-to-head comparisons need not line up into a ranking. Strategy A can beat strategy B on more days than not, B can beat C, and C can beat A, whenever one of them earns its money in rare large wins, as green does. Before choosing between strategies, decide whether you care about how often you win or how much you make, because under the first a cycle like this one means there may be no best choice at all. The limitation of the puzzle is that it is one roll; over many rolls with the total score counted, the mean matters more and green's 12 starts to pay.

pi/2, about 1.571. Rotate the sphere so the first point sits at the north pole; nothing changes, because the second point is uniform. On a unit sphere the surface distance is the polar angle theta of the second point. The northern and southern hemispheres are mirror images, so theta is as likely to be pi/2 - t as pi/2 + t, and its mean is pi/2.

Why can you put the first point at the pole?

Ask how far apart two random towns are on a perfectly round planet, and you can simply stand in one of them: the globe looks the same from every spot on it. Symmetry lets you fix one point anywhere, so the problem shrinks to one random point and its angle from the pole. On a sphere of radius 1, the distance along the surface between the pole and a point at polar angle theta is theta itself, measured in radians, so the question becomes: what is the average polar angle of a uniform point?

Where does sin theta come from? The circle of latitude at angle theta from the pole has circumference 2 pi sin theta, so a thin band there holds surface in proportion to sin theta: almost none near the poles, the most at the equator. Archimedes put it more neatly: the area of a band is proportional to its height along the axis, so cos theta is uniform between -1 and 1. The density (1/2) sin theta is a mirror image about pi/2, so the mean is pi/2 without doing the integral. Integration by parts confirms it, a numerical integral gives 1.5708, and a seeded simulation of 100,000 pairs gives 1.570.

If a uniform angle gives the same mean, why does the shape matter?

Because the mean survives by luck of symmetry and almost nothing else does. Choosing theta uniformly on 0 to pi crowds points near the poles. Ask for the chance the two points are within 60 degrees of each other and the correct answer is (1 - cos 60 degrees)/2 = 0.25, while the uniform angle says 0.33. Ask for the expected straight-line chord, 2 sin(theta/2), and the correct density gives 4/3, about 1.333, while the uniform angle gives 4/pi, about 1.273. The simulation gives 1.333 for the chord. This is the limitation of the shortcut: it answers this one question and must not be reused for the next.

Days 5 to 7, the run 8, -2, 3, which totals Rs 9 lakh. Walk through the days keeping the best total of a run ending today: either today alone or today added to yesterday's best run, whichever is larger. Record the largest value you see. The run 2, 4 looks attractive but totals only 6, and the -9 day makes it pointless to carry anything before it.

How do you find the best run without checking every start and end day?

Picture walking along a road with toll booths that either pay you or charge you. You may choose where to start and stop collecting. If the purse you have carried from earlier booths is in the red, the sensible move is to drop it and start fresh at the next booth. A run ending today is worth extending from yesterday only if the best run ending yesterday is positive; if it is negative it can only drag today down, so today starts a new run. That rule looks at each day once. Checking every pair of start and end days means 28 runs for seven days and 31,375 for a trading year of 250 days.

Why does the tempting run 2, 4 lose?

Because one later day beats it on its own. After the -9, the best run ending on day 4 is 6 - 9 = -3, so the rule drops the past and day 5 starts fresh at 8. The -2 on day 6 dips the run to 6, but the 3 on day 7 lifts it to 9. The best run can contain a losing day: 8, -2, 3 beats 8 alone because the day after the loss more than repays it. A candidate who stops a run at the first red day misses this, and the brute-force check over all 28 runs confirms 9 is the maximum.

What edge cases does the interviewer probe?

Three. If every day is a loss, the answer should be the least bad single day, so start the best at the first day's value, not at zero, or you will report an empty run worth 0. To report which days, store the start index whenever you restart and copy it when you record a new best. The same pass with the signs flipped finds the worst run, here -9, the single -9 day. The limitation on a desk is that the best run in hindsight is a selected statistic: a strategy that is judged by its best stretch will always look better than it trades.

About 65.5%. The gap A - B is normal with mean 10 - 8 = Rs 2 lakh and variance 3^2 + 4^2 = 25, so its standard deviation is Rs 5 lakh. A out-earns B when the gap is positive, and zero sits 2/5 = 0.4 standard deviations below the mean, so the probability is Phi(0.4), about 65.5%. The better trader loses about one month in three.

Why do the variances add when you subtract?

You and a colleague set off for the same meeting from different places, and each journey is uncertain by a few minutes. The gap between your two arrival times is more uncertain than either journey, not less, because either of you can be the late one. Subtracting an independent random amount adds its noise, so Var(A - B) = Var A + Var B = 9 + 16 = 25, and the gap's standard deviation is 5, not 1. The mean subtracts as you would expect, 10 - 8 = 2. The gap is normal because a difference of independent normals is normal.

How much does a longer comparison window help?

A lot, and at a predictable rate. Over a quarter of independent months the total gap has mean 6 and standard deviation 5 x sqrt(3), about 8.7, so A comes out ahead with probability 75.6%. Over a year the mean is 24 and the standard deviation 5 x sqrt(12), about 17.3, so the probability is 91.7%. The edge grows with the number of months and the noise with its square root, so the z-score grows with the square root of time. A risk manager who ranks traders on one month of P&L is ranking mostly noise.

What if the two traders' P&L are correlated?

Then the shared part cancels in the gap. With correlation 0.5, the variance is 9 + 16 - 2 x 0.5 x 3 x 4 = 13, a standard deviation of 3.61, and A wins with probability 71.0%. Positive correlation makes the comparison sharper because common market moves drop out of the difference; negative correlation does the opposite. The limitation is the normal assumption: real P&L has fat tails and skew, and if one trader earns through rare large months, the month-by-month win rate can disagree with the mean, so check the shape before trusting the 65.5%.

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.