Quant puzzles, solved step by step

I show heads 3/8 of the time; so should you; and the game is worth -1/8 to you per round. At my mix q, your heads pays 3q - 2(1 - q) and your tails pays -2q + (1 - q). Setting them equal gives q = 3/8, where both pay -1/8. By the same algebra your 3/8 mix makes my choices equal, so neither of us can improve: the table looks fair and is not.

Why do I randomise to make you indifferent, not to help myself?

Think of a penalty taker and a goalkeeper. If the taker shoots left more often than he should, the keeper dives left and gains; any pattern is exploited. In a zero-sum game, the mix that protects you is the one that leaves your opponent with nothing to exploit, which means making every one of their choices pay the same. So I do not pick my heads probability by looking at my own payoffs in isolation; I pick it so that your heads and your tails earn you equal amounts. Then it no longer matters what you do.

How do the two equations give 3/8 and -1/8?

Let q be my chance of heads. Your heads earns 3q - 2(1 - q) = 5q - 2 and your tails earns -2q + 1(1 - q) = 1 - 3q, and they are equal when 8q = 3, at q = 3/8. Plug back in: 5(3/8) - 2 = -1/8. For your side, let p be your chance of heads; my payoffs are the negatives of yours, and because the table is symmetric the same algebra gives p = 3/8. At those mixes, each cell's frequency is HH 9/64, TT 25/64 and the two mismatches 15/64 each: 3(9) + 1(25) - 2(30) = -8, over 64, is -1/8.

Why does a table with equal wins and losses favour me?

Because the cells are not played equally often. Your big win needs both of us on heads, and I can make that cell rare; the mismatches, where I win, happen 30 times in 64 at equilibrium. If I played heads half the time you could earn +1/2 by always showing heads, so a naive 50/50 from me would be a gift. In trading the same idea sets a market maker's quotes: they are placed so that the informed side has no choice that beats the others, not so that the market maker profits on any one trade.

200 miles. The car needs 100 / 50 = 2 hours to reach B, and the bird flies the whole time at 100 mph, so it covers 2 x 100 = 200 miles. Summing the zigzags gives the same answer: the first round trip is 133.3 miles, each later one is a third of the one before, and 133.3 / (1 - 1/3) = 200.

What is the question really asking you to count?

Think of a dog running back and forth between you and your front door while you walk home. You could trace every dash, or you could notice that the dog runs at a steady speed for exactly as long as your walk takes. Distance is speed times time, and the bird's flying time is fixed by the car, not by the zigzags, so the zigzag detail is a distraction. The car covers 100 miles at 50 mph in 2 hours; the bird flies at 100 mph for those same 2 hours. That is 200 miles, and it takes one sentence.

How do you check it by summing the zigzags?

The bird and car close the first 100 miles at a combined 150 mph, so they meet after 40 minutes, 33.3 miles from A. The bird flies back to B, 66.7 miles, arriving at 80 minutes, by which time the car is at 66.7 miles. Each round trip starts with the gap to the car one third of the previous gap, so the round trips form a geometric series with ratio 1/3. The first is 133.3 miles; the sum is 133.3 / (1 - 1/3) = 200. It agrees, and it shows why infinitely many turns still add to a finite distance.

Why do interviewers still ask a puzzle this well known?

Because the way you answer tells them more than the answer. A candidate who starts summing legs has reached for the first method that fits; a candidate who asks what quantity is fixed has found the invariant, and that is the habit the interviewer is hiring. The story about von Neumann summing the series in his head is part of the folklore; you get more credit for the one-line method and the series as a check. The same move, looking for a quantity that does not depend on the messy path, solves many expected-value and stopping questions on this page.

About 1, 3, 6 and 12 correct digits: the count roughly doubles each step. The update is x(next) = (x + 2/x)/2: 1.5 gives 17/12 = 1.41667, then 577/408 = 1.4142157, then 665857/470832 = 1.41421356237469. Each new error is about the old error squared divided by 2x, so if the error is 10^-k, the next is about 10^-2k. That is quadratic convergence.

Where does the update rule come from?

Think of guessing a side of a square room whose area is 2. If your guess is too big, 2 divided by your guess is too small, and the truth sits between the two. Newton's method for x squared minus 2 is exactly that: replace x with the average of x and 2/x. Formally, Newton follows the tangent of f(x) = x^2 - 2 down to zero, x - f(x)/f'(x) = x - (x^2 - 2)/(2x), which simplifies to (x + 2/x)/2. The averaging form is the one to use in your head.

How do you get the iterates without a calculator?

Keep fractions. From 3/2, the next value is (3/2 + 4/3)/2 = 17/12, then (17/12 + 24/17)/2 = 577/408, and the pattern continues: if x = p/q, the next is (p^2 + 2q^2)/(2pq). 17/12 is 1.41667, already right to 1.41. 577/408 is 1.4142157 against 1.4142136, right to 1.41421. The third step's fraction, 665857/470832, is too big for mental division, but you can predict its accuracy without doing it, which is the point of the question.

Why is it the digits that double, and when does that fail?

Subtract the root from the update and the algebra collapses to (x - root 2)^2 / 2x. Squaring an error of 10^-3 gives 10^-6, so each step doubles the number of correct digits once you are close. The errors here run about 0.09, 0.0025, 2 x 10^-6 and 1.6 x 10^-12. The doubling needs a good start and a simple root: far from the root, or where the slope is zero, Newton can creep or jump away. Bisection, by contrast, gains one binary digit per step whatever happens, which is why desk code often brackets with bisection and finishes with Newton when solving for implied volatility.

1/6. Let D be the distance to the nearest edge. D exceeds d exactly when the point lies in the inner square of side 1 - 2d, so P(D > d) = (1 - 2d)^2 for d up to 1/2. The expected value of a non-negative variable is the integral of its tail, and the integral of (1 - 2d)^2 from 0 to 1/2 is 1/6.

Why work with the chance of being far rather than the distance itself?

Think of a sandpit where a child stands at a random spot and the question is how far they are from the nearest edge. Writing the distance as min(x, 1 - x, y, 1 - y) and integrating a minimum of four things means splitting the square into four triangles. Asking instead when the point is farther than d from every edge has a one-picture answer: the point must lie in a smaller square, shrunk by d on every side. That square has side 1 - 2d, so its area, (1 - 2d)^2, is the tail probability. One formula replaces four cases.

How does the tail give the expectation?

For any non-negative random variable, the expected value equals the integral of the chance that it exceeds each level, E[D] = integral of P(D > d). Here that is the integral of (1 - 2d)^2 from 0 to 1/2. Substitute u = 1 - 2d and it becomes half the integral of u^2 from 0 to 1, which is 1/6. A simulation with 200,000 random points gives 0.1669, against the exact 0.1667.

How do you sanity-check 1/6 against simpler cases?

Build up the number of edges. The distance to one fixed edge averages 1/2, to the nearer of two opposite edges averages 1/4, and to the nearest of all four it falls to 1/6, so each added constraint pulls the minimum closer. The density of D is the slope of the tail, 4(1 - 2d), largest at the edge, which says most random points are near the boundary. That is the same reason most of the volume of a high-dimensional cube sits near its surface, a fact that matters when sampling scenarios in many risk factors at once.

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.

A^10 has 29,525 on the diagonal and 29,524 off it. A has eigenvalue 3 along (1, 1) and eigenvalue 1 along (1, -1). Writing A = Q D Q^T with D = diag(3, 1), the tenth power is Q D^10 Q^T, and only the numbers 3 and 1 get raised to the tenth. The entries are (3^10 + 1)/2 and (3^10 - 1)/2.

Why look for eigenvectors at all?

Think of a photocopier set to 300% on one axis and 100% on the other. Copy a copy ten times and you do not need to simulate every pass: that axis is 3 to the tenth times longer and the other is unchanged. An eigenvector is a direction the matrix only stretches, so applying the matrix ten times along it is just multiplying by the eigenvalue ten times. Symmetric matrices always have a full set of such directions at right angles, which is what makes this matrix easy.

Find them by inspection. Adding the two rows of A gives 3 in each, so A(1, 1) = (3, 3): eigenvalue 3. Subtracting gives 1, so A(1, -1) = (1, -1): eigenvalue 1. The trace is 4 and the determinant is 3, and 3 + 1 = 4 and 3 x 1 = 3, which confirms both in one line.

Is there an even faster route for this particular matrix?

Yes. Write A = I + J, where J is the all-ones matrix. J squared is 2J, so every power of J is a multiple of J, and (I + J)^n collapses to I + ((3^n - 1)/2) J. For n = 10 that is I + 29,524 J, which gives 29,525 on the diagonal and 29,524 off it: the same answer, and a good cross-check to say aloud. A brute-force multiplication in code agrees exactly.

Say why this matters on a desk. A covariance matrix with equal variances and one common correlation has exactly this shape, and its eigenvectors are the market direction and the spread directions. Powers of transition matrices in Markov chains are computed the same way, and the eigenvalue closest to 1 tells you how fast the chain forgets where it started.

Stop at the first draw of 0.8 or more; the game is worth 0.8 after all fees. Holding x, one more draw improves you by (1 - x)^2 / 2 on average, which equals the 0.02 fee at x = 0.8. Check: you expect 5 draws costing 0.10 in total, and the draw you keep averages 0.90, so the net value is 0.80.

How do you decide whether one more draw is worth it?

Think of hunting for a flat. Each viewing costs you an evening. If the flat in hand is already good, another viewing rarely beats it, and when it does it only beats it by a little. The value of one more look is the chance of beating what you hold times the average margin when you do, and you stop when that falls below the price of looking. With a uniform draw and a current value x, the chance of beating x is 1 - x and the average margin is (1 - x)/2, so the expected gain is (1 - x)^2 / 2.

Why are the threshold and the value the same number?

Because the game has no memory. After a disappointing draw you are back where you started, facing the same game worth V. You should accept a draw exactly when it beats what the fresh game is worth, so the threshold equals the value. This is the same logic as a reservation price: you walk away from any offer below what the next round is worth to you. The arithmetic check is worth saying: with threshold 0.8 a draw succeeds with probability 0.2, so you expect 5 draws and 0.10 of fees, and an accepted draw is uniform on 0.8 to 1 with mean 0.9. The net is 0.9 - 0.10 = 0.8. A seeded simulation of 200,000 games gives 0.800.

What happens as the cost changes?

The threshold is 1 - sqrt(2c), so it responds to the square root of the cost. A fee of 0.005, four times cheaper, only moves the threshold from 0.8 to 0.9. At a fee of 0.5 or more, the threshold hits zero and you take the first draw, because even the first draw's average of 0.5 barely covers what you paid. The limitation of the model is that draws are independent and the distribution is known; if you were learning the distribution as you drew, the first few draws would be worth more than this rule says.

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.