Quant puzzles, solved step by step

About 3.99. Give each face an indicator that equals 1 if that face appears at least once. A given face is missed on all six rolls with probability (5/6) to the 6th, about 0.335, so it appears with probability 0.665. The expected count of distinct faces is the sum of the six indicators' expectations: 6 x 0.665 = 3.99. No case listing is needed.

Why not list the cases?

You could work out the chance of exactly one, two, up to six distinct faces and average them. It works, but it needs Stirling numbersCounts of the ways to split a set of items into a given number of non-empty groups; they appear when counting surjections. or a lot of careful counting, and it is easy to slip. Linearity of expectation lets you ignore how the faces interact: the expected total of several indicators is the sum of their expectations, whether or not they are independent. Here the six indicators are clearly dependent, since seeing many faces leaves fewer rolls for the others, and it does not matter at all.

How does the indicator trick work step by step?

Think of a teacher counting how many of six friends turn up to a party. Instead of listing every guest list, she asks for each friend separately how likely that friend is to come, then adds. Write the count as I1 + I2 + ... + I6, where I_k is 1 if face k appears; take expectations; each E[I_k] is just the probability face k appears. Face k is missed on one roll with probability 5/6, on all six with (5/6) to the 6th, 0.335. So each indicator averages 0.665 and the total averages 3.99.

Where does this pattern reappear?

The same shape answers how many distinct birthdays a group of n people has, how many of n hash buckets get used, and how many different stocks a random sample of trades touches. For n faces and n rolls, the expected share of faces seen is 1 - (1 - 1/n) to the n, which tends to 1 - 1/e, about 63.2%, as n grows. Six faces give 66.5%, already close. The exact enumeration of all 46,656 rolls gives a mean of 3.9906, the same number.

Mean 2, variance 4, and the probability of exceeding 2 is e to the -1, about 36.8%. Each Z squared has mean 1 and variance 2, so the sum has mean 2 and variance 4. The sum is a chi-squared with two degrees of freedom, which happens to be exactly an exponential with mean 2. Its tail beyond t is e to the -t/2, so beyond 2 it is e to the -1.

How do you get the mean and variance without the distribution?

E[Z squared] is the variance of Z, which is 1. For the variance of Z squared you need the fourth moment: E[Z to the 4] is 3 for a standard normal. So Var(Z squared) = 3 - 1 = 2, and because Z1 and Z2 are independent the variances add, giving a mean of 2 and a variance of 4. Say the fourth moment of 3 aloud; it is the number interviewers check you know, and it is why normal kurtosis is quoted as 3.

Why is this particular sum exponential?

Think of a dart thrown at a board where both the horizontal and vertical errors are independent standard normals. The joint density depends only on the distance from the centre, so the dart's direction is uniform and all the information is in the radiusThe distance of the point (Z1, Z2) from the origin, the square root of Z1 squared plus Z2 squared.. Switching to polar coordinates, the chance that the squared distance exceeds t is e to the -t/2, which is the tail of an exponential with mean 2. At t = 2 the answer is e to the -1.

Why is 50% the tempting wrong answer?

Because 2 is the mean and people read the mean as the middle. For a right-skewed distribution the mean sits above the median, so less than half the mass lies beyond it; here the median is 2 ln 2, about 1.39. A simulation of 200,000 pairs gives a mean of 1.997, a variance of 3.96 and a tail share of 0.368, matching the exact results. The same polar trick is what powers the Box-Muller method for generating normal random numbers.

The call is worth 10. It pays 20 if the stock goes to 120 and 0 at 80. Half a share pays 60 or 40, so half a share with a loan of 40 pays 20 or 0, exactly the call. That portfolio costs 50 - 40 = 10 today. If the call traded at any other price, you could buy the cheap one and sell the dear one for a riskless profit, so no probability is needed.

How do you build the copy?

Match the swing first. The call's payoff moves by 20 between the two states while the stock moves by 40, so the copy needs 20/40 = 0.5 of a share: that ratio is the option's deltaHow much an option's value changes for a one-unit change in the underlying price; here, the number of shares that copies the option.. Half a share is worth 60 or 40 at the end, which is 40 more than the call in both states. Borrow 40 today, repay 40 at the end with zero interest, and the copy pays exactly 20 or 0.

Why does the chance of the up move not matter?

Think of a shop selling a bundle of two items that you can also buy separately. The bundle's price is pinned by the parts, whatever you think about how useful the items are. The call is a bundle of half a share and a loan; the share price already reflects everyone's views about the up move, so the option inherits them and adds none of its own. A 90% view is a reason to hold the stock itself, not a reason to pay more for the call than its parts cost.

What would you do if the call traded at 12?

Sell the dear thing and buy the cheap one. Sell the call for 12, buy half a share for 50 and borrow 40, a net cash inflow of 2 today; at the end the portfolio pays exactly what you owe on the call in either state. The 2 is kept whatever happens. The weight q = 0.5 that reproduces the price is called the risk-neutral probability, but it is a pricing weight backed out of the stock price, not a forecast. Say that distinction; interviewers listen for it.

1/11, about 9.1%. Any particular sequence with five reds and five blues has probability 5! x 5! / 11!, whatever the order, because the numerators just count up the reds and blues separately. There are 10 choose 5 = 252 such sequences, and 252 x 5! x 5! / 11! = 1/11. The same algebra gives 1/11 for every count from 0 to 10.

Why is this not ten coin flips?

Think of a new café and its first customers. If the first few visitors like it and bring friends, it fills up; if they do not, it stays empty. Early luck compounds. In this urn each red draw adds a red ball, so it raises the chance of the next red: the draws reinforce each other, and runs that start lopsided tend to stay lopsided. That spreads the count of reds far more widely than independent coin flips, which bunch around five.

Why does the order of the draws not matter?

Write out one sequence, say five reds then five blues. The chances are 1/2, 2/3, 3/4, 4/5, 5/6 for the reds, then 1/7, 2/8, 3/9, 4/10, 5/11 for the blues. Shuffle the order and the denominators are still 2 up to 11, while the red numerators still run 1 to 5 and the blue ones 1 to 5, so every arrangement of five reds and five blues has the same probability, 5! x 5! / 11!. Draws whose joint probability ignores order are called exchangeableA sequence of random variables whose joint distribution does not change when you reorder them, even though they need not be independent., and that is the property doing the work.

Is there a picture that makes 1/11 obvious?

Yes. This urn behaves exactly as if nature first picked a hidden red probability p uniformly between 0 and 1, and then flipped ten independent coins with that p. Averaging a binomial over a uniform p gives every count the same weight, 1/(n + 1). The urn is also a small model of momentum and of market share: an early lead makes further gains more likely, so the final split is highly uncertain, even though each step looks like a fair draw.

10 is more likely: 27 ways out of 216 against 25. The equally likely outcomes are the 216 ordered rolls, not the unordered combinations. A combination of three different faces covers six ordered rolls, one with a pair covers three, and a triple covers one. Total 9 includes 3 + 3 + 3, which counts once, and one fewer all-different combination, so it loses two orderings to 10.

Why are combinations the wrong thing to count?

Think of dealing two cards and asking whether a pair of kings or a king with a queen is more likely. There is one combination of each, but king and queen can arrive in either order while two kings are just two kings. Probability comes from counting outcomes that are equally likely, and with dice those are the ordered rolls: first die, second die, third die, 6 x 6 x 6 = 216 of them. Unordered combinations bundle different numbers of those outcomes, so counting combinations gives the wrong weights.

How do you count the orderings fast?

Classify each combination by its repeats. Three different faces give 3! = 6 orders, a pair gives 3 orders, one for each position of the odd die, and a triple gives 1. For 9: 1 2 6, 1 3 5 and 2 3 4 are all different, 18; 1 4 4 and 2 2 5 are pairs, 6; 3 3 3 is a triple, 1; total 25. For 10: 1 3 6, 1 4 5 and 2 3 5 give 18; 2 2 6, 2 4 4 and 3 3 4 give 9; total 27. So 10 comes up 12.5% of the time and 9 only 11.6%.

Is there a shortcut that avoids listing?

Yes: symmetry. Replacing each face x by 7 - x maps a total of t to 21 - t, so the distribution of three dice is symmetric about 10.5, and 10 and 11 are the two most likely totals, each 27/216. Anything further from 10.5, including 9, must be less likely or equal; a quick count confirms it is 25. Historically this is the question gamblers put to Galileo, who answered it by counting ordered outcomes, which is still the method.

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.

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.