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.

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.

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.