Quant puzzles, solved step by step

Do not spin: pulling straight away survives with probability 3/4, against 2/3 with a spin. The click tells you the hammer sat on one of the four empty chambers. Because the two bullets are adjacent, the four empties form a run, and only the last empty in that run is followed by a bullet. A spin throws that information away and resets you to 4 empties out of 6.

What does the click actually tell you?

Think of a row of six houses where two neighbours keep dogs. You knocked at a random house and no dog barked. If you now try the next house along, you are only in trouble if you had knocked on the one house sitting just before the dogs. The click narrows the hammer's position to the four empty chambers, and the question becomes how many of those four have a bullet immediately after them. With the bullets side by side, the empties run 3, 4, 5, 6 in firing order, and only chamber 6 hands over to a bullet.

Does the answer depend on the bullets being adjacent?

Completely, and that is the follow-up most interviewers ask. If the two bullets are not next to each other, the empties split into two runs, two empties now sit in front of a bullet, and pulling straight away survives only 1/2, worse than the 2/3 of a spin. The same count works for bullets one apart or directly opposite: in both cases two of the four empties are followed by a bullet. So the rule is not spin or do not spin; it is count the empties that border a bullet and compare with a fresh spin.

What is the general lesson for a trading interview?

A random reset destroys information, and information has value only if the structure of the problem lets you use it. Here the structure is clustering: the bullets sit together, so a safe chamber is likely followed by another safe one. Markets have the same feature in volatility: a calm day tends to be followed by a calm day, so conditioning on what just happened beats assuming each day is a fresh draw. Say that link in one line after the arithmetic.

1 - 2/e, about 26.4%. Convolving two exponential densities gives the total the density x e^-x, which starts at zero and peaks at one minute. Its area below one minute is 1 - 2/e. The same number drops out of the Poisson view: the total is under a minute exactly when at least two arrivals land in the first minute of a rate-one Poisson process.

Why does adding two waits change the shape so much?

Suppose you need two buses, one after the other, and each arrives on average a minute after you reach its stop. Catching the first bus quickly is common; catching both quickly is rare, because both have to cooperate. A single exponential wait is most likely near zero, but a sum of two is almost never near zero, so its density starts at zero and rises into a hump. That shift of mass away from zero is why the answer is much smaller than the 63.2% chance that one wait is under a minute.

Is there a way to get 1 - 2/e without integrating?

Yes, and it is the cleaner answer to give aloud. Exponential waits with mean one are the gaps between arrivals of a Poisson process with rate one per minute. The second arrival comes before one minute exactly when at least two arrivals land in the first minute, and the Poisson chance of zero or one arrival is e^-1 + e^-1 = 2/e. So the answer is 1 - 2/e, about 26.4%, with no calculus at all.

Sanity-check the size. Both waits being under a minute has probability (1 - 1/e)^2, about 40.0%, and the total being under a minute is a stricter event, so the answer must be smaller: 26.4% is. The limitation is the independence assumption; if the two waits were driven by the same traffic, they would move together and the total would be more spread out.

The two largest drawdowns are 130 to 80, 38.5%, and 120 to 90, 25.0%. Walk through the prices once, keeping the highest price seen so far. At each step the drawdown is one minus price over that running peak; the maximum drawdown is the largest value seen. For the n largest, close an episode each time a new high is set, record its trough, and sort the episodes.

What exactly is a drawdown measured from?

Think of a hiker measuring how far below the highest point reached so far she has dropped. A drop only counts from a summit already climbed, not from a peak further along the trail. A drawdown is the fall from the running maximum to a later price, so the order of the prices matters and the overall high and low cannot simply be paired. Here the 140 comes after the 80, so the tempting 42.9% never happened.

How do you get the n largest drawdowns rather than just the worst?

Split the path into episodes. An episode opens at a running peak and closes when the price makes a new high; its size is the fall from that peak to the lowest price inside it. Here the 120 episode closes when the price reaches 130, with a trough of 90, so it is 25%. The 130 episode closes at 140 with a trough of 80, 38.5%. The 140 episode never closes, so report it as open at 20.0%. Sorting the episodes gives the n largest in one pass plus a sort, O(N log N) at worst, and a heap of size n keeps it at O(N log n).

Say the edge cases, because the interviewer is testing code judgement as much as arithmetic. Two drawdowns from the same peak must not be counted twice: the fall to 80 and the later level of 110 belong to one episode. An episode still open at the end of the data is real risk and should be reported with a flag. And drawdown on a price series is not the same as drawdown on a strategy's cumulative P&L, where you would use the equity curve, not the price.

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.

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.

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.

400. Since 1,000 = 2^3 x 5^3, a number shares a factor with 1,000 exactly when it is divisible by 2 or by 5. There are 500 multiples of 2 and 200 of 5, but the 100 multiples of 10 sit in both lists, so 600 numbers share a factor and 400 do not. Euler's formula agrees: 1,000 x 1/2 x 4/5 = 400.

Which numbers share a factor with 1,000?

Picture a hall of 1,000 people where everyone wearing a red badge or a blue badge is asked to leave. To count who stays, you need the red-badge count, the blue-badge count, and how many wear both, because they would otherwise be counted out twice. Write 1,000 as 2^3 x 5^3: a number shares a factor with it exactly when it is divisible by 2 or by 5, so only two badges matter, and the powers 3 do not add any new conditions. Every multiple of 4 or 8 is already a multiple of 2, and every multiple of 25 or 125 is already a multiple of 5.

Why does the quick product formula work here?

Half of all numbers are odd, and among those, four in five are not multiples of 5. Because 1,000 is a multiple of 10, the numbers 1 to 1,000 contain exactly 100 full blocks of ten, and in each block exactly 4 numbers, 1, 3, 7 and 9, survive both tests, so 100 x 4 = 400. The strip of 1 to 20 in the figure shows the pattern repeating, 8 survivors in 20. The product is exact only when the range is a whole number of such blocks: for 1 to 1,234 it gives 493.6, while a direct count gives 494.

Where does a question like this lead in an interview?

Usually to powers and remainders. Euler's theorem says a number coprime to n, raised to the power phi(n), leaves remainder 1 when divided by n, and that is the engine behind last-digit puzzles: phi(100) = 40, so 3^40 ends in 01 and so does 3^400. It also leads to probability: the chance that two large random integers share no factor tends to 6/pi^2, about 0.608. The habit the question tests is factorising first: once you see only the primes 2 and 5 matter, a counting question becomes a two-circle Venn diagram.

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%.

OLS gives 0.8 and ridge gives 0.64. OLS minimises squared error and its slope is the sum of xy over the sum of x squared, 80/100. Ridge adds lambda times the slope squared to the loss, which puts lambda into the denominator: 80/(100 + 25) = 0.64. That is the OLS slope times 100/125 = 0.8, so ridge shrinks the slope towards zero but never to zero.

Where does lambda end up in the formula?

Think of a new analyst's forecast that you half trust: you do not discard it, you shade it towards zero, and the less data behind it the more you shade. Ridge does that mechanically. It minimises the squared errors plus lambda times the slope squared; setting the derivative to zero gives b = Sxy/(Sxx + lambda). Penalising the size of the slope acts exactly like adding observations whose x squared totals lambda and whose y is zero, data that say the slope is zero. With 100 of real evidence and 25 of make-believe evidence, the slope is 80/125 = 0.64.

Why would you want a slope that is biased towards zero?

Because a smaller, steadier estimate can be closer to the truth on average. Suppose the true slope is 0.5 and the noise variance is 25. OLS is unbiased but its variance is 25/100 = 0.25. Ridge at lambda 25 has variance 0.16 and a bias of -0.1, so its mean squared error is 0.17. Ridge trades a little bias for a larger cut in variance, and when the signal is weak relative to the noise that trade wins. In this one-feature case the best lambda is noise variance over slope squared, 100, which halves the slope and cuts the error to 0.125. In practice the truth is unknown, so lambda is chosen by cross-validation.

How does lasso differ?

Lasso penalises lambda times the absolute slope instead. In one dimension that subtracts lambda/2 from the cross-product rather than adding to the denominator: (80 - 12.5)/100 = 0.675 at lambda 25, and exactly zero once lambda reaches 160. Ridge scales coefficients down; lasso shifts them down and can set them to exactly zero, which is why lasso selects features and ridge does not. Both penalties depend on the scale of x, which is why the feature must be standardised first; with correlated features, ridge spreads the weight across them while lasso tends to keep one.