Quant puzzles, solved step by step

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.

Rs 4.20. Every roll pays its face, averaging 3.5, and with probability 1/6 the game then starts again, worth the same V. So V = 3.5 + V/6, which gives (5/6)V = 3.5 and V = 4.2. A second route agrees: the number of rolls averages 1/(5/6) = 1.2, and each averages 3.5, so 1.2 x 3.5 = 4.2.

Why write the game in terms of itself?

A pass that gets a free renewal each time you use it on a lucky day is worth its first use plus, on lucky days, another pass exactly like it. When a game can repeat with no memory, the value after the repeat is the value of the whole game, so one equation replaces an infinite sum. Here, after a six you are paid 6 and then face precisely the game you started with, worth V.

How do you set up the equation without slipping?

Condition on the first roll. With probability 5/6 it shows 1 to 5, averaging 3, and you stop. With probability 1/6 it shows 6: you collect 6 and then expect V more. So V = (5/6) x 3 + (1/6) x (6 + V), which simplifies to V = 3.5 + V/6, and V = 3.5 x 6/5 = 4.2. The form 3.5 + V/6 is worth saying: every roll pays 3.5 on average, and one time in six you get another go at the whole game.

Check by counting rolls. The chance a roll triggers another is 1/6, so the number of rolls averages 1/(1 - 1/6) = 1.2, and each roll averages 3.5 whatever came before, giving 1.2 x 3.5 = 4.2. Summing the series directly, k sixes and then a stop, also lands on 4.2, but the recursion gets there in two lines. The limitation to name if asked to price it: 4.2 is a fair value for one play; the payout has a long right tail, since two sixes in a row, one time in 36, already pay at least 13.

At least one six in four rolls is more likely, 51.8% against 49.1%, so take that side. The chance of no six in four rolls is (5/6)^4 = 625/1296, and the chance of no double six in 24 rolls is (35/36)^24. At even money the six earns about 3.5 paise per rupee staked; the double six loses about 1.7 paise and needs 25 rolls to become favourable.

Why does the obvious 2/3 come out the same for both bets?

The old gamblers' shortcut, sometimes called the rule of proportion, says that if one try succeeds with chance p, then n tries succeed with chance n times p. Four rolls at 1/6 gives 4/6, and 24 rolls at 1/36 gives 24/36, the same 2/3. This puzzle is the one the Chevalier de Mere is said to have brought to Pascal, because his winnings disagreed with the shortcut. Adding chances only works for events that cannot happen together, and two sixes in four rolls can. The shortcut counts a hand with two sixes twice, and that is why it breaks down completely after seven rolls, where it would claim 7/6.

A family checking the weather forecast makes the same slip: three days at a 40% chance of rain do not make a 120% chance of a wet weekend. The right question is the chance that it stays dry on all three days, 0.6 cubed, about 22%, so rain on at least one day is about 78%.

How do you compute each chance in your head?

Go through the complement. At least one success is one minus the chance that every try fails, and independent failures multiply. No six in four rolls is (5/6)^4 = 625/1296, which is just under one half, so the six wins 671/1296, or 0.5177. For the double six, use the approximation that (1 - 1/36)^24 is close to e to the power -24/36, which is e^(-2/3), about 0.513; the exact figure is 0.5086, so the double six comes up only 0.4914 of the time.

What edge does each side carry at even money?

At even money you win a rupee with probability p and lose a rupee otherwise, so the expected gain per rupee is p - (1 - p) = 2p - 1. The six gives 2 x 0.5177 - 1 = +0.0355, about 3.5 paise per rupee, while the double six gives -0.0172. That is a thin edge: over 100 bets of Rs 1 the expected profit is about Rs 3.55 with a standard deviation near Rs 10, so it takes thousands of bets before the edge shows reliably. The break-even for the double six is the smallest n with (35/36)^n below one half, which is 25 rolls, giving 0.5055. The limitation: the dice must be fair and independent; any bias in the dice swamps an edge this size.

e^pi is larger: about 23.14 against 22.46 for pi^e. Take logs of both and divide by e x pi, which turns the question into comparing ln e / e with ln pi / pi. The function ln x / x rises up to x = e and falls after it, so its value at e beats its value at any other number, pi included. Undo the steps and the order holds.

How do you turn two awkward powers into one comparison?

When two people race on different tracks, you compare them by converting to the same distance. Here the base and the exponent both differ, so convert each number into a common form. Take logarithms and divide by e x pi: e^pi against pi^e becomes ln e / e against ln pi / pi, the same function evaluated at two points. Logs and division by a positive number both preserve order, so whichever side wins the new comparison wins the original.

Why is ln x / x largest at e?

Differentiate: the derivative of ln x / x is (1 - ln x) / x squared. It is positive while ln x is below 1 and negative once ln x passes 1, so the function climbs until x = e and falls after. Its largest value, 1/e, occurs only at x = e, so ln pi / pi must be smaller, and therefore e^pi beats pi^e. Equivalently, x^(1/x) peaks at e with value 1.4447, while pi^(1/pi) is 1.4396.

There is a second proof that needs no calculus beyond one inequality. For any x other than zero, e^x is greater than 1 + x, because the exponential curve lies above its tangent line at zero. Put x = pi/e - 1, about 0.156: then e^(pi/e - 1) is greater than pi/e, so e^(pi/e) is greater than pi, and raising both to the power e gives e^pi greater than pi^e. Give that one if the interviewer asks for a proof without derivatives.

Notice how close the race is: the two values of x^(1/x) differ by only 0.0050, because pi sits near the flat top of the curve. That is why rough estimation is risky here and a proof is needed: 23.14 and 22.46 differ by about 3%. The same argument settles a whole family: for any two numbers a and b with e at most a, and a less than b, a^b is greater than b^a, which is why 3^4 = 81 beats 4^3 = 64.

15. The first six takes 6 rolls on average, so 5 rolls come before it. Each of those rolls is known not to be a six, so it is uniform on 1 to 5 and averages 3. Expected count times expected size gives 5 x 3 = 15. The check: all rolls including the six average 6 x 3.5 = 21, and taking off the final six leaves 15.

How many rolls come before the six?

Picture waiting at a stop where each minute a bus arrives with chance 1 in 6. On average you wait 6 minutes, and the sixth is the one where it comes. The number of rolls up to and including the first six is geometric with mean 1/p = 6, so the number strictly before it is 5. That is the first factor. Most candidates get this far; the loss comes in the second factor.

Why is each of those rolls worth 3 and not 3.5?

Because you are told something about them. Every roll before the stopping six is, by definition, not a six, so its distribution is the die conditioned on 1 to 5, which averages exactly 3. Using 3.5 gives 17.5, the most common wrong answer. It is the same slip as averaging the income of people who did not win a prize with everyone's income, prize winners included.

How do you check 15 a second way in the room?

Two checks, both fast. The recursion: with chance 5/6 the next roll is not a six, adds 3 on average and you are back where you started, so E = (5/6)(3 + E), which solves to 15. The full sum: Wald's identityFor a stopping time N that does not look into the future, the expected sum of N independent identical draws equals E[N] times the mean of one draw. applied to every roll including the six gives 6 x 3.5 = 21, and the last roll is always exactly 6, so the rest must average 15. Say both; the second one shows you understand why the conditional mean is 3.

V = S and V = K e^(-r(T-t)) satisfy it; V = S squared does not. The first is the stock itself and the second is a zero-coupon bond paying K at T, both traded assets that must earn r. S squared has gamma 2 and leaves (r + sigma squared) S squared unbalanced. Multiplying by e^((r + sigma squared)(T-t)) fixes it, which is the price of a claim paying S squared at expiry.

What is the equation actually saying?

Think of a household budget rule that any fair arrangement must obey: over one day, what you hold must earn the same as the same money in a savings account, once the risk has been hedged away. The Black-Scholes equation says that for a delta-hedged position, time decay plus the gamma term plus the financing of the hedge equals r times the value. Written in {term('greeks', 'Theta is the change in value with time, delta with the stock price, and gamma is the change in delta with the stock price.')}, it is theta + half sigma squared S squared gamma + r S delta = r V. A candidate price passes only if its greeks balance that line.

Why do the two that pass make sense as trades?

Anything that is itself a traded, self-financing asset must satisfy the equation, because the equation is only the statement that no hedged position earns more than r. V = S is just holding the stock: delta 1, no gamma, no decay, and the financing term rS matches rV. V = K e^(-r(T-t)) is a zero-coupon bond: it does not depend on S at all, and its value grows at exactly r as it approaches T. The stock and the bond are also the two pieces of the call price formula, which is why the check is worth a minute.

Why does S squared fail, and how do you repair it?

S squared has gamma 2, so the half sigma squared S squared gamma term adds sigma squared S squared, the delta term adds 2rS squared, and subtracting rV leaves (r + sigma squared) S squared with nothing to cancel it. A convex payoff gains from every move, so a fair price for it has to decay over time to pay for that gain, and S squared on its own has no decay. Try V = S squared times f(t): the equation forces f' = -(r + sigma squared) f, so the price of a claim paying S squared at T is S squared e^((r + sigma squared)(T - t)). At S = 100, r = 5%, sigma = 20% and one year, that is about 10,942, not 10,000, and the extra is the value of volatility.

1/sqrt(2 pi), about 0.399. Only the positive half contributes, and there you integrate z times the normal density. Because the derivative of the density is minus z times the density, the integral is simply the density's height at zero. So an at-the-money call on a normally distributed move is worth about 0.4 standard deviations of that move: roughly 0.4 x S x volatility x the square root of time.

Why is the answer not zero, and not one half?

Think of a shop that keeps the profit on good days and closes, losing nothing, on bad days. Its average day is better than the average of all days, because the bad days have been floored. max(Z, 0) throws away every negative outcome and keeps every positive one at its full size, so its mean is the positive half's contribution alone: the integral of z times the density from zero to infinity. That is not one half, which is only the chance of being positive; the size of each positive draw matters too.

How do you do the integral in one line?

Notice what differentiating the density gives. The derivative of e to the minus z squared over 2 is minus z times itself, so z times the density is the negative derivative of the density, and its integral from 0 to infinity is the density at 0 minus the density at infinity. The density at infinity is zero, and at zero it is 1/sqrt(2 pi). No tables, no substitution: the answer is 0.3989. Doubling it gives E[|Z|], about 0.798, which is the at-the-money straddle.

What does it say about an at-the-money option?

If the stock's move to expiry is roughly normal with standard deviation S x vol x sqrt(T), an at-the-money call pays the positive part of that move. So its value is about 0.4 x S x vol x sqrt(T), the rule of thumb option traders use to price at-the-money options in their heads. For a Rs 1,000 stock at 20% volatility and three months, sqrt(T) is 0.5 and the call is about 0.399 x 1,000 x 0.2 x 0.5 = Rs 39.9; the Black-Scholes value with zero rates is Rs 39.88. The rule loosens for long maturities and high volatilities, where the lognormal skew matters.

Stake 10% of the bankroll; that grows wealth by about 0.97% a bet, and twice Kelly grows it by only about 0.07%. Kelly is edge over odds: (2 x 0.4 - 0.6)/2 = 0.1. The growth rate is 0.4 ln(1.2) + 0.6 ln(0.9). At 20% the losses compound away almost the whole edge, and above about 20.4% the bankroll shrinks in the long run despite a positive expected value.

Why not bet as much as possible on a good bet?

Think of a shopkeeper with a profitable weekly sale who puts the entire shop's stock on it every week. The average week is good, but one bad week ends the business. With repeated bets, wealth multiplies, so what matters is the average of the log of each outcome, not the average outcome, and a big loss costs more in log terms than an equal gain earns. This bet has a clear edge: each rupee staked returns 0.4 x 2 - 0.6 = Rs 0.20 on average. The question is how much of that edge survives compounding at each stake size.

How do you get the Kelly fraction and the growth rate?

Stake a fraction f. A win multiplies wealth by 1 + 2f, a loss by 1 - f, so the growth per bet is g(f) = 0.4 ln(1 + 2f) + 0.6 ln(1 - f). Set the derivative to zero: 0.8/(1 + 2f) = 0.6/(1 - f), giving f = 0.1. The Kelly stake is the edge divided by the odds, (bp - q)/b = 0.2/2 = 10%. Plugging in, g = 0.4 x 0.1823 - 0.6 x 0.1054, about 0.97% a bet, so the typical path doubles its wealth roughly every 71 bets.

Why is overbetting so much worse than underbetting?

Near the peak the growth curve is close to a parabola, so the cost of a sizing error grows with its square. Half Kelly gives up only about a quarter of the growth, while twice Kelly gives up nearly all of it, and three times Kelly shrinks wealth at about 2.6% a bet. Real edges are estimated, not known, so a trader who thinks the win rate is 40% but faces 35% is already overbetting at the full 10%. That asymmetry is why desks size at a fraction of Kelly.

Pick THH; you win 7 times in 8. HHH can only win if the first three flips are all heads, which has chance 1/8. In any other run, the first HHH is preceded by a tail, and that tail with the next two heads spells THH, which is completed one flip before HHH. So THH wins every game except the one that opens with three heads.

Why is this not a fair race between two 1/8 sequences?

Think of two runners on the same track where one always starts one step ahead of the other on the only route to the finish. Their speeds are identical but the race is not even. In a race between patterns, what matters is not how often each appears but which one tends to appear first, and that depends on how the patterns overlap. THH is built from HHH's own first two heads with a tail placed in front, so it ambushes HHH whenever HHH has not already won at the start.

How do you prove 7/8 without a Markov chain?

Look at the first time HHH appears. If it does not start at flip 1, the flip immediately before it must be a tail, because otherwise an earlier HHH would already have appeared. That tail plus the first two heads of the HHH is THH, finished one flip before HHH. So HHH wins only if flips 1 to 3 are heads, chance 1/8, and THH wins otherwise: 7/8. The proof is one sentence, and interviewers want to hear it rather than a transition matrix.

What is the general lesson for the second mover?

This is Penney's gameA coin sequence race in which the second player, choosing after seeing the first, can always pick a sequence that wins more than half the time., and the second player always has an edge because the winning relation among three-flip sequences is not transitive: every sequence has another that beats it. The recipe: take the opponent's first two flips, and put in front of them the opposite of the opponent's second flip. Against HHH that gives THH at 7/8; against HTH it gives HHT at 2/3. In trading terms, a strategy that looks as good as any other in isolation can still lose systematically to one designed around it.

9 steps. Let E(k) be the expected steps left from state k. Each step costs one and lands on a neighbour with equal chance, so E(k) = 1 + (E(k - 1) + E(k + 1))/2, with E(3) = E(-3) = 0. The solution is E(k) = (3 - k)(3 + k): 9 from 0, 8 from plus or minus 1, 5 from plus or minus 2. A symmetric walk needs about the square of the distance to travel it.

Why is the answer not 3?

Think of someone pacing a narrow corridor, choosing left or right at random with every step. They do reach a wall eventually, but they spend most of their time doubling back. A random walk with no drift covers distance like the square root of time, so covering a distance of 3 takes about 3 squared steps, not 3. The answer 3 would only be right if the walk always moved in one direction. The structure to name in the room is a first-step recursion on the state, which turns the question into seven unknowns with simple links between them.

How do you solve the recursion quickly?

Use symmetry first: E(1) = E(-1) and E(2) = E(-2). Then three equations remain. From 2: E(2) = 1 + (E(1) + 0)/2. From 1: E(1) = 1 + (E(0) + E(2))/2. From 0: E(0) = 1 + E(1), because both neighbours are worth E(1). Substituting gives E(1) = 8 and E(0) = 9, and every value matches the product of the distances to the two walls. The second differences of E are all minus 2, which is why the answer is a parabola: each step must remove exactly one unit of expected time.

How do you check 9 a second way?

Use a martingaleA process whose expected next value, given everything so far, equals its current value: a fair game.. Squaring the position and subtracting the number of steps gives a fair game, because each step adds exactly 1 to the expected square. At the moment the walk stops its square is exactly 9, so the expected number of steps must also be 9. The same argument handles uneven walls: starting between -a and +b, the expected time is a times b. The cousin result is the hitting probability: from +1, the walk reaches +3 before -3 with probability 2/3, the distance to the far wall over the total width.