Quant puzzles, solved step by step

Rs 12.25. The two dice are independent, so the expected product equals the product of the expected faces: 3.5 x 3.5 = 12.25. A check on the 6 by 6 grid: row i averages 3.5i, and the six row averages, 3.5 to 21, average 12.25. If both numbers came from one die, the answer would be E[X squared] = 91/6, about 15.17, which is not the question.

Why can you multiply the averages?

Picture a shop whose daily takings are footfall times average spend, where the two have nothing to do with each other. Over a year, average takings are average footfall times average spend. When two quantities are independent, the average of their product is the product of their averages, because knowing one tells you nothing about how big the other will be. That fails the moment they move together: a crowded day with bigger spending lifts the product above the product of averages. Two dice thrown separately are the textbook independent pair.

How do you check 12.25 by brute force quickly?

Sum the grid by rows. Row i of the multiplication table sums to i x 21, so the whole grid sums to 21 x 21 = 441, and 441 / 36 = 12.25. That is the product rule made visible: the grid's total factors into the first die's total times the second die's total. Note too that the product is skewed: only 18 different values appear, most cells are small, and the big ones, 25, 30 and 36, pull the mean up, so {BELOW75} of the 36 outcomes pay less than the average.

What is the interviewer likely to ask next?

The usual next step changes the dependence. If the same die is used for both numbers, you are paid the square of one roll, and its expectation is 91/6, about 15.17, higher by exactly the variance of one die, 35/12. That gap is the covariance at work: E[XY] = E[X]E[Y] + Cov(X, Y). Then comes the price: if you would pay to play, you should quote around 12.25 and note that the payout's standard deviation is about 8.94, so a single play is very noisy relative to its mean.

The mean, 10, minimises squared error; the median, 3, minimises absolute error. Squared error grows with the square of a miss, so the single 40 pulls the best constant towards it. Absolute error charges every unit of miss equally, so the best constant sits in the middle of the pack. Choosing the loss is choosing which of the two statistics you estimate.

Why does each loss land on a different constant?

Picture five friends choosing a meeting point on a straight road: four live at the 1, 2, 3 and 4 km marks and one at the 40 km mark. If the goal is the smallest total travel, you meet at 3 km: moving towards the far friend saves them one km per km moved but costs the other four one km each. If the goal is the smallest total of squared travel, the far friend's 37 km squared, 1,369, dominates everything and the meeting point slides to 10 km. Absolute error balances the count of points on each side, which is the median; squared error balances the total distance on each side, which is the mean.

So which constant should you actually use?

It depends on what the 40 is, and that is the answer the interviewer wants to hear. If the 40 is a typing error or a one-off glitch in a price feed, the median is the honest summary: remove the 40 and the mean falls to 2.5, while the median barely moves. If the 40 is real, say the one big winning day in a strategy's P&L, the mean is the number that matters, because your total profit is the sum of the days and the sum is five times the mean. A robust estimator that ignores the big day would tell you the strategy earns 3 a day when it actually earns 10.

Where does this show up in a quant job?

Every regression makes this choice silently. Ordinary least squares minimises squared error and so fits conditional means; one extreme observation can swing the line. Least absolute deviation, or quantile regression at the 50% level, fits conditional medians and shrugs off the same extreme point. In practice, desks winsorise or clip returns before fitting squared-error models, or switch to a Huber loss that is squared near zero and linear in the tails. The limitation is that none of these fixes is free: every one of them throws away some of the information in genuine large moves.

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.

110 paths avoid the block, so the probability is 110/210 = 11/21, about 52.4%. All shortest paths use 6 rights and 4 ups in some order: C(10,4) = 210. Paths through (3,2) combine 10 ways in with 10 ways out, 100 in all. Subtract, and 110 survive.

How do you count all the shortest paths?

A shortest path is a string of 10 moves with exactly 6 rights and 4 ups, like a ten-letter word made of R and U. Choosing which 4 of the 10 slots are ups fixes the path completely, so there are C(10,4) = 210 shortest paths. That is the whole sample space, and every one of those strings is equally likely by the question's rule.

Why multiply for the paths through the blocked point?

Think of a trip from home to the office with a stop at a coffee shop. If there are 10 routes to the shop and 10 routes from the shop to the office, there are 10 x 10 = 100 full trips, because every first half pairs with every second half. Paths through (3,2) split the same way: C(5,2) = 10 in and C(5,2) = 10 out, so 100 of the 210 pass through the block. Subtract and 110 remain.

The grid method in the figure is the check, and it is also what you would code. Each point's count is the count from the left plus the count from below, because the last step into any point came from one of those two neighbours. Setting the block to zero removes every path through it automatically, and the same method handles several blocks, where the subtraction formula needs inclusion and exclusion.

Does the answer change if the walker flips a coin at each step?

Yes, and interviewers often ask this next. If the walker flips a fair coin for right or up at each step, any visit to (3,2) happens on move five, and a walker still able to get there has not yet touched the top or right edge, so all five of those moves were free coin flips. The chance of standing on (3,2) after five flips is C(5,2)/2^5 = 10/32 = 5/16, so the coin-flip walker avoids the block with probability 11/16, about 68.8%, well above 11/21. Choosing uniformly among complete paths is not the same as flipping coins: conditioning on the end point (6,4) pulls paths towards the diagonal that leads there, and (3,2) sits on it. Say which model the question means before you answer.

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.

Two thirds of days are calm. In the long run the fraction of days moving from calm to stressed must equal the fraction moving back, so calm share x 0.2 = stressed share x 0.4. Calm days are therefore twice as common as stressed days, 2/3 against 1/3. The same answer comes from spell lengths: calm spells average 5 days and stressed spells 2.5, and 5 / 7.5 = 2/3.

Why can you balance flows instead of solving equations?

Think of two rooms at a party. Every few minutes, one in five people in the kitchen wanders to the lounge, and two in five people in the lounge wander back. Once the crowd settles, the numbers crossing each way must match, or one room would keep filling up. In a two-state chain the long-run shares are fixed by one equation: the share of days leaving calm must equal the share of days leaving stressed. Calm leaves at rate 0.2 and stressed at rate 0.4, so calm must hold twice as many days.

How do you check it a second way?

Use spell lengths. A calm spell ends each day with probability 0.2, so it lasts 1/0.2 = 5 days on average; a stressed spell ends with probability 0.4, so it lasts 2.5 days. The chain alternates calm spell, stressed spell, calm spell, so the calm share is 5 out of every 7.5 days, which is 2/3. Two methods, one answer, in under a minute.

How fast does the chain forget where it started?

The transition matrix has a second eigenvalue of 0.8 + 0.6 - 1 = 0.4, and any gap between today's probabilities and the long-run split shrinks by that factor each day. After five days the starting state explains only 0.4^5, about 1%, of the gap, so the answer does not depend on how the week began. A trader would add the limitation: real regimes are not memoryless, and a stress spell that has already lasted a month is not as likely to end tomorrow as one that started yesterday.

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.

76. Split 100 into 4 x 25. Any power of 4 is 0 mod 4. Mod 25, Euler's theorem applies because 4 and 25 share no factor, and 3000 is a multiple of phi(25) = 20, so 4^3000 is 1 mod 25. The numbers below 100 that are 1 mod 25 are 1, 26, 51 and 76, and only 76 is divisible by 4.

Why does the obvious Euler shortcut fail?

Euler's theorem says a to the power phi(n) is 1 mod n, and phi(100) = 40, so it is tempting to say 4^3000 = (4^40)^75 ends in 01. The theorem needs the base and the modulus to share no factor, and 4 and 100 share a factor of 4, so it does not apply. A quick sense check kills 01 anyway: every power of 4 is divisible by 4, and a number is divisible by 4 exactly when its last two digits are, which 01 is not.

How do you split the problem so the theorem does apply?

Think of a clock with 100 hours as two smaller clocks running together, one with 4 hours and one with 25. Knowing where both small clocks point fixes the big one exactly. Mod 4 the answer is 0, since 4^3000 is a multiple of 4; mod 25 the answer is 1, since 4 and 25 share no factor and 3000 is a multiple of phi(25) = 20. Now list the numbers below 100 that are 1 mod 25: 1, 26, 51, 76. Only 76 is a multiple of 4. This step is the Chinese remainder theorem, and naming it earns credit.

What is the fastest check if you have a pencil?

Just list the endings. Multiply each ending by 4 and keep the last two digits: 04, 16, 64, 56, 24, 96, 84, 36, 44, 76, and then 76 x 4 = 304, which ends in 04, so the cycle has length 10. Because 76 x 76 = 5,776 also ends in 76, every power of 76 ends in 76, and 4^3000 = (4^10)^300 must end in 76. On a multiple-choice test, this listing takes about thirty seconds and needs no theorem at all.

The new divisor is 2. Before the split the prices sum to 300, and 300 / 3 is 100. After a 3 for 1 split the 150 stock trades at 50, the sum is 200, and only a divisor of 2 keeps the index at 100. A cap-weighted index needs no adjustment at all, because a split triples the share count as it cuts the price to a third, leaving market value unchanged.

Why must the divisor change when nothing about the company changed?

Cut a pizza into twelve slices instead of four and you have not made more pizza. A stock split does the same to a company: three times the shares, each worth a third. A price-weighted index adds up share prices, so a split drops the sum even though no value was lost, and the divisor must be cut to stop a fake fall in the index. Solve for it by keeping the index level fixed: 200 divided by the new divisor must equal 100, so the divisor is 2.

What else changes in a price-weighted index after the split?

The weights. In a price-weighted index a stock's weight is its price over the sum of prices, so the expensive stock dominates whatever the size of the company. Before the split C carried 50% of the index; after it, C carries only 25% and B, untouched, jumps to 50%. A 10% rise in C used to add 5 index points; now it adds 2.5. Nothing about C's business changed; the index simply started caring less about it, which is the main criticism of price weighting.

How do the other common methods treat the split?

A market-cap-weighted index sums price times shares, and a 3 for 1 split multiplies shares by 3 while dividing price by 3, so the stock's market value, its weight and the index are all unchanged; no divisor adjustment is needed for a split. Cap-weighted divisors still change for events that alter total market value without a price move, such as share issuance, buybacks or a constituent being replaced. An equal-weighted index is also untouched by a split, since weights are reset to equal at each rebalance, but it has to trade at every rebalance to get back to equal, which costs money.