Derivatives Foundation puzzles, solved step by step

Roll while the pot is below 20 and stop once it reaches 20 or more. One more roll loses the pot with probability 1/6 and otherwise adds a face of 2, 3, 4, 5 or 6, which sum to 20. The expected change is (20 minus pot) over 6: +3.33 from an empty pot, +0.83 at 15, zero at 20 and negative beyond. The number of rolls so far is irrelevant; only the pot matters. Played this way, the expected bank from an empty pot is about 8.14.

What does one more roll actually buy you?

Think of a street game where you can keep picking envelopes that each add a few rupees to your winnings, but one envelope in six says lose everything. Whether to pick again depends on how much you already hold, not on how many envelopes you have opened. The expected change from one more roll is the chance of adding, five sixths, times the average addition, 4, minus the chance of ruin, one sixth, times the pot you would lose. That is (20 minus pot) over 6, and it is positive exactly while the pot is under 20.

Why is the one-step rule the whole answer here?

In many stopping problems you cannot trust a one-step look: a roll that loses value today might open a bigger gain tomorrow. Here the gain from one more roll only falls as the pot grows, so once rolling stops being worth it, it never becomes worth it again, and the one-step rule is optimal. You can check it by valuing the whole game for each stopping threshold: stopping at 20 (or 21, which is equivalent because the roll at exactly 20 is worth zero) gives an expected bank of about 8.14, stopping at 15 gives 7.85 and at 25 gives 8.00. Say the word monotone if you know it; say the reason either way.

What would a desk add to the textbook answer?

That the rule maximises expected value and nothing else. A player who cannot afford to lose the pot, or who is paid on a target rather than on the average, would stop earlier, and a trader who sizes positions knows that expected value is only the first thing to check. The limitation is the same in the puzzle and on the desk: expected value is right for a game you can play many times, and this game is played once.

For 10 the chance is 0.2893, and for large targets it settles at 2/7, about 0.286. The total lands on n only by landing on one of the six numbers before it and then rolling the exact gap, so p(n) is the average of the previous six values, starting from p(0) = 1. Running that recursion gives p(10) = 17,492,167/60,466,176. In the long run the total advances 3.5 per roll, so it lands on one number in 3.5.

How do you set up the recursion?

Think of climbing a staircase by jumping one to six steps at a time, each jump picked at random. To stand on step 10 you must at some point stand on one of steps 4 to 9 and then make exactly the right jump. The running total equals n only if it first equals one of n minus 1 down to n minus 6 and then the next roll is exactly the gap, and those six routes cannot both happen, so p(n) is the sum of p(n minus k) times 1/6 for k from 1 to 6. In words, each value is the average of the six before it, with p(0) = 1 because you start at zero and p of a negative number = 0.

Why does it settle at 2/7?

If you walk down a long street taking steps that average 3.5 paving stones, then over a kilometre you will have stepped on about one stone in every 3.5. A long run of rolls moves the total forward 3.5 per roll on average, so the totals visited are a share 1/3.5 = 2/7 of all the numbers passed, and far from the start every number is equally likely to be one of them. That is the renewal argument, and it gives the limit without any recursion. It also explains why the answer is not 1/6: the total does not get one try at each number, it passes every number and either lands on it or steps over it.

Why the hump at 6, and what is the desk point?

Small totals have many routes compared with their distance from zero: you can reach 6 in one roll, or in two, three, up to six rolls. For n from 1 to 6, p(n) = (1/6)(7/6) to the power n minus 1, so it grows each step and peaks at 0.360 at six; after that the averaging takes over and damps the swings. The interview point, often set as a coding task, is dynamic programming: one pass, six additions per number, no enumeration of paths. The limitation to say out loud is that the 2/7 limit needs a fair die and nothing that depends on the total so far; a rule such as skip your turn above 50 breaks it.

A one-year call struck at 100 plus a six-month put struck at 100. At the decision date you take the larger of the call and the put. Write that as the call plus max(P - C, 0). Put-call parity with zero rates and no dividend says P - C = 100 - S at that date, so the extra piece is max(100 - S, 0) paid at six months: a six-month put. At 20% volatility that is 7.97 + 5.64 = 13.60, against 15.93 for a one-year straddle.

Why is a chooser worth less than a straddle but more than a single option?

Booking a restaurant table for a date six weeks away while keeping the right, at four weeks, to switch it to a different restaurant is worth more than a fixed booking but less than holding two bookings to the end. The chooser is the same: it beats any one plain option because you choose with six months of news in hand, but it is cheaper than a straddle because you must give up one leg at six months. A straddle keeps both the call and the put for the full year; the chooser keeps only whichever looks better at the half-way point, and then lives or dies with it. At 20% volatility the one-year straddle is 15.93, a one-year call alone is 7.97, and the chooser sits in between at 13.60. The question is whether you can find that middle number without a model of the choice itself.

How does put-call parity turn the choice into plain options?

At six months you hold max(C, P), where C and P are the values then of the call and the put with six months left. Split it: max(C, P) = C + max(P - C, 0). Now use parity. With no dividend and zero rates, a put minus a call with the same strike and expiry is worth the strike minus the stock, so P - C = 100 - S at the decision date. The awkward piece max(P - C, 0) becomes max(100 - S, 0), which is exactly the payoff of a put struck at 100 that expires at six months, so the chooser is a one-year call plus a six-month put. The call is always in the portfolio because you either keep it or, by parity, the put you switch into is that call plus a short forward, and the short forward's value is what the six-month put pays you when you switch.

How do you check the answer, and what changes with rates and dividends?

Test the two ends of the decision date. If the choice had to be made today, the chooser is just whichever is dearer now, and at the money with zero rates the call and put cost the same, 7.97; the formula gives a call plus a put expiring today at the money, which is worth zero, so it agrees. If the choice is made at one year, you simply take whichever pays, which is a straddle, and the formula gives a call plus a one-year put, which agrees too. A simulation of 40,000 paths to six months, taking the better of call and put on each, gives 13.61 against 13.60. With rates or dividends the same split works, but parity puts a discounted strike on the put: it is struck at K times the discount factor from six months to a year, adjusted for the dividends paid in that window. The limitation to say: the decomposition relies on European options and on parity holding exactly; an American chooser, or one on a stock with a borrow cost, needs a model.

It pays exactly 20 at any stock price, so 19.20 today implies a one-year rate of 20/19.20 - 1, about 4.17%. The 90 call and short 90 put make a long forward at 90; the short 110 call and long 110 put make a short forward at 110. The stock cancels and 110 - 90 = 20 is left. Buying the box is lending at 4.17% and selling it is borrowing at that rate, so it suits a lender whose other return is lower, or a borrower whose funding costs more.

Why does the payoff not depend on the stock?

Agree to buy a scooter from one friend for 90 and to sell it to another for 110, both next year, and you will make 20 whatever scooters cost by then. The box is the same pair of agreements: long the 90 call and short the 90 put is a promise to buy at 90, and short the 110 call and long the 110 put is a promise to sell at 110, so the stock comes in and goes out and only the gap between the strikes, 20, is left. Seen as spreads, the call spread pays from 0 below 90 up to 20 above 110, and the put spread pays the mirror image, 20 below 90 down to 0 above 110. Wherever the stock ends, the two add to 20.

How do you turn the price into a rate?

You pay 19.20 today and receive 20.00 in a year with no market risk, so it is a deposit. The simple rate is 20 / 19.20 - 1 = 4.17%, and the continuously compounded rate is ln(20 / 19.20) = 4.08%. A box is a zero-coupon bond built from options, and its price is the strike gap times the discount factor, so any box that trades away from the market's interest rate is mispriced. Prices consistent with this rate, for an illustrative stock at 100 with 20% volatility, are a 90 call at 16.11, a 110 call at 5.69, a 110 put at 11.29 and a 90 put at 2.51: 16.11 - 5.69 + 11.29 - 2.51 = 19.20. Volatility does not enter the box price at all, because every volatility effect in the calls is cancelled by the puts.

When is it attractive, and what can go wrong?

Compare 4.17% with your own rates. If cash would otherwise earn 3.5%, buying the box lends at a better rate: 19.20 at 3.5% grows to only 19.87, against 20 from the box. If your funding costs 5%, selling the box borrows more cheaply: you receive 19.20 today and owe 20, while 20 owed at 5% would have raised only 19.05. Whether a box is cheap or dear is never a property of the box alone; it depends on the rate you can otherwise lend or borrow at, which is why boxes are a funding trade. The risks to name are practical. American options can be exercised early, which breaks the box, so use European index options. Four bid-offer spreads and fees can eat the 0.80 of interest. And the counterparty, or the clearing house, must still be there in a year.

The typical player ends with about 3% of the starting capital, while the average outcome is about 13,781 times the start. With 60 wins and 40 losses the stack is multiplied by 1.5 to the 60 and 0.5 to the 40, which is 0.033. Betting half the stack is 2.5 times the Kelly fraction of 20%, and anything beyond twice Kelly turns a positive-edge game into a losing one for the person actually playing it.

Why can every bet be favourable and the typical result still be a loss?

Think of a shopkeeper who doubles the stock every time a line sells and halves it every time it does not. A good line sells six times in ten, yet after a hundred cycles the shelf is nearly bare, because growth is a chain of multiplications and the order of operations does not rescue you. Expected value is an average across many imaginary players, but your capital follows one path, and a path is a product of factors, not a sum. Betting half the stack makes each win a factor of 1.5 and each loss a factor of 0.5. Sixty wins and forty losses, the most likely split, multiply to 0.033. The arithmetic mean of 13,781 is carried by the rare paths with 80 or 90 wins, which almost nobody lives on.

What is the right quantity to look at instead of the mean?

Look at the expected growth rate of the logarithm of capital, because the log turns products into sums and the law of large numbers then works. Each bet changes log capital by 0.6 x log(1.5) + 0.4 x log(0.5), which is -0.034, a negative number, so the typical player loses about 3.4% of capital per bet on average. Over 100 bets that is a factor of exp(-3.4), the 0.033 you saw above. The chance of finishing below the start is 76%, so the mean is a number you will almost never meet.

Where does the sizing go wrong, and what would fix it?

The Kelly fraction for an even-money bet is 2p minus 1, here 20% of capital, and it maximises the growth rate at +0.0201 per bet, which turns 100 bets into a typical multiple of about 7.5. Growth is zero at exactly twice Kelly, 40% here, and negative beyond it, so a half-stack bettor is on the wrong side of the hill even though the coin is in their favour. The limitation is that the Kelly rule assumes you know p exactly and will play many times; with an uncertain edge, desks size below Kelly on purpose, often at half of it.

Offer nothing. Every positive offer loses money on average. If you offer b and the owner accepts, you learn the value is below b, so its average is b/2, worth 1.5 x b/2 = 0.75b to you. You pay b, so each accepted deal loses 0.25b, and the deal is accepted a fraction b/100 of the time. Expected profit is minus 0.0025 b squared, negative for every b above zero. The 1.5 multiplier is not enough to overcome what acceptance tells you.

Why is the owner saying yes bad news for you?

A friend sells you their old scooter for any price you name, but only if your price beats what they privately think it is worth. If they take Rs 20,000 instantly, you have just learned the scooter is worth less than that to someone who knows it well. Acceptance is information: it tells you the true value lies below your bid, so the only companies you ever buy are the ones worth less than you paid, and the ones worth more walk away. Averaging over all possible values, as if you bought every company, is the mistake; you must average only over the values at which the owner says yes.

How do you set up the expected profit cleanly?

Condition on acceptance, then multiply by its probability. Given a bid b that is accepted, the value V is uniform on 0 to b, so E[V | accepted] = b/2, the company is worth 1.5 x b/2 = 0.75b to you, and the profit on an accepted deal is 0.75b - b = - 0.25b. Acceptance happens with probability b/100, so expected profit is - 0.25b x b/100 = - b squared over 400. At b = 50 that is minus 6.25 crore; at b = 100 it is minus 25 crore. The derivative is negative everywhere above zero, so the maximum is at b = 0.

At what multiplier does a bid start to make sense?

Replace 1.5 with a general m. Given acceptance, the company is worth m x b/2 to you against the b you pay, so the deal breaks even when m/2 = 1, that is m = 2. Unless the company is worth more than twice as much to you as to the owner, acceptance always costs you, and at exactly double every bid is a wash. That is the winner's curse in its purest form: the party with less information loses whenever the informed party decides whether to trade. The limitation is the uniform prior and the single offer; with a floor on the value, or a negotiation that reveals information, positive bids can be profitable.

About 0.17, exactly one sixth. For n equally weighted stocks with the same volatility and the same pairwise correlation rho, index variance is stock variance times (1/n + (1 - 1/n) rho). Index variance over stock variance is 0.15 squared over 0.30 squared = 0.25, so 0.25 = 0.1 + 0.9 rho and rho = 0.15/0.9 = 1/6. The large-n shortcut of 0.25 overstates it because it ignores the 1/n term.

Why does the index have less volatility than its members?

Ten shopkeepers in one market each have noisy daily takings, but the market's total takings are steadier, because one shop's bad day is often another's good day. Only the part of the noise they share, the weather or a festival, survives the averaging. Index variance has two parts: the stocks' own noise, which averages away as 1/n, and the common movement, which survives in proportion to the correlation. With ten stocks at 30% and no correlation at all the index would still have 9.5% volatility, and that floor is why the implied correlation is below the naive 0.25.

How do you derive the formula on the spot?

Write the index as the average of n returns and expand its variance. There are n variance terms, each sigma squared over n squared, and n(n - 1) covariance terms, each rho sigma squared over n squared, so index variance is sigma squared times (1/n + (n - 1) rho / n). Set that equal to 0.15 squared with sigma = 0.30 and n = 10: 0.0225 = 0.09 x (0.1 + 0.9 rho), so 0.25 = 0.1 + 0.9 rho and rho = 1/6. The derivation takes four lines, and the interviewer wants to hear the count of covariance terms, n(n - 1), said out loud.

What does a desk do with implied correlation?

It compares it with realised correlation and trades the gap. Selling index options and buying single-stock options is a short correlation position, called a dispersion trade, and it pays when the stocks move more independently than the 1/6 the market has priced in. The limitation of the puzzle is its symmetry: real indices have unequal weights and unequal volatilities, so the implied correlation is a weighted average that can differ from the simple formula, and the implied figure also moves with the skew of the index options, which the single number cannot show.

Yes. Buy the 90 call, sell two 100 calls, buy the 110 call, and you are paid 1 to own a payoff that is never below zero. The butterfly costs 14 - 16 + 1 = -1, so you receive 1 today. At expiry it pays nothing below 90, rises to 10 at 100 and falls to nothing above 110, never negative. Call prices must be convex in strike: the 100 call cannot exceed (14 + 1)/2 = 7.5, and it trades at 8.

Where does the 7.5 come from?

Picture three houses on one street at 90, 100 and 110 square metres, priced at 14, 8 and 1. The middle house should be worth no more than the average of its neighbours if each extra metre is worth less than the last; if it is priced above the average, you sell it and buy the two neighbours. A call's price falls as the strike rises, and it falls at a decreasing rate, so the price at the middle strike must sit on or below the straight line between its neighbours. The line from 14 at 90 to 1 at 110 passes through 7.5 at 100. The market says 8. That half point is the mispricing.

How do you prove the payoff is never negative?

Walk the stock price up. Below 90 nothing is in the money, payoff zero. Between 90 and 100 only the 90 call pays, so the payoff is the stock price minus 90, rising to 10. Between 100 and 110 the two short 100 calls start paying out, and the slope turns to 1 - 2 = -1, so the payoff falls from 10 back to zero at 110. Above 110 the long 110 call kicks in and the slope is 1 - 2 + 1 = 0: flat at zero. The payoff is zero or positive everywhere, you were paid 1 to hold it, so you have a riskless profit of at least 1 and up to 11.

Check the other bounds too, out loud, so the interviewer sees you are not pattern matching. A call spread can never be worth more than the gap between its strikes: the 90/100 spread costs 6 and the 100/110 spread costs 7, both under 10, so those pass. Prices fall as strike rises: 14, 8, 1, pass. Only convexity fails. The limitation: in a real screen you trade at bids and offers, not mids, and a half point of theoretical edge can vanish inside the spread. The structure is an arbitrage at these prices; whether you can execute it is a separate question.

232 of the 1,024 paths touch 104, about 22.7%. A path ending at or above 104 must have touched it: that needs at least 7 up days, 176 paths. A path that touches 104 and ends below it can be reflected in the 104 line after its first touch, and the reflection ends above 104, needing at least 8 up days: 56 paths. The pairing is one to one, so the count is 176 + 56 = 232.

Why is the reflection a fair way to count?

Imagine a walker who reaches a wall at some point and then wanders on. Flip every step after the first touch, up for down, and you get another walker who also touched the wall at the same moment and is now the same distance on the other side of it. Each flipped step is as likely as the original, so the two walks have equal probability. Reflection pairs every path that touches 104 and finishes below it with exactly one path that finishes above 104, so the hard count becomes an easy count of endpoints. Paths that finish at exactly 104 are their own partners and are counted once.

How do the binomial counts fall out?

After 10 days of plus or minus 1, the position is 100 plus ups minus downs, which is 100 + 2 x ups - 10. Ending at or above 104 needs ups of at least 7: C(10,7) + C(10,8) + C(10,9) + C(10,10) = 120 + 45 + 10 + 1 = 176. Ending strictly above 104 means at or above 106, since the endpoint is always even, so ups of at least 8: 45 + 10 + 1 = 56. The probability of touching is (176 + 56)/1,024 = 232/1,024, about 22.7%, against only 176/1,024 = 17.2% for ending at or above 104. The gap is the 56 paths that visited the barrier and came back.

Say why a derivatives desk cares. A knock-out or one-touch option pays on exactly this event, and the reflection principle is how the closed-form barrier prices are derived in continuous time. The limitation is in the symmetry: reflection needs the up and down steps to be equally likely and the barrier to sit on the lattice. With a drift, or in continuous time with a non-zero rate, the reflected path is not equally likely and a correction factor appears, which is why barrier formulas carry that extra power of the barrier over the spot.