Derivatives Foundation puzzles, solved step by step

About 12 billion, with an honest range of 10 to 14 billion. A floor plate of 4,000 square metres at 3.8 metres a floor gives 1.52 million cubic metres over 100 floors; take half as usable, 0.76 million. A golf ball is 4.27 centimetres across, about 41 cubic centimetres, so 24,500 would fit in a solid cubic metre. Spheres leave gaps: at a packing factor of 0.64 that is 15,700 per cubic metre, and 0.76 million times 15,700 is 11.9 billion.

Why build a chain instead of guessing a number?

If someone asks how much rice a storeroom holds, you do not guess a tonnage. You measure the room, decide how much of it is shelving, and know how much a sack holds. An estimation question is marked on whether each link in the chain is stated, sized and defensible, not on the final figure. Here the chain is floors, floor plate, floor height, usable fraction, ball volume, packing factor. Say the six links before you fill any of them in, so the interviewer can follow and correct a single link rather than the whole answer.

Where does the packing factor come from, and why does it matter so much?

Pour marbles into a jar and shake it: they settle at roughly 64% of the jar's volume filled, with the rest air. Stack them by hand in the tightest possible pattern and you reach about 74%. Tip them in gently without shaking and you can be as low as 56%. The floor plate and height are things you can look up or pace out, but the packing factor is a physical assumption you must own, and it moves the answer from 10.4 to 13.8 billion on its own. Say which one you are using and why: balls poured into a building are a shaken random pile, so 0.64 is the defensible middle.

What do you say when the interviewer pushes on the usable fraction?

Lifts, stairs, structural columns, ducts and ceiling voids take a large slice of a tower, and nobody knows it to the percent. Give a range and show which way it pushes the answer: 40% usable takes you to about 9.5 billion, 60% to about 14 billion, so the final figure is somewhere between 10 and 15 billion whatever you assume. A candidate who reports a single number to three figures has missed the point of the question. A candidate who says 12 billion, give or take a third, with the reasons, has answered it.

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.

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.

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.

About 8,400 people, with an honest range of roughly 5,558 to 12,994. Of 45 floors, take 40 as ordinary office floors after plant rooms, lobby and trading floors. A floor plate of 3,000 square metres with 70% usable gives 2,100 square metres of desk space per floor. At 10 square metres per person that is 210 people a floor, and 40 floors give 8,400. The widest assumption is the space per person, 8 to 12 square metres, which alone moves the answer by half.

Why build from the floor rather than from the bank?

If you wanted to know how many people a wedding hall holds, you would not guess from the size of the family; you would pace the hall and think about chairs per row. An estimate is only as good as the link the listener can check, and anyone who has worked on an office floor has a feel for how many desks it holds. So the chain is office floors, floor plate, usable share, square metres per person. Say the four links before any number, so the interviewer can argue with one of them rather than with the whole answer.

Which assumption should you spend your time on?

The one with the widest range. Floors run 38 to 42, a ratio of 1.11; the plate 2,700 to 3,300, a ratio of 1.22; the usable share 65% to 75%, 1.15; but space per person runs 8 to 12 square metres, a ratio of 1.5, so it moves the answer most. Check it against a floor you know: a trading floor packs people at 6 to 8 square metres, a floor of meeting rooms and offices spreads them at 15 or more. If the interviewer gives you one fact, ask for that one.

What would you add about occupancy?

That desks and people are different counts. With hot-desking, a floor of 210 desks might be home to 250 or 300 people who are not all in on the same day, so a question about who works in the building can give a larger answer than a question about who is in it. Say which one you are answering. The limitation to state: the usable share and the space per person are guesses from general experience, not measurements, and a real number would come from the building's floor plans and the badge-in data.

No. The sample standard deviation is biased low, by about 1.3% at n = 20, and for a volatility estimate that is small next to the sampling noise. The square root is concave, so by Jensen's inequality the expected value of the root is below the root of the expected value. For normal returns the exact factor is c4(n): the expected sample standard deviation is 0.9869 times the true one at n = 20. On the same 20 points the estimate's own standard error is around 16%, so the bias is not what you should worry about.

Why does an unbiased variance give a biased standard deviation?

If you average the areas of several square plots and then take the square root, you do not get the average of their side lengths; the big plots pull the area average up more than they pull the side average. Unbiasedness is a statement about averages, and averages do not pass through a curved function: for a concave function like the square root, the average of the outputs is below the output of the average. That is Jensen's inequality, and it is the whole answer. The variance is unbiased; its root is not, and the direction is down.

Does it matter for a 20-day volatility estimate?

Not much, and saying why is the second half of the marks. The bias is 1.3%, but the standard error of a standard deviation from 20 observations is roughly 1 over the root of 2 times 19, about 16% of the true value, so the noise is more than ten times the bias. You can multiply by 1/c4 to remove the bias if you like, but you cannot remove the noise without more data, and 20 daily returns is simply a short window. On a volatility desk the honest answer is that a 20-day estimate of 16% could easily have been 13% or 19%.

What assumption does the exact factor need?

Normal returns. The direction of the bias, downward, holds for any distribution, because it comes from concavity alone; the size of the factor c4 depends on the distribution, and fat-tailed daily returns change it. Say the general result first, Jensen, then the normal-case number, then the limitation. That order shows you know what is a theorem and what is a model.

Let the other player choose first, then take the die that beats theirs. Red beats green 5 times in 9, green beats blue 5 in 9, and blue beats red 5 in 9. The three dice form a cycle like rock, paper, scissors, so no die is best on its own; the advantage belongs to whoever picks second. If you must pick first, no choice does better than 4 in 9 against a wise opponent, and you should say so rather than pretend one die is stronger.

How can three dice with the same average not have a best one?

Three cricket teams can each beat one of the others and lose to the third; a league table would show them level, and still no team is the best. Winning a roll depends only on which die shows the higher face, pair by pair, and pairwise comparisons do not have to line up in a single order the way averages do. Red and blue average 5 and green averages 6, and yet green loses to red 5 times in 9: every head-to-head is lopsided, 5 to 4, in a circle, and the highest average sits inside it. Green's 12 wins by a mile and its 1 loses by a mile, and a roll pays nothing for the margin. The question tests whether you check the comparison that matters instead of the summary that does not.

How do you check a pair quickly in the room?

Write one die's faces across and the other's down and count the cells where the first is higher. Red against green: 2 beats only the 1; 6 beats 1 and 5; 7 beats 1 and 5; that is 1 + 2 + 2 = 5 of 9. Green against blue: 1 beats nothing, 5 beats 3 and 4, 12 beats everything, again 5 of 9. Blue against red: 3 and 4 each beat the 2, and 8 beats 2, 6 and 7, again 5 of 9. Three counts, under a minute, and the cycle appears.

What is the trading lesson the interviewer is after?

That the order of moves can be worth more than the thing being chosen. The second mover has a guaranteed 5 in 9; the first mover, against someone who knows the cycle, has at best 4 in 9, so you should pay to move second and never volunteer to move first. That is the same instinct as quoting after you have seen the other side's interest rather than before. The limitation to state: the edge is only 5 to 4, so over a few rolls luck dominates, and a one-roll bet on it is a small edge with a large variance.

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.

5/9, about 55.6%. Put A's arrival time across and B's up a 60 by 60 square; every pair of times is a point, all equally likely, so probability is area. They meet when the times are within 20 minutes, the band either side of the diagonal. They miss in two corner triangles, each with legs of 40 minutes and area 800 of 3,600, which is 2/9. So the meeting chance is 1 minus 4/9 = 5/9.

Why turn two arrival times into a square?

Throw a dart at a square board without aiming and the chance it lands in any patch is just that patch's share of the board. Two independent arrival times, each spread evenly over the hour, behave exactly like that dart: A's time picks a position across, B's time picks a position up. With two independent uniform times, every pair of arrivals is a point in a 60 by 60 square, equally likely anywhere in it, so a probability becomes an area you can see. The event they meet is the set of points where the two times differ by less than 20 minutes, a band hugging the diagonal.

Why is it easier to compute where they miss?

The band is an awkward six-sided shape; the regions outside it are two clean triangles. In the top-left triangle B arrives more than 20 minutes after A, so A has gone; in the bottom-right one, A is the late one. Each triangle has legs of 60 minus 20 = 40 minutes, so its area is 40 x 40 / 2 = 800 square minutes out of 3,600, which is 2/9, and the two together are 4/9. The clause about leaving at 2 pm changes nothing, because no one can arrive after 2 pm anyway; it only stops the question from looking ambiguous.

How does the answer move with the waiting time?

Not in a straight line, and that is a common follow-up. Doubling the wait from 10 to 20 minutes takes the meeting chance from about 31% to about 56%, not from one third to two thirds, because the miss region shrinks as a square. The table runs the formula for four waits. The trading version is two orders that must arrive within a latency window to match: halving the gap you can tolerate does more than halve the matches, and a picture of the square is the fastest way to see by how much. The limitation to state is the uniform assumption; real arrivals bunch near the hour, which raises the meeting chance.