Quant puzzles, solved step by 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.

Fire into the air. B and C each target the other, the bigger threat, so while both live nobody shoots at you. Aiming in the air gives survival of 2/3 x 3/7 + 1/3 x 1/3 = 25/63, about 39.7%. Aiming at C gives 31.2%, because a hit leaves you in a duel with B shooting first. Aiming at B gives 26.5%, because a hit leaves C, who never misses, to shoot you.

Who does everyone else aim at?

Start with the stronger players, because their choices fix yours. B aims at C, because if B shot you instead, C would kill B next turn for certain; C aims at B, the more dangerous of the two remaining threats. So while all three are alive, nobody is shooting at you. Think of two large firms in a price war while a small competitor stays out of it: the small firm's best move is often to let the giants weaken each other.

How do you work out the two-player duels?

Against B with you shooting first, you win if you hit now, or if both miss and the same duel restarts. Call your survival x: x = 1/3 + (2/3)(1/3)x, so x = 3/7; if B shoots first, you must survive B's first shot, 1/3 of the time, giving 1/7. Against C you get exactly one shot, since C never misses: 1/3 if you shoot first, 0 if C does. Now combine. In the air: B hits C two times in three, giving you the 3/7 duel; otherwise C kills B and you get your one shot at C, 1/3. Total 25/63.

What is the general lesson?

In a game with several players, weakening one rival can hurt you if it frees the strongest remaining player to turn on you. Your best shot is the one that keeps the others focused on each other. Say the limitation too: the answer depends on the hit rates and the order. Change the order of shooting, or let C aim at you, and the tree changes; the interviewer will often change a number or the order to see whether you rebuild the tree or repeat the slogan.

10 flips for HTH, against 8 for HTT. Track how much of the pattern you currently hold: nothing, H, or HT. For HTH, a tail after HT wrecks everything and you restart from nothing. For HTT, a head after HT breaks the pattern, but that head is itself a fresh start, so you keep an H. Solving the three expected-wait equations gives 10 and 8.

Why do equal probabilities give unequal waits?

Think of a combination lock where a wrong digit sometimes resets you to zero and sometimes lets you keep part of your progress. Two combinations can be equally likely to be dialled at random yet take different times to reach. Each three-flip window is HTH or HTT with the same 1/8 chance, but the windows overlap, and HTH occurrences tend to arrive in clusters, such as HTHTH, which spaces out the first appearance. The waiting time depends on where a near miss leaves you.

How do you set up the equations?

Let E0, E1 and E2 be the expected remaining flips when you hold nothing, H and HT. Each flip costs one and moves you to the next state with probability one half each way, so each state's wait is 1 plus the average of the two states it can move to. For HTH: E0 = 1 + (E1 + E0)/2, E1 = 1 + (E1 + E2)/2, E2 = 1 + (0 + E0)/2. Solving gives E2 = 6, E1 = 8, E0 = 10. For HTT only the last equation changes, to E2 = 1 + (0 + E1)/2, and the answers become 4, 6 and 8.

Is there a shortcut an interviewer will accept?

Yes, the overlap rule, which comes from a fair-bet argument known as the ABRACADABRA methodA martingale argument in which gamblers arriving each flip bet on the pattern, used to compute expected waiting times for patterns.. For a fair coin, the expected wait is the sum of 2 to the k over every k where the first k flips of the pattern equal the last k. HTH matches itself at length 3 and at length 1, the single H, giving 8 + 2 = 10. HTT matches only at length 3, giving 8. HHH matches at 1, 2 and 3, giving 14. Derive the states first, then offer the rule as the check.

f'(x) = x to the x times (ln x + 1), and the minimum is at x = 1/e, about 0.368, where f is about 0.692. Take logs: ln f = x ln x. Differentiating, f'/f = ln x + 1, so f' = x to the x (ln x + 1). The derivative is zero when ln x = -1, that is x = 1/e, negative before it and positive after, so this is a minimum.

Why do both standard rules fail?

The power rule, n x to the (n - 1), treats the exponent as fixed; the exponential rule, a to the x times ln a, treats the base as fixed. In x to the x both the base and the exponent move, so neither rule applies on its own, and each one gives half of the right answer. Indeed the correct derivative is the sum of the two: x times x to the (x - 1), which is x to the x, plus x to the x ln x. That sum is a quick check on your final answer.

How does taking logs make it routine?

Think of converting a messy multiplication into addition before doing it, the way a slide rule does. Write ln f = x ln x; the right-hand side is a product of two simple functions, and the product rule gives ln x + x times 1/x = ln x + 1. The left side differentiates to f'/f by the chain rule, so f' = f (ln x + 1). This is logarithmic differentiationDifferentiating the logarithm of a function instead of the function, then multiplying back; useful when the variable sits in an exponent., and it works for any function of the form g(x) to the h(x).

How do you confirm it is a minimum and state the value?

Check the sign of ln x + 1, since x to the x is always positive. For x below 1/e, ln x is below -1 and the slope is negative; above 1/e it is positive, so the function falls and then rises: a minimum. The value is (1/e) to the (1/e) = e to the (-1/e), about 0.692. Also say what happens at the edges: as x shrinks towards zero, x ln x tends to zero, so x to the x tends to 1, and at x = 1 it is exactly 1 again.

They leave in the order 2, 4, 1, 3. Person 1 takes 4 and rejoins with 3; person 2 takes 3 and leaves; person 3 takes 4 and rejoins with 6; person 4 takes 2 and leaves. Then person 1 takes 3 and leaves, and person 3 needs two more visits. The shortcut: each person leaves in round amount divided by 4, rounded up, and ties go to whoever stood first.

How do you simulate it cleanly?

Model the line as a queueA first-in, first-out list: items join at the back and leave from the front, as in a real line. of pairs, person and amount still wanted. Pop the front, subtract the lesser of the cap and what they still want, and if anything is left push them onto the back; otherwise record them as leaving. The four people take seven visits in all. This is the answer most interviewers expect first, and it is correct, but its cost grows with the total number of visits, which is the sum of each amount over the cap.

Is there a faster way than simulating?

Yes. Think of a canteen that serves one plate per person per pass: someone wanting three plates leaves on the third pass, whatever the others want. Person i leaves in round ceiling(a_i / k), and within a round the queue keeps its original order, so the exit order is simply the people sorted by their round number, ties broken by starting position. Here the rounds are 2, 1, 3 and 1, which sorts to 2, 4, 1, 3. That costs n log n, however large the amounts are, instead of the number of visits.

Why does queue order survive between rounds?

Everyone still waiting after a round rejoins in the same relative order they were served, because the queue is first in, first out. So round two serves the survivors of round one in their original order, and so on. That invariant is what lets you replace the simulation with a sort, and saying it out loud is what separates an answer that works from one you can defend. If a very large cap or tiny amounts made most people finish in round one, the sort still costs n log n, and a counting sort on round numbers can make it linear.

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.

90%. Each die starts at one in three. The chance of two sixes is 1/36 for the fair die, 1/4 for the loaded die and zero for the die with no six. Weight each by its prior: 1/108 for the fair die, 1/12 for the loaded die, nothing for the third. The loaded die's weight is nine times the fair die's, so its probability is 9/10.

What does the die that never shows six do to the answer?

A neighbour tells you a red car blocked the gate this morning. If one of your suspects owns only a blue scooter, that suspect is out, however likely they looked before. A hypothesis that cannot produce the evidence gets zero weight afterwards, no matter what its prior was. The no-six die could never give two sixes, so it drops out, and the question becomes a contest between the fair die and the loaded die, which started level at one third each.

How much does each surviving die's likelihood count?

Now compare how easily each remaining die produces what you saw. The fair die gives two sixes 1 time in 36. The loaded die gives a six half the time, so two in a row 1 time in 4. With equal priors, the posterior odds are just the ratio of the likelihoodsThe probability of the observed evidence under each hypothesis, before any prior is applied.: 1/4 against 1/36, which is 9 to 1. Nine parts in ten is 90%.

Check it by counting, the safer habit under pressure. Imagine 108 rounds of drawing a die and rolling it twice, 36 rounds with each die. The fair die gives two sixes once, the loaded die 9 times, the no-six die never. Of the 10 double sixes, 9 came from the loaded die, and the prior of one third cancels because every die got the same number of rounds. The loaded die's other faces, 1 in 10 each, never enter, because only sixes were seen.

161/36, about Rs 4.47. The larger face equals k in 2k minus 1 of the 36 equally likely outcomes: 1, 3, 5, 7, 9 and 11 cells for k from 1 to 6. Multiply each value by its count and add: 1 + 6 + 15 + 28 + 45 + 66 = 161. Divided by 36, that is 4.47, almost a full point above a single die's 3.5.

Why is the answer well above 3.5?

When two friends each suggest a restaurant and you always go with the better rated one, your average dinner beats either friend's average. Taking the larger of two draws pulls the result toward the top, because a low result survives only if both draws are low. A payout of 1 needs both dice on 1, one cell in 36. A payout of 6 needs just one six, and 11 cells in 36 contain at least one.

How do you count the cells without listing all 36?

Count the outcomes where the larger face is at most k: both dice must be at most k, which is k squared cells. The cells where the larger face is exactly k are k squared minus (k - 1) squared, which is 2k - 1. That is the L-shaped band in the grid: a new row and a new column, sharing one corner cell. The bands are 1, 3, 5, 7, 9 and 11, and they add to 36, which is the check that nothing was double counted.

A second route helps when the interviewer changes the dice. Add up the chance that the payout reaches each level: the payout is at least k unless both dice are below k, so the sum of 1 minus (k - 1) squared over 36, for k from 1 to 6, is 6 minus 55/36, which is 161/36 again. Two methods landing on the same fraction is the check worth saying out loud. By symmetry the smaller face averages 7 minus 4.47, about 2.53, and with three dice the larger face rises to 4.96.

Median about Rs 8.2; 95% interval roughly Rs 1.5 to Rs 45. Log returns add, so after 30 years the log of wealth has mean 30 x 0.07 = 2.1 and standard deviation 0.16 x √30 = 0.88. The median is e to the 2.1, about 8.2. The band is e to the power 2.1 plus or minus 1.96 x 0.88. In rupees it is lopsided, and the mean, about Rs 12, sits above the median.

Why work in log returns rather than percentage returns?

Pay rises compound: 10% and then another 10% is 21%, not 20%. Logs turn that multiplication into addition. Log returns add across years, so the 30-year log return is a sum of 30 yearly pieces, and a sum of independent pieces is close to normal. With mean 0.07 and standard deviation 0.16 a year, the sum has mean 2.1 and variance 30 x 0.16 squared, so a standard deviation of 0.16 x √30, about 0.876. The spread grows with the square root of time, not with time.

Why is the band so lopsided, and where does the mean sit?

Symmetric in the exponent means lopsided in rupees. Going 1.96 standard deviations down divides the median by e to the 1.72, a factor of 5.6; going the same distance up multiplies by 5.6. Dividing and multiplying by the same factor leaves Rs 6.7 of room below the median and Rs 37.3 above it. The same skew separates mean from median. The mean of a lognormalA variable whose logarithm is normally distributed; it is always positive and skewed to the right. variable is e to the power (mean plus half the variance), about Rs 12.0 here, because a few very good paths pull the average up while most paths finish below it.

Close with the limits. The 7% and 16% are illustrative inputs, not a claim about any real index. The calculation assumes independent years and constant volatility; real markets have fat tails and calm and stormy regimes, so treat the band as a floor on the true uncertainty. What the interviewer is testing is whether you scale the mean with t and the volatility with √t, and exponentiate only at the end. One useful extra: the chance the Rs 1 is worth less than Rs 1 is the chance the log falls below zero, about 0.8%.

86/99, about 86.9%. Track only the current streak: none, me on 1 or 2, you on 1 or 2. Each game I win with probability 1/3, you win with 1/6, and a draw, 1/2, resets the streak. Write my chance of taking the match from each state in terms of the others, solve the five equations, and the start state comes out at 86/99.

What is the state, and why is the running score not it?

A door lock that opens after three correct codes in a row does not care how many wrong codes came before the last mistake. The only thing that matters for the rest of this match is the current streak, so the states are: no streak, my streak of 1 or 2, and your streak of 1 or 2. Games won earlier, draws played, games elapsed: all irrelevant once the streak is known. That is the Markov propertyThe future depends on the past only through the present state., and spotting it turns an infinite tree of game sequences into five numbers. Per game, I win with probability 1/2 x 2/3 = 1/3, you win with 1/2 x 1/3 = 1/6, and the rest are draws.

How do the equations go, and how do you solve them quickly?

Let x be my chance from the start, a1 and a2 from my streaks, b1 and b2 from yours. Every equation reads the same way: play one more game, three things can happen. A draw always returns to the start, a win for me always moves to a1 or one step up, and a win for you always moves to b1 or one step up. From a2 my next win ends the match in my favour; from b2 your next win ends it against me.

Solve by substituting: b2 is in terms of a1 and x, which gives b1 in the same terms; feed that into a2 and a1, and the start equation leaves one unknown. The answers: x = 86/99, a1 = 29/33, a2 = 10/11, b1 = 28/33, b2 = 8/11. Check the ordering: the further I am ahead, the higher the value, and every value sits between 0 and 1. Your chance is 13/99. A quick cross-check: three wins in a row is (1/3) cubed for me and (1/6) cubed for you, a ratio of 8 to 1, which would suggest about 89%; landing within two points of that rough race is a sign no term was dropped.