Derivatives Foundation puzzles, solved step by step

Reroll any 1, 2 or 3; keep a 4, 5 or 6; the game is worth 4.25. With a Rs 1 fee, keep a 3 as well and the value falls to 3.83. A fresh roll is worth 3.5, so you reroll only faces below it. The value is (4 + 5 + 6)/6 + (3/6) x 3.5 = 4.25. With the fee, a fresh roll nets 2.5, so a 3 is now worth keeping, and the value is (3 + 4 + 5 + 6)/6 + (2/6) x 2.5 = 23/6.

Why is the threshold the value of a fresh roll?

You are offered a mango from a basket; you can keep the one in your hand or swap it blind for another. You swap only if the one you hold is worse than the average mango. The reroll replaces a known face with the average of an unknown one, so you take it exactly when the face you hold is below that average, which is 3.5 for a fair die. Faces 1, 2 and 3 are below, so reroll them; 4, 5 and 6 are above, so keep them. There is no face equal to 3.5, so there is no tie to argue about.

How does the fee change the decision and the value?

The fee lowers what a fresh roll is worth, from 3.5 to 2.5, and the threshold moves with it. A 3 was worth rerolling for free, since 3 is below 3.5, but with the fee a 3 beats the 2.5 a reroll now nets, so you keep it, and only a 1 or a 2 is rerolled. The value becomes (3 + 4 + 5 + 6)/6 + (2/6) x 2.5 = 3 + 0.833 = 3.83. Check the alternative: keeping only 4 and above with the fee gives 2.5 + (3/6) x 2.5 = 3.75, which is worse, so the threshold really does move. The fee costs you 0.42 of value in total, less than the Rs 1 charge because you pay it only a third of the time.

What is the general pattern a desk is looking for?

Backward induction. Value the last decision first, then use that value as the threshold for the decision before it; with two rerolls the second-stage value of 4.25 becomes the bar for the first roll, so you keep only a 5 or a 6 and the game is worth (5 + 6)/6 + (4/6) x 4.25 = 4.67. This is how an American option is priced on a tree, with exercise now compared against the continuation value. The limitation is that the puzzle has a known distribution; real stopping problems have to estimate the continuation value, and a wrong estimate moves the threshold.

6 flips for HH and 4 flips for HT. Track one state: whether the last flip was a head. For HH, a tail after a head sends you back to the start, so E(start) = 1 + E(one H)/2 + E(start)/2 and E(one H) = 1 + E(start)/2, giving 6. For HT, a head after a head leaves you still holding a head, so E(one H) = 1 + E(one H)/2 = 2 and E(start) = 2 + 2 = 4.

Why do two patterns with the same probability take different times?

Two queues at a counter: in one, a mistake sends you to the back; in the other, a mistake keeps your place. Both queues move at the same speed, but one is a much longer wait. Waiting for a pattern is a race with restarts, and what matters is how much progress a failure destroys: a tail after a head destroys everything for HH, while a head after a head destroys nothing for HT. That asymmetry, not the probability of the pattern, sets the expected time. It is also why HT and TH take 4 while HH and TT take 6.

How do you set up and solve the equations in the room?

Two states, two equations, each saying one flip plus the average of what remains. For HH: from the start, a head takes you to one H and a tail keeps you at the start, so E0 = 1 + E1/2 + E0/2; from one H, a head finishes and a tail returns you to the start, so E1 = 1 + E0/2; substitute to get E1 = 4 and E0 = 6. For HT the second equation changes to E1 = 1 + E1/2, since a head keeps you at one H, giving E1 = 2, and the first equation gives E0 = 2 + E1 = 4. The first equation is the same in both problems; only the failure branch differs.

What is the quick check, and where does this matter beyond coins?

There is a general rule: the expected time to a pattern is the sum of 2 to the k over every length k at which the pattern overlaps itself. HH overlaps itself at lengths 1 and 2, so 2 + 4 = 6; HT overlaps only at its full length 2, so 4. The same arithmetic prices a bet on which of two patterns appears first, and it shows up on a desk whenever a signal needs a run of confirmations: a rule that resets on any contradicting tick waits far longer than one that keeps partial progress. The limitation is the fair coin; with a biased coin the overlap rule still holds but the powers of two become products of the relevant probabilities.

Open around 35, say 33 bid at 37 offer. After the king and the 3, requote around 36.9, say 35.5 at 38.5. Each card averages 7, so five average 35. Once a 13 and a 3 are known, 16 points are fixed and three cards remain from a 50-card deck whose average is now (364 - 16)/50 = 6.96, so the total is 16 + 3 x 6.96 = 36.88. Three unknown cards carry less spread than five, so the market can tighten.

Why is the opening fair value simply five times seven?

Five friends each pick a sweet from a jar without looking; the expected total weight is five times the average sweet, even though each pick changes what is left for the next. Linearity of expectation holds whether or not the draws are independent, so the expected total of five cards is 5 x 7 = 35 and the dealing-without-replacement detail changes only the spread, not the centre. The standard deviation of the total is about 8.0, slightly below the independent-draw figure because the finite deck pulls the cards apart, and a market four wide around 35 is a reasonable opening quote.

What does the reveal change, and what do people get wrong?

Two things move. The two revealed cards swap their expectation of 7 each for their actual values of 13 and 3, which lifts the total by 2, and the remaining deck has lost a high card and a low card, so its average drops from 7 to 6.96, which trims 0.12 off the three unknown cards. The careful centre is 36.88; the quick answer of 16 + 21 = 37 is off by only 0.12, and in the room 37 with a note that the deck is slightly poorer is a fine answer. What is not fine is leaving the market at 35, or widening it when the uncertainty has fallen from five cards to three.

How should the width change, and what trade do you expect next?

Width tracks the remaining uncertainty. The standard deviation of the total falls from about 8.0 to about 6.2 once only three cards are unknown, so a market that was 4 wide can go to 3 wide without taking more risk per trade. The next step is usually a trade: if the interviewer lifts your 38.5 offer, you are short at above fair and hold; if they hit 35.5 you are long below fair. The limitation is that the puzzle assumes a fair deck and honest reveals; in the real version of this game the counterparty may have seen a card you have not, and a run of trades in one direction is the tell.

The probability is 1/6, not 1/5. By symmetry the game ends on a 2 one third of the time. A first roll of 1 happens one sixth of the time and leaves all three evens unseen, so 2 is last with probability 1/3. The joint probability is 1/18, and 1/18 divided by 1/3 is 1/6. The naive 1/5 treats the five possible first rolls as equally likely given the ending, and they are not.

Why does the ending change the odds of the start?

Think of a race where you learn only who finished last. If you are told that runner C came last, the start line was still fair, but some starting arrangements make C last more often than others, and you should update towards those. The die works the same way. A first roll of 4 leaves only two evens unseen, so 2 comes last half the time; a first roll of 1 leaves three evens unseen, so 2 comes last only one third of the time. The ending is twice as consistent with a first-roll 4 as with a first-roll 1, and a first-roll 2 is ruled out entirely, because a 2 that has already appeared cannot complete the set.

How do you set up the Bayes calculation in thirty seconds?

Group the first roll into three cases rather than six, because 1, 3 and 5 are interchangeable and so are 4 and 6. Odd first roll: probability 1/2, and then 2 ends the game with probability 1/3, giving a joint weight of 1/6. First roll 2: joint weight 0. First roll 4 or 6: probability 1/3, then 2 ends the game with probability 1/2, joint weight 1/6. The weights add to 1/3, which is the unconditional chance of ending on a 2, as symmetry says they must. Given the ending, the first roll was odd with probability 1/2 and was 4 or 6 with probability 1/2, so each odd face carries 1/6 and each of 4 and 6 carries 1/4.

The sanity check the interviewer wants to hear: 3 x 1/6 + 2 x 1/4 = 1, so the posterior weights over the five possible first rolls add up. Then say the general point. An odd roll is a wasted roll that tells you nothing about which even finishes last, which is why its posterior weight is simply its prior, 1/6, unchanged. The information in the ending all goes into shifting weight from the 2, which is now impossible, onto 4 and 6.

A 30% performance fee holds revenue at 4% of assets at the mean, but the swap is not neutral once returns vary. Two and twenty earns 2 + 0.2 x 10 = 4; one and thirty earns 1 + 0.3 x 10 = 4. The performance fee pays 20% of max(return, 0), which is a call on the fund's return struck at zero. Investors have written it, and its value rises with volatility, so moving fee from fixed to performance raises what the manager expects to collect.

How do you make the two schedules equal at the mean?

A shopkeeper who swaps a fixed monthly rent for a share of sales asks one question first: at my usual sales, what share leaves the landlord no worse off? Do the same here, in percent of assets. Two and twenty collects 2 fixed plus 20% of a 10% gain, 4% of assets; cutting the fixed fee to 1 leaves 3 points to be earned from a 10% return, which needs a 30% performance fee. That is the arithmetic the interviewer wants first, and it is one line: 1 + 10x = 4, so x = 0.3.

Why is the performance fee an option, and who has written it?

The manager receives 20% of the gain when the fund is up and nothing when it is down. That is the payoff of a call on the fund's return with a strike of zero: convexA payoff that bends upward, so the average of the payoff over a spread of outcomes is higher than the payoff at the average outcome. in the return, floored at nothing. Investors are on the other side; they have granted the manager that call and are paid for it only through the management fee they do not have to pay. Because a call is worth more when the underlying is more volatile, the performance fee is worth more than its value at the mean return, and the more of the fee you move into performance, the more the schedule is worth for the same expected return.

Put numbers on it. With that spread the expected performance fee under two and twenty is 0.2 x 14.0% = 2.79%, so the manager expects 4.79% of assets, not 4%. Under one and thirty it is 1 + 0.3 x 14.0% = 5.19%. The swap that looked neutral at the mean adds about 0.40% of assets a year in expected revenue. Say the limitation too: real schedules carry hurdles and high-water marks, which raise the strike and cut the option's value, and the fee is charged on the net of the management fee, which shaves a little off both sides.

76. Multiplying by 4 and keeping only the last two digits gives 04, 16, 64, 56, 24, 96, 84, 36, 44, 76, and then 76 x 4 = 304 returns to 04. The cycle has length 10 starting at 4^1, and 3000 is a multiple of 10, so 4^3000 ends like 4^10, in 76. Check: 76 x 76 = 5,776, so 76 reproduces itself under squaring, which is what a power of 4^10 must do.

Why do only the last two digits matter at each step?

When you work out what time it will be 3,000 hours from now, you do not count the hours; you note that the clock face has 24 positions and ask where 3,000 lands on it. Last two digits are a clock with 100 positions. The last two digits of a product are fixed by the last two digits of the factors alone, so the sequence of 4^n mod 100 can only visit 100 states and has to fall into a cycle. Walk it: 04, 16, 64, 56, 24, 96, 84, 36, 44, 76, and 76 x 4 = 304 brings you back to 04. Ten steps, then it repeats.

How do you reduce 3000 without miscounting the start of the cycle?

The cycle begins at 4^1 = 04, not at 4^0 = 01, because 01 is never revisited: once a power of 4 is a multiple of 4 it stays one, and 01 is not. So the positions n = 1, 11, 21 and so on share last digits 04, and the positions n = 10, 20, 30 and so on share 76. 3000 is a multiple of 10, so it sits in the same slot as 10, and 4^10 ends in 76. Candidates who start counting from n = 0 land one step off and say 44.

A second route, worth one sentence: split 100 into 4 and 25. Any 4^n with n at least 1 is 0 mod 4, and 4^10 = 1,048,576 is 1 mod 25, so 4^3000 is 1 mod 25. The number under 100 that is 0 mod 4 and 1 mod 25 is 76. Two methods agreeing is the finish an interviewer wants. The limitation: the cycle trick is only this short because 4 is small; for a base like 7 the cycle mod 100 is 4 steps, for 3 it is 20, and you should check rather than assume.

About 10 million a day, so quote 8 million bid, 12 million offered. Take a population of roughly 67 million, births around 1.1% a year, so about 737,000 babies, children in nappies for about two and a half years, giving 1.84 million children, at five or six nappies a day each: 10.1 million. Running every input at its low gives 5.2 million and at its high 17.6 million, so the quote sits inside that range with room to be wrong.

How do you turn a guess into a market?

If a friend asks you to bet on how many samosas a canteen sells a day, you would not name one number. You would count the seats, guess the sittings, guess how many order a samosa, and then say a range you would bet either side of. A market is that range with a price on each end. Decompose the quantity into inputs you can bound, carry a low and a high through each step, and let the spread of the outputs set the width of your quote. Population times birth rate gives babies a year; times years in nappies gives children in nappies; times nappies per child per day gives the answer. The population and birth rate are the kind of round figures you carry in your head; the interviewer accepts any sensible anchor and marks the structure.

How wide should the market be?

Not as wide as the full range. Every input at its low or every input at its high is unlikely, because the errors are mostly independent and partly cancel. A quote about one third as wide as the all-low to all-high range, centred on the point estimate, is tight enough to be a real market and wide enough that you expect to be inside it. Here the range runs from 5.2 to 17.6 million, so 8 at 12 is a working quote. Say the limitation: the tree ignores adult incontinence products and children over three, both of which push the true figure up, so if anything you would skew the market higher rather than lower.

Then expect the interviewer to trade. If they lift your offer at 12 million, ask yourself what they know: perhaps the adult market, perhaps a higher nappies per day. Move your quote up and tighten, do not freeze it. If they hit your bid, do the reverse. The exercise is not the answer; it is whether you update your market when someone trades against you, which is what market making is.

One half. By the time the last passenger boards, seats 2 to 99 are always taken, because each owner either sat in theirs or found it taken. The free seat is either seat 1 or seat 100. Every passenger who chooses at random, the first and each one displaced after him, faces seat 1 and seat 100 with exactly the same chance, so the game is equally likely to settle on either. The answer is 1/2 for any plane with two or more seats.

Why do only two seats matter?

Think of a cloakroom where one guest hangs his coat on a random hook. Each later guest uses their own hook if it is free and a random free hook if not. Every hook except two has an owner who will turn up and either use it or find it used. The last passenger can only end up in seat 1 or seat 100, because every seat in between has an owner who boards earlier and leaves it occupied. So the whole question is which of those two seats is still empty when the last passenger walks down the aisle.

Why are the two endings equally likely?

Follow the chain of displaced people. Passenger 1 picks at random. If he takes seat 1, nobody is ever displaced and the last passenger gets seat 100. If he takes seat 100, the last passenger is shut out. If he takes some other seat k, everyone up to k sits normally and passenger k inherits the problem: a random choice among the free seats, which still include both seat 1 and seat 100. Every random chooser in the chain picks seat 1 and seat 100 with equal probability, and the chain stops the moment either is taken, so the two endings carry the same total weight, 1/2 each. The other seats only pass the problem along; they never decide it.

Check it on the smallest case aloud: with two seats, passenger 1 takes his own or the other with equal chance, so 1/2. Running the recursion for every plane from 2 to 100 seats returns exactly 1/2 each time. A useful extension the interviewer often asks next: passenger j, for j from 2 to n, gets their own seat with probability (n - j + 1)/(n - j + 2). Passenger 2 is almost always fine; passenger 99 is fine two times in three. The limitation is in the rules: if displaced passengers preferred seats near the front, the symmetry between seat 1 and seat 100 breaks and so does the half.

Switch: either other closed door wins with probability 3/8, against 1/4 for staying. Your door had a 1/4 chance and the host's action cannot change it, because he can always open an empty door among the other three. The remaining 3/4 sat on three doors and now sits on two, so each carries 3/8. Switching does not make you a favourite, it only beats staying, 37.5% against 25%.

Why does your own door keep its 1/4?

A friend who knows the answers to a quiz looks at three questions you did not attempt and crosses out one wrong option. She would have done that whether your own answer was right or wrong, so her crossing out says nothing about your answer. Information only moves probability onto or off your door if the action could have depended on what is behind it, and the host can open an empty door among the other three every time. So your door stays at 1/4, exactly where it started.

How does Bayes give 3/8 for each of the other doors?

Say the host opens door 4. If the prize is behind your door 1, he chooses among doors 2, 3 and 4 at random, so he opens door 4 with chance 1/3. If it is behind door 2, he must choose between 3 and 4, chance 1/2; the same for door 3. If it is behind door 4, he never opens it. Multiply each by the prior of 1/4: 1/12, 1/8, 1/8 and 0, which add to 1/3. Divide through and the posterior is 1/4 for door 1, 3/8 for door 2, 3/8 for door 3 and nothing for door 4. The host is more likely to open door 4 when the prize sits behind door 2 or 3 than when it sits behind yours, and that difference is the information.

Then generalise, which is what the question tests. With n doors and one opened, staying wins 1/n and each other closed door wins (n - 1)/(n(n - 2)). With three doors that is 2/3, the familiar answer; with 100 doors it is 99/9,800, barely above the 1/100 for staying. The edge from switching shrinks as the host reveals a smaller share of the doors. The limitation is the host's rule: if he opened a door at random and it merely happened to be empty, he would carry no information about the other doors and switching would gain nothing.

You pay 3.30; the most you can make is 1.70, the most you can lose is 3.30, and you break even at 98.30. You buy the 95 call at the offer, 7.40, and sell the 100 call at the bid, 4.10. At expiry the spread is worth nothing below 95, the stock minus 95 between the strikes, and 5 above 100. So the profit runs from minus 3.30 to 5 - 3.30 = 1.70, and is zero when the stock is 95 + 3.30 = 98.30.

Which side of each quote do you deal on?

At a currency counter at the airport there are two rates on the board: the one they buy at and the one they sell at, and you always get the worse one for you. Option screens are the same. If you want to trade now, you buy at the offer and sell at the bid, so a two-leg trade pays the spread on both legs. Buy the 95 call at 7.40, sell the 100 call at 4.10, net debit 3.30. At the mid prices, 7.20 and 4.30, the same trade would cost 2.90. The 0.40 difference is the cost of crossing two bid-offer spreads of 0.40 each, half of each.

How do you read off the maximum profit, loss and break-even?

Walk the stock price up. Below 95 both calls expire worthless, so you lose the 3.30 you paid. Between 95 and 100 only the long call pays, one for one, so the profit rises from minus 3.30. Above 100 the short call pays out as fast as the long call pays in, so the value is capped at the strike gap of 5. A call spread can never be worth more than the gap between its strikes, so the most you can make is 5 minus what you paid, 1.70, and you break even where the stock has risen 3.30 above the lower strike, at 98.30.

Two checks to say aloud. First, the price passes the no-arbitrage bounds: a 95/100 call spread must cost between 0 and 5, and 3.30 does. Second, the risk-reward: you risk 3.30 to make 1.70, which only makes sense if you think the stock finishes above 98.30 more than 3.30/5 = 66% of the time. The limitation: this is the payoff at expiry. Before expiry the spread's value moves with volatility and time, and you could unwind it, but again only by selling the 95 at its bid and buying the 100 at its offer.