Quant puzzles, solved step by step

8 cubes have 3 painted faces, 12 have 2, 6 have 1 and 1 has none; the chance the top face is painted is exactly 1/3. Corners carry three, edge middles two, face centres one, and the core none. Picking a random cube and rolling it picks a random small face out of 27 x 6 = 162. The painted ones are the big cube's surface, 6 x 9 = 54, so the probability is 54/162 = 1/3.

Where do the 8, 12, 6 and 1 come from?

Think of a Rubik's cube: its pieces are corners, edges and centres, plus a hidden core. A small cube's painted faces equal the number of outer walls it touches: a corner touches three, an edge middle two, a face centre one, the core none. A cube has 8 corners, 12 edges with one middle piece each, and 6 faces with one centre each. That accounts for 8 + 12 + 6 = 26 cubes; the 27th is the core.

Why is the roll probability exactly one third?

Do it the long way first: weight each cube type by its share of cubes and its share of painted faces. 8/27 x 3/6 + 12/27 x 2/6 + 6/27 x 1/6 + 1/27 x 0 = (24 + 24 + 6)/162 = 54/162. Then notice the shortcut: a random cube with a random face up is a uniform pick from all 162 small faces, and the painted small faces are exactly the big cube's surface, 6 x 9 = 54. That gives 1/3 without any case split, and it works for any size: an n x n x n cube gives 6n squared over 6n cubed, which is 1/n.

Interviewers use the second part to see whether you look for the structure before the arithmetic. Counting faces instead of cubes turns a four-case weighted average into one division. Say both routes: the case split proves you can count, the face count proves you can see.

25/48, about 0.521. For a stick broken into n pieces, the expected k-th smallest piece is (1/n)(1/n + 1/(n - 1) + ... ) with k terms. For n = 4 the sorted pieces average 3/48, 7/48, 13/48 and 25/48, which add to 1. The longest is (1/4)(1 + 1/2 + 1/3 + 1/4) = 25/48, more than twice the average piece of 1/4.

Why is the longest piece so much longer than a quarter?

Cut a sheet of dough at three random spots and the pieces are rarely even: one is usually a big slab and one a sliver. Random breaks produce uneven pieces, and the longest piece collects the unevenness, so its average sits far above the average piece. The average piece is always 1/4; the question asks about the largest of four correlated lengths, which is an {term('order statistic', 'The k-th smallest value in a sample, for example the minimum, the median or the maximum.')}.

Where does the harmonic formula come from?

Start with the shortest piece. The chance that all four pieces exceed x is (1 - 4x) cubed: take x off every piece and the three breaks must fit into the remaining length 1 - 4x. Integrating that from 0 to 1/4 gives an expected shortest piece of 1/16. Then the key fact: the step from each sorted piece to the next adds on average (1/n) times 1 over the number of pieces still longer. After the shortest, three pieces remain longer, so the next piece averages 1/16 + (1/4)(1/3); then add (1/4)(1/2), then (1/4)(1). The longest piece is (1/4)(1/4 + 1/3 + 1/2 + 1) = 25/48.

The step rule comes from the fact that the pieces behave like independent exponential lengths scaled to total 1, and the gap between successive minima of exponentials is memoryless. You can name that in the room rather than prove it. The check that the formula is right: the four sorted averages add to exactly 1, and a simulation of 100,000 sticks gives 0.521 for the longest. For n pieces in general, the longest averages (1/n) times the n-th harmonic number, which grows like (ln n)/n.

Bluff with the Q one time in three, and call with the K one time in three. You always bet the A and always check the K; I always call with the A and fold the Q. A Q bluff risks 1 more to win the 2 antes, so I must defend two thirds of hands facing it: the A gives half, the K calling one time in three gives the rest. The game is worth 1/18 per hand to you.

Which hands are easy, and where is the real decision?

Clear away the obvious hands first. With the A you always bet, because you win whether I call or fold. With the K, betting only gets called by the A and folds out the Q, which you beat anyway, so you check. As the caller, I always call with the A and always fold the Q. The whole game comes down to two mixed choices: how often you bluff with the Q, and how often I call with the K.

How do the indifference conditions fix both frequencies?

A teacher who spot-checks homework faces the same logic: check every paper and time is wasted, never check and everyone copies, check at the right rate and copying stops paying. In equilibrium each player mixes at the rate that makes the other indifferent between their two options. Your Q loses 1 by checking. Bluffing loses 2 against my A, and against my K loses 2 if I call and wins 1 if I fold. The two are equal only when I call with the K one time in three. My K loses 1 by folding; calling loses 2 against your A and wins 2 against a bluff, which is worth -1 only when you bluff one time in three.

Check against the pot-odds rule. A bluff risks 1 extra to win the 2 antes, so the caller must defend 2/(2 + 1) = 2/3 of the hands facing it; the A already covers half, and the K calling one time in three covers the other sixth. That two thirds is the number people misremember as the K's calling rate. Put the strategies together and the bettor, who acts first with more information about their own hand, earns 1/18 of a chip per hand. The limitation: with a bigger bet or more cards the ratios change, but the method, indifference on both sides, carries over.

A drift shared by both assets for the whole month can outweigh daily noise that moves them in opposite directions. Within a month the drift is constant, so only the noise shows and the correlation is negative. Summed over 21 days the drift's covariance grows with 21 squared but the noise's only with 21. With noise correlation -0.5 and drift standard deviation 0.3% a day, monthly correlation is +0.48.

What does a three-month example look like?

Picture two shops in the same market street. On any one day, a customer who buys from one did not buy from the other, so their daily takings move against each other. But in festival months the whole street is busy and in the rains the whole street is quiet, so their monthly takings rise and fall together. Correlation at one horizon says nothing on its own about another, because a slow common factor and fast opposing noise can sit in the same data.

Make it numerical with five-day months. In month 1 both assets drift at -0.8% a day, in month 2 at +0.2%, in month 3 at +1.2%. On top, asset A gets daily noise of +0.6, -0.3, 0, +0.3, -0.6 and asset B gets -0.3, +0.3, 0, -0.3, +0.3, which move in opposite directions. Within each month the correlation is -0.95. Summed over each month, the noise nets to zero, so both assets return -4%, +1% and +6%: identical, a monthly correlation of +1. Even all 15 days pooled show +0.71, because the month-to-month swing in drift is larger than the noise.

Why does summing more days push the correlation positive?

Write each daily return as the month's drift m plus noise. Over n days the drift adds up to n times m, while the noise adds up to a sum of n separate shocks. The drift's contribution to covariance scales with n squared, the noise's only with n, so the longer the horizon the more the shared drift wins. Take noise with standard deviation 1% a day and a within-month correlation of -0.5, and a drift whose standard deviation across months is 0.3% a day. For a 21-day month the drift adds 39.69 to the covariance and the noise subtracts 10.5, for a monthly correlation of 29.19/60.69 = 0.48.

The crossover sits where n times 0.09 equals 0.5, about 5.6 days, so even weekly returns of five days would still show a slightly negative correlation, -0.03. A hedge sized on daily correlation can therefore fail at a monthly horizon, which is why a desk measures correlation at the frequency it actually holds risk. The other mechanisms worth naming are mean reversion in the spread between the two assets and stale prices that lag by a day; both also make correlation depend on frequency. The limitation of the example is the assumption that drift is constant inside a month and noise is independent from day to day.

Yes: sell the 95 call at 9.80 and buy the 100 call at 4.50, collecting 5.30 for a position that can never cost more than 5. A 95/100 call spread pays between 0 and the strike gap of 5 at expiry, so its price must sit between 0 and 5. The market lets you sell it for 5.30, which locks in at least 0.30 per spread, more if the stock ends below 100.

What is a 95/100 call spread worth at most?

Think of two coupons for the same shirt: one lets you buy it for Rs 950, the other for Rs 1,000. The first is worth more, but never by more than Rs 50, because the most it can save you over the second is the Rs 50 difference in price. Long the 95 call and short the 100 call pays the stock's rise above 95, capped once it reaches 100, so at expiry it is worth between 0 and the strike gap of 5. Anything that is certain to pay no more than 5 cannot be worth more than 5 today; with interest it is worth at most 5 discounted, slightly less.

Which side of each quote do you trade at?

This is where the question is really won or lost. You sell at the bid and buy at the offer, so the spread you can sell is worth 9.80 - 4.50 = 5.30 to you, not the mid of 5.60. 5.30 is still above 5, so the bound is broken at prices you can actually deal at. Buying the spread would cost 10.20 - 4.30 = 5.90 for something worth at most 5, a certain loss, so only one direction works. Check the stock price cases: below 95 both calls expire worthless and you keep 5.30; at 97 you owe 2 on the short call and keep 3.30; at 100 or above you owe exactly 5 net and keep 0.30.

What could stop the arbitrage from paying?

If the calls are American and the short 95 is exercised early, exercise the 100 call too: you pay the stock price minus 95 and receive the stock price minus 100, a net 5, and you already hold 5.30. The real frictions are fees, the margin the short call ties up, and the risk that the quote vanishes after you trade one leg. Trade both legs together as a spread order. On a real screen a 0.30 bound violation lasts seconds, which is why the interviewer is testing whether you can see it fast and name the side, not whether such quotes are common.

36 passengers started. Work backwards from 16 and undo each step with its inverse. Growing by a third multiplies by 4/3, so undo it by multiplying by 3/4: 16 becomes 12. Undo half getting off by doubling: 24. Then 3/4 again: 18. Double again: 36. Forwards it checks: 36, 18, 24, 12, 16.

Why work backwards instead of setting up an equation?

Retracing your route to find a dropped wallet works because you know where you ended up. Here you know the final count and every step, so the cheapest route is to run the film in reverse. Each step is a multiplication, so each can be undone by multiplying by its reciprocal, starting from the 16 and moving toward the depot. An equation also works, 4x/9 = 16, but the backward chain shows every intermediate count, which lets you check that each one is a whole number of people.

What is the inverse of growing by a third?

This is the step that catches people. Growing by a third means the new count is 4/3 of the old one. To undo a rise of a third you multiply by 3/4, which removes a quarter of the new number, not a third of it. Taking a third off 16 gives 10.67, which is not a whole person and is a red flag on its own. Multiplying by 3/4 gives 12, and 12 grown by a third is 16 again, so the step checks.

The whole-number check also tells you what starting counts are possible at all. Every intermediate count, x/2, 2x/3, x/3 and 4x/9, must be a whole number, so x must be a multiple of 18. A free consistency check is worth saying aloud: 36 is a multiple of 18, and every count on the way, 18, 24 and 12, is whole. The same structure appears on a desk whenever a number passes through several percentage changes: a price up 10% and then down 10% ends at 99% of where it began, and undoing a change always means dividing by the factor, never subtracting the percentage.

5/9, about 55.6%. Put your arrival time on one axis and your friend's on the other, so every outcome is a point in a unit square. You meet when the two times differ by at most a third of an hour, a band along the diagonal. The two corner triangles outside the band each have legs of 2/3, area 2/9, so the band is 1 - 4/9 = 5/9.

Why turn two arrival times into a square?

Think of two people trying to catch each other at a tea stall with no phones. Nothing about the answer depends on who is you and who is the friend; it depends only on the pair of times. With two independent uniform times, every pair is equally likely, so the pair is a point spread evenly over a square and any probability is simply an area. The event you meet becomes a region: the set of points where the two times are within 20 minutes of each other. That region is a diagonal band, because the line x = y is where you arrive together.

How do you get the area without any integration?

Count the region you do not want. The two corners where one person arrives more than 20 minutes after the other are right triangles with both legs 40 minutes long, which is 2/3 of the side. Each has area (2/3) x (2/3) / 2 = 2/9, together 4/9, so the band is 5/9. Complements are the fastest route here, as they are for most geometric probability questions, because the leftover pieces are usually triangles.

What does the general formula tell you that the number does not?

Read 2w - w squared term by term. The 2w is the naive answer of either person waiting, and the minus w squared removes the double count and the clipping at the edges of the hour. It also tells you how waiting time buys certainty: to meet half the time each person must wait about 17.6 minutes, and to be sure each must wait the whole hour. The same picture prices any tolerance between two random arrivals, such as two orders landing in the same matching window of an auction.

One eigenvalue of 2.6 and four of 0.6, so the first principal component explains 52%. Write the matrix as 0.6 times the identity plus 0.4 times a matrix of ones. The all-ones vector is an eigenvector with eigenvalue 0.6 + 5 x 0.4 = 2.6; any vector whose weights sum to zero is killed by the ones matrix and has eigenvalue 0.6. The trace check: 2.6 + 4 x 0.6 = 5.

What structure should you spot before touching a determinant?

Think of five students whose marks all move together when the paper is hard, plus their own good and bad days. There is one shared shock and five private ones. An equicorrelation matrix is exactly that: R = (1 - rho) I + rho J, where J is the matrix of all ones, so its eigenvectors are those of J and you never need a characteristic polynomial. J sends the all-ones vector to 5 times itself and sends any vector whose entries sum to zero to zero. Those two facts give every eigenvalue.

How do the eigenvalues fall out, and how do you check them?

Apply R to the all-ones vector: each row sums to 1 + 4 x 0.4, so the equal weight portfolio has eigenvalue 1 + (n - 1) rho = 2.6. Apply R to any vector with weights summing to zero, such as long asset 1 and short asset 2: the rho J part vanishes and only (1 - rho) = 0.6 is left, and there are four independent such vectors. The eigenvalues must add to the trace, the sum of the diagonal, which is 5: 2.6 + 2.4 = 5.

What does the answer say about a real portfolio?

The first component is the market: equal weights, and its share rises towards rho as you add assets. With 50 assets at the same correlation the first eigenvalue is 1 + 49 x 0.4 = 20.6, 41.2% of the total, while each of the other 49 stays at 0.6. Diversification removes the private shocks but never the common one. The same formula gives a limit: the smallest eigenvalue 1 - rho is always fine, but 1 + (n - 1) rho must stay positive, so five assets cannot all share a correlation below -0.25.

36 for 10, 45 for 11, and (n - 1)(n - 2)/2 in general. Write n as a row of n stars. Splitting it into three positive parts means placing two bars in two different gaps among the n - 1 gaps between stars. Each choice gives exactly one ordered triple, so the count is n - 1 choose 2: 9 choose 2 = 36 and 10 choose 2 = 45.

Why turn the sum into a row of stars?

Think of ten sweets in a line to be shared among three children in order, each getting at least one. You do not need to decide amounts; you only need to decide where to cut the line. Every ordered split of n into three positive parts is exactly one choice of two cut points among the n - 1 gaps between items, and every choice of two gaps gives a valid split. That one-to-one match is the whole argument, and it is what the interviewer wants to hear you state.

What if the interviewer meant something slightly different?

Ask two quick questions before you answer: are zeros allowed, and does order matter. The word numbers hides three different questions, and each has a different count. If zeros are allowed, add one to each part first so they become positive and sum to n + 3: for 10 that is 12 choose 2 = 66. If order does not matter, the count for 10 drops to 8 unordered triples, and for 11 to 10, which you would list rather than compute. Asking which one is wanted takes five seconds and is part of the answer.

How do you check 36 without the formula?

Fix the first number and count the rest. If the first part is a, the other two must sum to 10 - a, which can be done in 9 - a ordered ways. For a from 1 to 8 that is 8 + 7 + ... + 1 = 36. The counts for each total are the triangular numbers 1, 3, 6, 10 and so on, which is the same formula read another way.

1/6, the same as with no information. An odd first roll says nothing about the order in which 2, 4 and 6 first appear, so it is independent of 2 finishing last. A first roll of 4 or 6 raises the chance 2 is last from 1/3 to 1/2, so conditioning on that ending shifts weight onto 4 and 6, which rise to 1/4 each. The odd faces keep 1/6 each; 1/5 wrongly spreads the weight evenly.

Why does the ending tell you anything about the start?

Suppose you hear that a friend reached a party last. That makes it a little more likely they left home late, because leaving late and arriving last go together. It says nothing about whether they wore a blue shirt, which has no bearing on arrival order. Conditioning on an outcome reweights every starting state by how likely that state makes the outcome, and a state that does not affect the outcome keeps its original probability. Here the outcome is 2 finishing last among the evens; the question is which first rolls make that more or less likely.

How do the numbers work out with Bayes?

Odd rolls never change which new even appears next, so only the order of first appearances matters, and without information it is a random ordering of three: 2 is last with chance 1/3. If the first roll is 4, then 2 and 6 are left to race, and each is equally likely to show first, so 2 ends last with chance 1/2. Now weigh: each face has prior 1/6. The joint chance of first roll 1 and 2 last is 1/6 x 1/3 = 1/18; of first roll 4 and 2 last, 1/6 x 1/2 = 1/12. The total is 1/3, so first roll 1 has posterior (1/18)/(1/3) = 1/6 and first roll 4 has (1/12)/(1/3) = 1/4.

Where does the 1/5 intuition go wrong?

It treats the information as simply ruling out one face and renormalising the rest. Ruling out an outcome and conditioning on an event are the same thing only when every remaining outcome makes the event equally likely, and here they do not. A check: the posteriors 1/6, 1/6, 1/6, 1/4, 1/4 and 0 add to 1, while five faces at 1/5 would give 4 and 6 the same weight as 1. On a desk this is the error of reading a trade's outcome as if it said nothing about which signal triggered it.