Derivatives Foundation puzzles, solved step by step

4/7 of days trend, about 57.1%, and a trend lasts 3.3 days on average. In the long run the flow out of trending must equal the flow in: 0.3 times the trending share equals 0.4 times the choppy share, so the shares stand 4 to 3. A trend ends on any given day with probability 0.3, so its expected length is 1/0.3, and a choppy spell lasts 1/0.4 = 2.5 days.

Why must the two flows balance?

Picture two rooms at a party with a door between them. Each minute, 30% of the people in room A wander into B and 40% of those in B wander into A. The crowd settles when the two queues through the door carry the same number of people; otherwise one room keeps filling. A stationary split is one where the number of days leaving each state equals the number entering it, and that single equation fixes the split. Days leaving trending: 0.3 times the trending share. Days entering it from choppy: 0.4 times the choppy share. Set them equal and the ratio is 4 to 3.

Why is the average streak 1/0.3 and not something longer?

A trend that has lasted five days is no more likely to end tomorrow than one that started today: the chain has no memory beyond yesterday. Each trending day ends the run with probability 0.3, independent of its age, so the run length is a geometric count with mean 1/0.3 = 3.33 days. That is the same reason the expected number of rolls to a six is 6. The two answers also check each other: the share of a state equals how often a run of it starts, times how long it lasts, and 4/7 against 3/7 is exactly 3.33 against 2.5 scaled by the same start rate.

What does the chain say about tomorrow, given today?

This is where the puzzle connects to trading. Today's state carries real information: after a trending day, tomorrow trends with probability 0.7, well above the unconditional 57%. After a choppy day it is only 0.4. The long-run split tells you nothing about tomorrow; the transition row for today's state does. Say that distinction out loud, because an interviewer who hears 57% quoted as a one-day forecast knows you have confused the stationary distribution with a conditional one. The limitation to add: a two-state chain with fixed probabilities is a toy, and real regime persistence drifts over time.

25.98% at a correlation of 0.5, and 21.21% at zero. Basket variance is the sum of the two weighted variances plus twice the weighted covariance: 0.25 x 0.09 + 0.25 x 0.09 + 2 x 0.25 x 0.5 x 0.09 = 0.0675, whose square root is 25.98%. At zero correlation the cross term vanishes, leaving 0.045, whose root is 21.21%. The basket is less volatile than either stock because they do not move in step.

Why is the basket calmer than the stocks inside it?

Two commuters who each arrive late by a random ten minutes rarely arrive late together; the average of their lateness swings less than either does alone. Volatilities do not add; variances do, and the cross term that joins them is scaled by the correlation, so anything below a correlation of 1 cuts the basket's swing below its parts. At a correlation of 1 the two stocks are one stock and you get 30% back. At minus 1 they cancel exactly and the basket is flat.

How do you do it in your head?

Use the shortcut in the formula: with two equal stocks the basket volatility is 30% times the square root of (1 plus rho) over 2. At rho 0.5 that is 30% times the root of 0.75, about 0.866, giving 26.0%; at rho 0 it is 30% times the root of 0.5, about 0.707, giving 21.2%. Say the structure first, then the number, so a slip in the arithmetic does not look like a slip in the thinking.

What is the limitation you should name?

The formula treats correlation as a fixed number, and it is not. Correlations between stocks tend to rise in a sell-off, which is exactly when a basket holder wants the diversification, so the 25.98% is a fair-weather figure. On a derivatives desk that is why basket options and dispersion trades are priced with a correlation assumption that is marked, stressed and hedged rather than looked up once. Say that the answer depends on the correlation you assume, and that the assumption is the risk.

7.50. Three moves of plus or minus 10 end at 130, 110, 90 or 70, reached by 1, 3, 3 and 1 of the eight equally likely paths. The 100 call pays 30 at 130, 10 at 110 and nothing below. Its value is the average payoff: (1 x 30 + 3 x 10) over 8, which is 60 over 8, or 7.50. With zero rates and symmetric moves, the real-world probabilities are already the pricing probabilities, so no discounting and no adjustment is needed.

Why is counting paths all the tree needs?

Think of three coin tosses where you get a sweet for each head. The chance of exactly two heads is not one in four; it is three in eight, because there are three orders in which two heads can arrive. The tree recombines, so the value of an end node is its payoff weighted by how many of the eight paths reach it, and the path counts are the binomial coefficients 1, 3, 3, 1. Nothing else in the problem carries information: the step size fixes the end prices and the counts fix the weights.

Where does the risk-neutral machinery go?

In a general tree you would replace the real probabilities with the risk-neutral ones, chosen so that the stock's expected growth equals the interest rate. Here rates are zero and the moves are symmetric, so the stock already has zero expected drift and the risk-neutral probability is the same one half you were given. Say that out loud: it shows you know the shortcut is a coincidence of the setup, not a rule. If the up move were 10 and the down move 5, one half would no longer price the stock and you would have to solve for the probability that does.

What sanity checks do you say before the number?

Two quick ones. The call cannot be worth more than the expected value of the stock above the strike ignoring the max, which is zero here, so the call is worth exactly the expected positive part, and that is what 7.50 is. And put-call parity with zero rates says the 100 put must also be 7.50, which you can confirm from the lower nodes: (3 x 10 + 1 x 30) over 8. Giving the put price unprompted, and showing it matches, is the cheapest way to prove the tree was right.

The stock ends at about 28% of its start; the three-times fund at roughly 8 millionths of its start, effectively zero. Each up-and-down pair multiplies the stock by 1.1 x 0.9 = 0.99, and 125 pairs give 0.99^125 = 0.285. The leveraged fund moves 30% each way, so each pair is 1.3 x 0.7 = 0.91, and 0.91^125 is about 7.6e-06. The arithmetic average return is zero in both cases; the compounded return is not.

Why does a zero average return lose money?

Take a 100 rupee note to a shop that marks everything up 10% in the morning and discounts 10% in the afternoon. The afternoon discount is taken off a bigger number, so the price ends at 99, not 100. A gain and a loss of the same percentage do not cancel, because the loss acts on the larger base; the pair costs the square of the move, 1% for a 10% swing. Repeat that 125 times and the 1% losses compound to a 72% fall. This is volatility drag: the gap between the average return and the compounded return.

Why is three times the move so much worse than three times the loss?

The drag per pair is the square of the move. Tripling the move multiplies the drag by nine, not three: the stock loses 1% per pair, the fund loses 9%. That is why a daily-rebalanced leveraged fund in a choppy, sideways market bleeds even when the underlying ends flat. The general rule you can quote: over many periods the compounded growth rate is roughly the average return minus half the variance, and leverage multiplies the variance by the square of the leverage.

What would you say to a client who holds the three-times fund?

That the product tracks three times the daily move, exactly as promised, and that this is not the same as three times the return over a year. A leveraged fund is a tool for a view on the next day or week; held through a sideways year it loses to its own rebalancing. The limitation to state: the alternating path is the worst case for drag, and a strongly trending market can make a leveraged fund return more than three times the underlying. The drag is about path, not just direction.

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.

About 0.45. Buying the 99 call and selling the 101 call pays 0 below 99, 2 above 101 and a straight ramp between. Divide that by the width of 2 and the payoff is 0 below 99, 1 above 101 and a ramp through 100: a digital with its edge smoothed over two points. Its cost is (5.60 minus 4.70) over 2, which is 0.45. The narrower the spread, the closer the ramp sits to the step, and the price converges to the digital.

Why does a call spread stand in for a digital?

A light switch is a step: off or on. A dimmer that goes from fully off to fully on over a tiny turn of the knob is, for every practical purpose, the same switch. A call spread over its width is a dimmer: it ramps from 0 to 1 across the two strikes, and as the strikes close in on 100 the ramp becomes the step. So the digital is the limit of a scaled call spread, and a traded call spread gives you a price for it without any model.

Is 0.45 the digital's price or an approximation, and which way is it off?

It is an approximation to the slope at 100 taken from two points either side. Because the spread is centred on 100, the first-order error cancels and what remains is small, of the order of the curvature of the call price between 99 and 101. If the digital were struck at 99 instead, the same spread would overstate it, because the call price is convex in strike and the ramp sits above the step on that side. On a desk you would quote the digital from the tightest spread the market will show you, and hedge it with that spread, so the approximation is also the hedge.

What does 0.45 say about the market, and what is the limitation?

A digital paying 1 above 100 at 0.45, with rates near zero, means the pricing probability of finishing above 100 is about 45%, slightly below one half. That is a risk-neutral probability, not a forecast, and it is pulled down by the skew: with a steeper put skew, out-of-the-money calls are cheaper in volatility terms and the slope in strike is steeper, which moves the digital. Say that a flat-volatility formula would miss this, and that the call spread picks the skew up automatically because it uses the two traded prices.

You lose everything about 2.0% of the time, 0.52 to the sixth power, and the expected result is about minus Rs 265. Rs 63,000 covers exactly six bets: 1, 2, 4, 8, 16 and 32 thousand. A win at any of them recovers every earlier loss and nets Rs 1,000, which happens 98.0% of the time. Six losses in a row cost all Rs 63,000. Weighted, 980 of expected winnings against 1,246 of expected loss leaves minus Rs 265.

Why does a plan that wins 98 times in 100 still lose money?

Picture a friend who sells phone insurance to classmates for Rs 50 a month. Month after month nobody drops a phone, and the Rs 50 notes pile up; it feels like free money until the month three phones go into a pond. A win rate tells you how often you are paid, not how much you are paid against how much you can lose, and the expected value needs both. Doubling after every loss builds exactly that shape: Rs 1,000 collected almost every time, and Rs 63,000 handed back rarely. The rare branch is 63 times the size of the common one, so a 2% chance of it more than cancels a 98% chance of the small win.

How do you lay out the six bets in the room?

Write the ladder down before computing anything. The stakes are 1, 2, 4, 8, 16 and 32 thousand, which add to 63 thousand exactly, so the seventh bet of 64 thousand can never be placed. If the first win comes at bet k, it pays 2 to the power k minus 1 thousand, and the losses before it add to one thousand less than that, so every winning session nets exactly plus Rs 1,000. There are only two outcomes, and the table shows how quickly the chance of reaching each rung falls: by the sixth bet you are staking Rs 32,000 to recover Rs 31,000 of losses and win one more thousand.

Is there a faster way to see the sign without the ladder?

Yes, and it is the one a trader reaches for first. Every rupee placed on red loses 4 paise on average, whatever happened on the previous spin, because the wheel has no memory. The expected result of any staking plan is the edge per rupee times the expected total amount staked: here minus 4% of Rs 6,633, which is minus Rs 265, the same figure as the ladder. Doubling raises the amount you put down when you are losing; it cannot change the sign of the edge. On a fair 50/50 wheel the same plan has an expected value of exactly zero, with the same lopsided shape.

Why does a desk interviewer care about a roulette plan?

Because the shape is the shape of selling far out-of-the-money options, or of adding to a losing position to get back to flat. Both produce a long run of small gains and a rare large loss, and a good-looking track record says almost nothing about the tail. Repetition makes the rare branch common: play 50 sessions and the chance of at least one ruin is 1 minus 0.98 to the 50th, about 63%. The limitation to state is that the plan assumes no table limit; a casino maximum bet cuts the ladder short and makes ruin more likely, not less.

About 1% a day, Rs 10. With roughly 256 trading days in a year and volatility growing with the square root of time, daily volatility is annual volatility divided by 16, so 16% a year is 1% a day. A delta-hedged long option earns half its gamma times the square of each day's move and pays theta every day; the two cancel when the move equals the implied daily move. Using 252 days gives 1.008%, still Rs 10.

Where does dividing by 16 come from?

Walk randomly on a straight road, one step forward or back each second, and after 100 seconds you are typically about 10 steps from where you began, not 100, because most steps undo each other. Price moves add up the same way. Volatility scales with the square root of time, and the square root of 256 trading days is 16, so an annual volatility divided by 16 is the standard deviation of one day's move. Traders call this the rule of 16. It makes 16% implied volatility the cleanest number on the screen: 1% a day, Rs 10 on this stock.

Why does that daily move decide whether the hedged option makes money?

Once the delta is hedged, the option's daily P&L is two pieces. Gamma pays you half gamma times the square of the move, in either direction; theta charges you a fixed amount for the day passing. For a one-month at-the-money option here, gamma is 0.0087 per rupee and theta is 0.435 a day. A Rs 10 move earns 0.5 x 0.0087 x 100 = 0.435, exactly the theta, because the option's price was built so that theta pays for a move of one implied standard deviation. A flat day loses 0.435; a Rs 20 day makes 1.31.

What does the quick answer leave out?

Three things, and naming one earns the follow-up. First, the breakeven is on the average squared move, not the average move. If the stock's moves are normal with a standard deviation of Rs 10, its average absolute move is only about Rs 7.98, so a stock that typically moves Rs 8 a day is already moving enough to break even. Second, gamma changes as the stock drifts away from the strike and as expiry nears, so the Rs 10 holds for an at-the-money option on the day you measure it. Third, hedging once a day adds noise to the P&L even when realised volatility exactly matches implied. The rule of 16 is a desk shortcut, not a pricing model.

Exactly the same. Each glass ends with 100 ml, so whatever wine is missing from the wine glass has been replaced, millilitre for millilitre, by water, and the missing wine can only be in the water glass. With a 10 ml spoon and a thorough stir, the return spoon carries back 0.91 ml of wine and 9.09 ml of water, leaving 9.09 ml of water in the wine and 9.09 ml of wine in the water.

Why does the first spoon feel like it settles the question?

Because it is pure wine going one way and a diluted mixture coming back, so it feels as though more wine travelled. Think instead of two cricket teams of eleven who swap some players and still field eleven each. Each glass ends with exactly 100 ml, so every millilitre of wine that left the wine glass and did not come back has been replaced by a millilitre of water: the two foreign amounts must be equal. The number of team A players now in team B is the number of team B players now in team A, however the swaps were done.

What do the millilitres actually look like?

Take a 10 ml spoon. After the first transfer the water glass holds 100 ml of water and 10 ml of wine, 110 ml in all, so a stirred spoonful from it is 10/110 wine. The return spoon carries 0.91 ml of wine and 9.09 ml of water, so 9.09 ml of wine stays behind in the water glass and 9.09 ml of water arrives in the wine glass. The arithmetic confirms the argument, but the argument came first and did not need the spoon size, the stirring or any division.

Why do the interviewer's variations not change the answer?

Interviewers vary the story: no stirring, five spoonfuls back and forth, a ladle instead of a spoon. As long as both glasses end at their starting volume, the answer is equal, because the argument uses only the totals. The limitation to say out loud: if the return spoon is a different size from the first, the glasses end at different volumes and the amounts differ, so check the volumes before using the shortcut. The desk lesson is the bookkeeper's: in a closed system, look at the totals before tracking every transfer, the same way a net position check catches a booking error faster than replaying every ticket.

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.