Derivatives Foundation puzzles, solved step by step

Skew the whole quote up, to about 50 at 52, keeping the width of 2. The short is a risk you want to shed, so make your bid attractive to sellers and your offer less attractive to buyers. A simple rule moves the mid 0.05 per lot of inventory, so 20 lots short moves it up 1. Widening to 48 at 52 also stops the buying, but it pushes sellers away too, so the short stays on your book and you earn less while you wait.

What is the position telling you to do?

A fruit seller who has run short of mangoes by mid-morning does not shut the stall; she raises the price she pays suppliers and nudges up the price to customers, so more mangoes arrive and fewer leave. Inventory changes what you want to trade next, not what the contract is worth: with fair value still 50, a short of 20 lots means your next trade should be a buy, so the quote should lean towards buying. You have used 80% of the risk limit, and five more lifts would put you at it. The question is testing whether you separate the two numbers a market maker carries: the fair value, which has not moved, and the price at which you want to trade, which has.

How far should you skew, and what does it cost?

A common rule moves the mid in proportion to the position: here 0.05 per lot, so 20 lots short lifts the mid by 1 and the quote becomes 50 at 52. The bid now sits at fair value, so you buy back with no edge, and the offer is 2 above fair, so a buyer who still lifts it pays you well for adding to the risk. Skewing trades some edge for risk reduction, and the closer the position is to the limit, the more edge you should be willing to give up. In a toy model where a quote d away from fair trades with probability 0.6 x e to the minus d each period, the unchanged quote earns 0.44 a period and never reduces the short; the full skew earns 0.16 but buys back a net 0.52 lots a period, about 39 periods to flatten; the half skew sits between at 0.38 and 0.23. The numbers are illustrative; the direction is not.

Why not simply widen?

Widening to 48 at 52 makes the offer as unattractive as the skew does, but it also moves the bid two points below fair, so sellers go elsewhere. In the toy model both sides fill 0.08 of the time and the expected change in the position is zero: a wider quote stops new risk arriving but does nothing about the risk you already hold, and the 20-lot short keeps moving with the market while you wait. Widening has its place, when you think fair value is uncertain or that the people lifting you know something, because then both sides are dangerous. Say that distinction, and then say the last resort: if skewing does not bring sellers fast enough, hedge the short in a related market, or cross the spread and buy, rather than drift up to the limit.

54 people. Each stop halves the load and then adds a third of what is left, so the load is multiplied by 1/2 x 4/3 = 2/3 at every stop. Undoing three stops means dividing 16 by 2/3 three times: 16 to 24 to 36 to 54. The forward check works: 54 to 27 to 36, then 18 to 24, then 12 to 16.

Why work backwards rather than guess a start?

If someone tells you a shirt was marked down by half, then marked up by a third, and now costs Rs 16, you do not guess the original tag; you undo the two moves in reverse order. Each stop is a pair of multiplications, and multiplications are undone by dividing in the opposite order, so the end number walks back to the start without any trial and error. The growth of a third comes last at each stop, so it is undone first: divide by 4/3, which is multiply by 3/4. Then undo the halving by doubling. From 16: times 3/4 is 12, times 2 is 24. Twice more gives 36 and then 54.

What is the one-line version an interviewer wants to hear?

Collapse each stop to a single factor. Losing half is x 1/2 and gaining a third is x 4/3, so a stop is x 2/3, and three stops are x 8/27. The start is 16 x 27/8 = 54. Saying that in one breath shows you saw the structure rather than the arithmetic, which is what the question is for. The limitation is that it only works because every stop has the same two moves; a route where the fractions change needs the step-by-step walk back.

Where do the whole-number checks help you?

The bus carries people, so every intermediate count must be a whole number, and that is a free check. If the backwards walk ever produces a fraction, you have undone the moves in the wrong order. Undoing the halving first from 16 gives 32, then times 3/4 gives 24, which happens to be whole here, but on the next stop it would give 48 then 36, and the order mistake would have cost you nothing visible until the end, where 72 x 3/4 = 54 again by luck. Order matters in general even when the numbers hide it, so say the order out loud.

4 seconds. By symmetry, the two corners next to the start are the same state, call it adjacent. From the start you always move to adjacent in one step. From adjacent, half the time you reach the target and half the time you return to the start. So E(start) = 1 + E(adjacent) and E(adjacent) = 1 + E(start)/2, giving E(adjacent) = 3 and E(start) = 4.

Why collapse four corners into three states?

If you are lost in a town with a river on one side, what matters is how far you are from the river, not which street you are on. The square is the same: standing at either corner next to the start, the token's future looks identical, one coin flip from the target and one from the start. Grouping corners by their distance from the target turns a four-state chain into a three-state line, and a line is solved with one equation per unknown. You have two unknowns, the expected time from the start and from an adjacent corner, because the target itself takes zero time.

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

Each equation says the same sentence: one step, plus the average of what is left from where you land. From the start, every step lands on an adjacent corner, so E(start) = 1 + E(adjacent); from an adjacent corner, half the steps finish and half return, so E(adjacent) = 1 + (1/2) x 0 + (1/2) x E(start). Substitute the first into the second: E(adjacent) = 1 + (1/2)(1 + E(adjacent)), so E(adjacent)/2 = 3/2, E(adjacent) = 3, and E(start) = 4. Saying the sentence before the symbols is what keeps the equations honest.

What is the check, and what does the general case look like?

Check it with the geometric picture. After the first step you are adjacent, and from there each attempt either finishes in one step or costs two steps, back to the start and out again, before you are adjacent once more. Each attempt succeeds with chance 1/2, so E(adjacent) = (1/2) x 1 + (1/2) x (2 + E(adjacent)), which again gives 3, and the first step makes it 4. On a cube the same method with four distance classes gives 10 steps to the opposite vertex; on a general graph the method is the same, but the number of distance classes grows and the arithmetic stops being mental.

Yes. Sell the three-month call at 6.50, buy the six-month call at 6.10, and lock in 0.40 today with no risk. On a non-dividend stock, a call with more time to expiry is worth at least as much as the same strike with less, because at three months the six-month call is still worth at least its intrinsic value, which is exactly what the expiring call pays out. The position never owes money, so the 0.40 is a pure profit.

Why must the longer call be worth at least the shorter one?

A ticket that lets you buy a train seat any time in the next six months is worth at least as much as one that expires in three; you can always do with the long ticket exactly what you would have done with the short one, and then keep it. At the three-month expiry the short call pays max(S - 100, 0), and the six-month call at that moment is worth at least max(S - 100, 0) too, because a call on a non-dividend stock is never worth less than its intrinsic value and is usually worth more. So holding the long and being short the short can never leave you with a negative balance at three months, and the 0.40 you took in today is yours.

What happens at three months in each case?

Walk through both branches. If the stock is above 100, the short call is exercised against you, you deliver stock for 100, and you still hold a six-month call worth more than S - 100, so you exercise or sell it and finish ahead; if the stock is at or below 100, the short call expires worthless and you are left holding a live six-month call for free. Either way you keep the 0.40 from today and own something worth zero or more. The limitation is the non-dividend assumption: with a large dividend before six months, early exercise of an American three-month call could matter, and the comparison needs the dividend's present value subtracted.

Why would a market maker see this quote and what does it say about the vol surface?

The three-month price of 6.50 corresponds to an implied volatility of about 33% with zero rates, and at that volatility a six-month call would be worth around 9.18. A quote of 6.10 for the six-month implies a term structure of volatility so inverted that no volatility at all could justify it, which is why this is an arbitrage and not merely a view on calendar spreads. Real quotes rarely break the bound outright; what you see instead is a long-dated call a few ticks above the floor on an illiquid name, and the question is then whether the bid-ask spread swallows the edge before you can trade both legs.

Show heads 3/8 of the time, and do not play unless you are paid at least Rs 0.125 a round. If you show heads with probability p, your expected payoff is 5p - 2 when I show heads and 1 - 3p when I show tails. I will pick whichever is lower, so you choose p to make the lower line as high as possible, which is where they cross: p = 3/8, value - 1/8. Any other p lets me push you below that.

Why is the answer a mix rather than a single face?

Two children playing odds and evens learn fast that any pattern is punished: show heads every time and the other child shows tails every time. In a game where my best reply depends on what you do, any fixed choice is exploited, so you protect yourself by randomising in a ratio that leaves me with nothing to exploit. That ratio is found by making me indifferent between my two replies. If you show heads a fraction p of the time, my heads earns you 3p - 2(1 - p) = 5p - 2 and my tails earns you - 2p + (1 - p) = 1 - 3p. They are equal at p = 3/8.

How do you know minus 1/8 is the most you can guarantee?

Look at the lower of the two lines across all p. To the left of 3/8 the heads line is lower and rising; to the right the tails line is lower and falling, so the lower envelope peaks exactly at the crossing, and that peak is your guaranteed value. I have the same calculation from my side: if I show heads a fraction q of the time, you are indifferent when 3q - 2(1 - q) = - 2q + (1 - q), which again gives q = 3/8, and at that mix I hold you to - 1/8 whatever you do. Both sides landing on the same number is the minimax theorem, attributed to von Neumann, at work in a two by two table.

What is the desk version of this question?

Quoting against a counterparty who sees your pattern. A market maker who always leans the same way after a fill is the child who always shows heads, and the counterparty who notices earns the difference, so randomised sizing and skew are the trading-floor form of the 3/8 mix. The limitation of the puzzle answer is that it assumes I play optimally; against an opponent who shows heads half the time out of habit, your best reply is pure heads, with an expected 0.5 x 3 - 0.5 x 2 = + Rs 0.50 a round, and the game becomes worth playing. Ask who you are playing before you quote the value.

78.0% that it is the two-headed coin, and 89.0% that the next flip is heads. Prior odds are 1 to 9. The two-headed coin gives a head with certainty and a fair coin with chance 1/2, so each head multiplies the odds by 2; after five heads the odds are 32 to 9, which is 32/41. The next flip is heads with chance 32/41 x 1 + 9/41 x 1/2 = 73/82.

Why work in odds rather than probabilities?

A doctor who sees the same symptom five mornings running does not recompute the whole diagnosis each day; each new observation multiplies the odds of the condition by one fixed factor. In odds form, Bayes' rule is a multiplication: posterior odds equal prior odds times the likelihood ratio of each observation, and here every head has the same ratio of 1 to 1/2, which is 2. So the odds on the two-headed coin go 1:9, 2:9, 4:9, 8:9, 16:9, 32:9. Convert at the end: 32 divided by 32 plus 9 is 32/41. Doing it in probabilities means dividing by a different normaliser five times, which is where people slip.

Why is the next flip not simply 78% heads?

Because the fair coin also produces heads. The next flip is heads if the coin is two-headed, with chance 32/41, or if the coin is fair and lands heads, with chance 9/41 times 1/2, and the two routes add to 73/82, about 89%. Candidates who answer 78% have confused the probability of the hypothesis with the probability of the outcome. The gap between the two is the fair coin's half chance of a head, weighted by the 22% chance you are holding a fair coin.

How many heads would it take to be nearly sure, and what does a desk take from this?

Each head doubles the odds, so after 10 heads the odds are 1,024 to 9, about 99.1%, and after 5 you are only at 78%. Evidence that is merely consistent with a hypothesis moves you slowly when the alternative also produces it often, which is why five good months from a new trading strategy prove far less than people feel they do. The limitation is the prior: if the bag held 99 fair coins and one two-headed, five heads would leave you at 32 to 99, still under 25%, and no amount of looking at the flips alone tells you the composition of the bag.

About Rs 1,00,000 profit, and you need to sell about 20,000 shares. P&L from gamma is one half of gamma times the move squared: 0.5 x 2,000 x 10 squared = Rs 1,00,000. The delta picked up during the move is gamma times the move, 2,000 x 10 = 20,000 shares long, which you sell to get back to neutral. A Rs 10 fall would earn the same amount and leave you 20,000 shares short to buy back.

Why does a delta-neutral book make money on a move?

A cyclist at the bottom of a valley is on flat ground, but every metre up either slope gets steeper. Delta neutral means the P&L is flat at the current price only; gamma is how fast the slope changes, so as the stock moves the book acquires delta in the direction of the move and earns on it the whole way. With gamma of 2,000 shares per rupee, after the first rupee you are 2,000 shares long, after the fifth 10,000, after the tenth 20,000. The P&L is the area under that rising delta, a triangle with base 10 and height 20,000, which is 1,00,000.

What is the arithmetic, and where does the one half come from?

Expand the book's value as a Taylor series in the stock price. The first-order term is delta times the move, zero here; the second-order term is one half of gamma times the move squared, 0.5 x 2,000 x 100 = Rs 1,00,000; and the new delta is the derivative of that, gamma times the move, 20,000 shares. The one half is the same one half as in the area of a triangle: delta started at zero and finished at 20,000, so on average it was 10,000 shares over the Rs 10 move. The limitation is that a jump also changes implied volatility and burns a day of theta, both ignored here.

What happens if you rehedge and the stock comes back?

You sell 20,000 shares at Rs 510. If the stock then falls back to Rs 500, the options give back their Rs 1,00,000 but the short stock earns 20,000 x Rs 10 = Rs 2,00,000, so you net Rs 1,00,000 from the round trip. That is what long gamma means in practice: each rehedge locks in half of gamma times the move squared, and a stock that moves a lot and comes back pays you twice. The cost is theta, the daily decay you pay for holding the options, and the trade only works if realised movement is larger than the implied volatility you paid for.

The call is worth 20/3, about 6.67, and the 70% does not matter because the call can be copied with stock and cash. Hold 2/3 of a share and borrow 60: at 120 the copy is worth 80 - 60 = 20, at 90 it is worth 60 - 60 = 0, matching the call in both states. The copy costs 66.67 - 60 = 6.67, so the call must too. The real-world odds are already in the stock price, which the copy uses.

How do you copy the call?

If a shop sells a gift box of two items for more than the items cost separately, you buy the items and skip the box; the box's price is pinned by what goes in it. Options work the same way. Find a mix of stock and cash that pays exactly what the call pays in every state, and the call must cost what the mix costs, whatever anyone believes about the odds. The call pays 20 or 0, a swing of 20, while the stock swings from 120 to 90, a swing of 30. So hold 20/30 = 2/3 of a share. At 120 that is 80, which is 60 too much, and at 90 it is 60, also 60 too much: borrow 60 today and repay it in either state.

Where did the 70% go?

It is in the stock price. A stock that goes up 70% of the time to 120 and is still priced at 100 today is one the market demands a return on, because it is risky. The copy buys the stock at that price, so it inherits whatever the market thinks of the odds and the risk. The option is priced relative to the stock, not relative to anyone's forecast, so the real-world probability cancels out of the answer. What does appear is a different probability, q, the one that makes the stock earn the risk-free rate: 100 = q x 120 + (1 - q) x 90, so q = 1/3. Discounting the call's payoff at q gives 1/3 x 20 = 6.67, the same answer by another route.

Then show the arbitrage, since that is what makes the answer binding. If the call traded at 8, sell it and buy the copy for 6.67: you pocket 1.33 today and the two positions cancel in both states. If it traded at 5, do the reverse. The limitation is the one-step world: real prices take many values, so the copy has to be rebalanced as the stock moves, which is where the Black-Scholes model comes from, and where trading costs and jumps make the copy imperfect.

Rs 11. A first tail on flip n pays 2^n with probability 1/2^n, so each of flips 1 to 10 contributes exactly Rs 1, Rs 10 in all. Reaching flip 11 or later has probability 1/1,024 and pays the cap of Rs 1,024, adding one more rupee. Without the cap every flip keeps adding Rs 1 and the expected value is infinite, yet nobody would pay much to play, which shows expected value alone cannot price a bet on outcomes the payer cannot honour.

Why does every flip add exactly one rupee?

Think of a raffle where each ticket in the next bundle is half as likely to win but the prize is twice as large. Every bundle is worth the same to you. Here the first tail on flip 1 pays Rs 2 half the time, on flip 2 pays Rs 4 a quarter of the time, on flip 3 pays Rs 8 an eighth of the time. The payout doubles exactly as the probability halves, so each flip contributes 2^n x 1/2^n = Rs 1 to the expected value, and the fair price is simply the number of flips before the cap bites, plus the capped tail. Ten flips reach Rs 1,024; anything later is paid at the cap, worth 1,024 x 1/1,024 = Rs 1.

What goes wrong without the cap?

Remove the cap and the sum is 1 + 1 + 1 + ... with no end, an infinite expected value. Yet the game pays Rs 8 or less seven times in eight, and more than Rs 11 only one time in eight. The infinite value lives entirely in outcomes so rare and so large that no house could pay them, so the cap is not a detail; it is the price. Raising the cap barely moves it: a cap of Rs 1 crore makes the game worth only about Rs 24.19, because each doubling of the cap adds one rupee. That is the market maker's answer to the paradox: price what the counterparty can actually pay.

Mention the other classic resolution, then put it in its place. Economists answer the paradox with diminishing utility: a doubling of wealth is worth less to you than the first rupee, so a risk-averse player pays little. True, but on a desk the binding constraint is usually the counterparty's balance sheet, not your utility. The limitation of the Rs 11 is the same as any fair value: it is an average over many plays. Played once, the price feels steep because 87.5% of the time you get Rs 8 or less.