Quant puzzles, solved step by step

About 39.8%. The first card can be anything. The second must come from one of the three other suits: 39 of the 51 cards left. The third must avoid both suits already seen: 26 of the 50 left. Multiply: 39/51 x 26/50 = 1,014/2,550, just under 40%. Drawing with replacement would give 3/4 x 1/2 = 37.5%.

Why does the first card cost nothing?

Picture three guests arriving at a party with four dress colours in the wardrobe, and you want no two to match. The first guest cannot clash with anyone. A condition about the cards differing only bites from the second card on, so the first card contributes a factor of 1 and you start counting from card two. Candidates who write 13/52 for the first card have fixed a particular suit and then have to multiply by the number of suit orders to recover, which is where the slips happen.

How do you check it a second way in the time?

Count unordered hands. Choose which three suits appear, 4 ways, then one card from each, 13 cubed, and divide by all three-card hands, 52 choose 3. That is 4 x 2,197 = 8,788 over 22,100, which is the same 0.3976. In a speed round you do not have time for both, but knowing the counting route exists lets you sanity-check the product: 0.765 x 0.52 is a little under 0.40.

What is a speed round actually testing?

Thirty questions in forty-five minutes cannot all be worked in full. The skill being tested is choosing the shortest correct route and estimating the product well enough to pick from the options. Here, 39/51 is about 0.76 and 26/50 is 0.52; 0.76 x 0.52 is about 0.40, which eliminates every other option before you finish the exact fraction.

Take X equal to -1, 0 or 1 with probability 1/3 each, and Y = X squared. Y is fixed by X, so they are as dependent as variables can be. But E[X] = 0 and E[XY] = E[X cubed] = (-1 + 0 + 1)/3 = 0, so the covariance E[XY] - E[X]E[Y] is zero. Correlation only measures straight-line association, and this relationship is a V.

What does correlation actually measure?

Think of a thermostat that runs the air conditioner hard on very hot days and the heater hard on very cold days. Energy use is clearly driven by temperature, but a straight line through the data is flat: high use at both ends, low in the middle. Correlation measures only how well a straight line summarises the relationship, so any symmetric U or V shape can have zero correlation while being completely determined. Independence is the stronger claim that knowing X tells you nothing about Y at all.

How do you show the covariance is zero in one line?

Write the definition and let symmetry do the work. Cov(X, Y) = E[XY] - E[X]E[Y], and with Y = X squared the first term is E[X cubed], which is zero for any X symmetric about zero; the second term has E[X] = 0 in it. With the three-point version you can even list the products: -1 x 1, 0 x 0 and 1 x 1 add to zero. Yet P(Y = 0 given X = 0) is 1 while P(Y = 0) is 1/3, which is dependence in plain sight.

Why does a quant interviewer care?

Because models quietly substitute zero correlation for no relationship. A delta-hedged option book gains or loses roughly with the square of the underlying's move, so its daily P and L can show near-zero correlation with the market while being entirely driven by it. The same holds for a volatility strategy or any payoff with a kink. The one case where zero correlation does mean independence is when the pair is jointly normal, which is worth adding before the interviewer asks.

91/216 of a rupee, about 42 paise. A fair price for a ticket paying Rs 1 is the probability of winning. The fastest route is the complement: the chance of no six on three dice is 5/6 x 5/6 x 5/6 = 125/216, so the chance of at least one six is 1 - 125/216 = 91/216, or 0.421. Adding 1/6 three times gives 50 paise and overcounts.

Why is the price just a probability?

If a raffle pays Rs 100 and you win one time in four, playing many times earns you Rs 25 a ticket on average, so Rs 25 is the break-even price. A ticket paying Rs 1 on some event is worth exactly the probability of that event, because that is its average payout. Trading firms phrase probability questions as prices on purpose: it makes you answer in the units a desk uses, and it sets up the next question, which is where you would quote a bid and an offer.

Why is at least one a signal to use the complement?

At least one six covers exactly one six, exactly two, or three, and each needs its own count. The opposite event, no six at all, is a single clean case: every die avoids six, and independent dice multiply. So the complement takes one line. Adding 1/6 + 1/6 + 1/6 fails because the three events overlap: a roll of 6, 6, 2 is counted once for the first die and again for the second. With ten dice the same mistake would give a probability above 1.

What does a trader add after the number?

A fair value is the centre of a market, not the market itself. A market maker quotes a bid below 42 paise and an offer above it, and the width depends on how confident they are in the number and how much risk one ticket adds to their book. Here the fair value is exact, so a tight market such as 40 bid, 44 offer is defensible. Saying that sentence turns a probability answer into a trading answer, which is what the question format is inviting.

3/32 = 0.09375, 7/16 = 0.4375 and 11/64 = 0.171875. Every denominator here is a power of two, so the unit fraction is a chain of halvings: 1/2 = 0.5, 1/4 = 0.25, 1/8 = 0.125, 1/16 = 0.0625, 1/32 = 0.03125, 1/64 = 0.015625. Find the rung, then multiply by the numerator. Each decimal ends exactly, because 2 divides a power of 10.

Why do powers of two give clean decimals?

Think of cutting a one-metre ribbon in half again and again: 50 cm, 25 cm, 12.5 cm, 6.25 cm. Each cut adds at most one digit to the length. A fraction terminates in decimal exactly when its denominator has no prime factors other than 2 and 5, so every power-of-two fraction ends, and 1 over 2 to the n has exactly n decimal places. That tells you before you start that 11/64 will have six digits after the point.

How do you do 11/64 without losing a digit?

Split the numerator into pieces you already know. 11/64 is 8/64 + 2/64 + 1/64, which is 1/8 + 1/32 + 1/64: 0.125 + 0.03125 + 0.015625 = 0.171875. Or take 11 x 0.015625 as 10 x 0.015625 plus one more, 0.15625 + 0.015625. Either way you add numbers you have memorised instead of dividing. 7/16 works the same way as 1/2 - 1/16, 0.5 - 0.0625 = 0.4375.

Why do trading firms test this?

Some bond and futures markets have long quoted prices in 32nds and 64ths of a point, and option deltas and odds come up as fractions all day. A trader who converts 3/32 at the speed of reading reacts to a price while a slower colleague is still dividing. The same round usually mixes in products such as 38 x 42, which is 40 squared minus 2 squared, 1,596: the test is spotting structure that turns long arithmetic into one step.

Both are worth Rs 1. The call pays 0, 0, 0, 1, 2 and 3 on faces 1 to 6, which sum to 6, so its average payoff is 6/6 = Rs 1. The put pays 3, 2, 1, 0, 0 and 0, also summing to 6, so it is worth Rs 1 too. They match because the die is symmetric: face x and face 7 - x are equally likely, which turns one payoff into the other.

Why is the price just an average of payoffs?

If a friend offers you a game where you win the number of rupees shown on a die above 3, playing a hundred times earns about Rs 100: some rolls pay nothing, some pay 1, 2 or 3. With no interest and no risk premium in a dice game, the fair value of any payoff is its average over the equally likely outcomes. That is exactly how an option is priced in the simplest world: list the states, write the payoff in each, weight by the probabilities and add.

Why is 0.50 the tempting wrong answer?

Because it plugs the average face, 3.5, into the payoff: 3.5 - 3 = 0.5. An option's value is the average of the payoff, not the payoff at the average, and because the payoff is floored at zero, the average payoff is always at least the payoff at the average. The floor throws away the losing faces, which is the whole point of owning an option. That gap, here Rs 0.50, is what traders pay for, and it grows with how spread out the outcome is: this is convexityA payoff that curves upward, so averaging over uncertain outcomes gives more than the payoff at the average outcome. in a single roll.

Where is put-call parity in this?

Take a call and a put with the same strike, 3. Owning the call and selling the put pays max(X - 3, 0) - max(3 - X, 0) = X - 3 on every face, so its value must be the average of X - 3, which is 0.5. The put struck at 3 pays 2, 1, 0, 0, 0, 0, worth 1/2, and 1 - 0.5 = 0.5 as parity requires. Saying this unprompted shows the interviewer you see a dice game as a model of a real options book, which is why the question is asked.

The next term is 312211. Each term describes the one before it, read aloud as runs of equal digits. 1 is one 1, written 11. 11 is two 1s, written 21. 21 is one 2 and one 1, written 1211. 111221 is three 1s, two 2s and one 1, written 312211. The sequence is known as look-and-say.

Why does no arithmetic rule fit?

Try the usual moves first, as you would with any number series: differences, ratios, squares. 11 minus 1 is 10 and 21 minus 11 is 10, but 1211 minus 21 is 1,190, and the pattern dies. When the gaps jump like that, stop treating the terms as quantities. These terms are strings of digits, not numbers, and the rule works on the digits one run at a time. Think of reading a phone number to someone on a bad line: you say double two, triple five. Each term here is the previous one said that way and then written down.

How do you produce the next term without slipping?

Mark the runs first, then speak each run as a count followed by its digit. The slip people make is merging two runs of the same digit that have something else between them. In 111221 the three 1s at the front and the single 1 at the end are separate runs, so the reading ends with one 1, not four 1s. Bracket the runs on paper or in your head and read left to right: three 1s, two 2s, one 1 gives 312211. One more turn for practice: 312211 reads one 3, one 1, two 2s, two 1s, which writes as 13112221.

What can you say about the sequence beyond the next term?

Two facts show you understand the rule rather than just ran it. First, starting from 1, no digit above 3 ever appears. Neighbouring runs always hold different digits, so four equal digits can never sit in a row, no run is longer than three, and no count above 3 is ever written. Second, the terms grow at a steady rate: the 40th term has 63,138 digits, and each term is about 1.30 times as long as the one before, a ratio John Conway studied. Run the rule in a short loop to see both, which is also the coding follow-up many firms ask next.

On a timed screen, the point of a question like this is speed at dropping a wrong frame. Candidates who spend a minute hunting for a formula lose the minute; candidates who ask what else the digits could be doing find the rule in seconds.

Go second. The player facing 1, 5, 9, 13, 17 or 21 matches loses against good play, and 21 is on that list. Whatever your opponent takes, take 4 minus that, so every round removes exactly 4. They then face 17, 13, 9, 5 and finally 1, and are forced to take the last match.

How do you find the losing positions?

Start from the end of the game, the way you would plan the last few stops of a journey before the first. With 1 match in front of you, you must take it and lose. With 2, 3 or 4, you take enough to leave exactly 1 and win. With 5, every move leaves 2, 3 or 4, each a winning spot for the other player, so 5 loses. A position is losing when every move from it hands your opponent a winning position, and here that happens every 4 matches: 1, 5, 9, 13, 17, 21.

Why does answering with 4 minus k always work?

Your opponent can take 1, 2 or 3; you can always take 3, 2 or 1 in reply. The pair of moves removes exactly 4 matches whatever they chose, so you control the count at the end of every round. From 21 the rounds end at 17, 13, 9 and 5, and then your opponent faces a single match. The number 4 is the maximum take plus one; that is where the period comes from.

A short check you can say aloud: the game has only 21 positions, and marking each as winning or losing from the bottom up, a position wins if any move reaches a losing one, reproduces the list 1, 5, 9, 13, 17, 21. If the pile had been 20, you would go first and take 3 to leave 17. This type of game has a backward inductionSolving a game by working out the best move at the last step first, then the step before, back to the start. solution, and the interviewer mainly wants to hear you build it from the end.

Centre it at 50 and quote around 46 at 54. The fair value is exactly 50. The standard deviation is √(100 x 0.5 x 0.5) = 5, so settlement lands between 45 and 55 about 73% of the time. A market 4 either side of fair earns 4 per lot on any trade, loses on a single sale at 54 only 18% of the time, and leaves room to move the quote if the other side seems to know something.

Where does the centre come from, and what sets the width?

A shopkeeper selling mangoes by the dozen knows the fair price; the margin he adds depends on how much the price of the next crate can swing and on whether the buyer knows something he does not. The centre of your market is the expected value, and the width is a choice about risk and information, scaled by how much the outcome can move. Here the expected value is 100 x 0.5 = 50, and nobody can know more than you about fresh flips of a fair coin, so the width is about risk alone.

How do you justify 46 at 54 rather than 49 at 51?

Use the standard deviation as the ruler. The count has variance 100 x 0.5 x 0.5 = 25, so a standard deviation of 5. A quote 4 either side of fair earns 4 on each lot traded, against a settlement that typically moves 5, so every trade has an edge worth a large fraction of its risk. If someone buys at 54, you lose only if the count finishes at 55 or more, about 18% of the time. A tight 49 at 51 earns 1 per lot and a sale at 51 loses whenever the count reaches 52, about 38% of the time. Tighter wins more trades and earns less on each; in an interview game, start around one standard deviation wide and tighten as you learn.

Then say how you would react to trades, because that is the follow-up. If the interviewer lifts your 54 again and again, either they are testing your nerve or they know something, perhaps that the coin is not fair or that some flips are already done. Repeated one-way trading is information: move your market toward it and cut your size, rather than defending 50. The limitation of the simple answer is exactly that it assumes nobody knows more than you.

56 x 56 = 3,136 and 73 x 74 = 5,402. For 56 squared, split it as 50 + 6: 2,500, plus two strips of 300, plus 36. For 73 x 74, anchor both on 70: 4,900, plus 70 x 7 = 490, plus 3 x 4 = 12. A second route checks each: (60 - 4) squared = 3,136, and 73.5 squared minus a quarter = 5,402.

What is the shortcut for squaring a two-digit number?

Tiling a floor that is 56 tiles on each side, you would lay the big 50 by 50 block first, then two thin strips along the edges, then a small corner. Splitting a number into a round base plus a small part turns one hard product into one easy square and a few small ones: (a + b) squared = a squared + 2ab + b squared. For 56: 2,500 + 2 x 300 + 36 = 3,136. You can also go down from the next round number: (60 - 4) squared = 3,600 - 480 + 16, again 3,136.

What changes when the two numbers differ, as in 73 x 74?

When two numbers share a tens digit, anchor both on it. For (70 + 3)(70 + 4), the product is 70 squared, plus 70 times the sum of the units, plus the product of the units: 4,900 + 490 + 12 = 5,402. The midpoint route gives the same: numbers equally spaced around 73.5 multiply to 73.5 squared minus the square of the half gap, 0.25, and 73.5 squared is 4,900 + 490 + 12.25. Another quick path: 73 x 74 = 73 squared + 73 = 5,329 + 73.

Timed tests reward a fixed routine more than cleverness. Pick one decomposition, say the partial products in order, and check with a second route only if time allows. The last digit is a free check: 6 x 6 ends in 6 and 3 x 4 ends in 2, so 3,136 and 5,402 pass. A good habit is to sanity check the size too: 56 squared must sit between 50 squared, 2,500, and 60 squared, 3,600.

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.