Derivatives Foundation puzzles, solved step by step

About 4.5%. Count 100,000 trades. One in a thousand is suspicious, so 100 are, and 95 of those alert. The other 99,900 are normal, and 2% of them, 1,998, alert anyway. Alerts total 2,093, of which 95 are genuine, so the probability that an alerted trade is suspicious is 95 over 2,093, about 4.5%. The 95% hit rate is not the answer; the base rate is what decides it.

Why does a 95% accurate system give a 4.5% answer?

A smoke alarm that goes off for 2% of toast is a fine alarm in a house that is never on fire; nearly every ring will be toast. When the thing you are looking for is rare, even a small false-alarm rate applied to the huge normal pile produces more alerts than the true cases produce. Here 2% of 99,900 normal trades is 1,998, twenty times the 95 genuine alerts. The system is not bad; the base rate is low, and that is what the question is testing.

What is the fastest way to say it in the room?

Do not write Bayes' formula; count a round number of trades. Say: in 100,000 trades, 100 are suspicious and 95 alert; 99,900 are normal and 1,998 alert; 95 over 2,093 is about 4.5%. Three sentences, no algebra, and every number is checkable by the person listening. The odds form is just as quick: prior odds 1 to 999, likelihood ratio 0.95 over 0.02, about 47.5, so posterior odds 47.5 to 999, roughly 1 to 21.

What does the desk do with a 4.5% answer?

It decides what the alert is for. A 4.5% hit rate is fine for a filter that sends trades to a human for a second look, and useless for an automatic block, because 95% of blocked trades would be legitimate business. That is the trade-off every surveillance, fraud and risk-limit system lives with: a lower threshold catches more of the 100 but drags in more of the 99,900. Say the limitation too: the 2% and 95% are themselves estimates from past data, and a system tuned on last year's patterns can drift.

You finish short 4, you lose Rs 2 on the settlement, and your expected profit when you stopped trading was Rs 6. Sold 5, bought 3, sold 2 is a net short of 4. Cash is +37.5 - 19.5 + 16 = +34. Settling at 9 costs 4 x 9 = 36, so the result is 34 - 36 = -2. Against the fair value of 7 the short would have cost 28, leaving +6: the edge of 2.5 + 1.5 + 2 captured on the three trades.

Why keep three numbers in your head and not one?

A shopkeeper who sells umbrellas at a markup has a profit on each sale, a stock count, and a worry about whether it rains. Three separate things. A market maker tracks the same three: edge per trade against fair value, net position, and the exposure to the final number, and the game checks that you never let one of them slip. Say the fair value, 7, first. Then say each trade's edge as you do it: +2.5 on selling 5 at 7.5, +1.5 on buying 3 at 6.5, +2 on selling 2 at 8. Then say the position: short 4.

Was the loss a mistake?

No. Every trade was done at a better price than fair value, so the trading was right; the dice came in high. Expected profit of +6 is what you controlled, and the realised minus 2 is what the dice did; an interviewer wants to hear you separate the two without being asked. The sum of two dice has a standard deviation of about 2.4, so a short of 4 carries a one-standard-deviation swing of nearly 10, far larger than the 6 of edge. The real question is whether a short of 4 was more risk than you wanted to carry against 6 of edge.

What would you have done differently in the game?

Skewed the quote as the short grew. After selling 5, you are short 5 and should lower both your bid and your offer so that the next trade is more likely to be a buy that cuts the position, which is exactly what buying 3 at 6.5 did. Selling 2 more at 8 added to the short again; at that point a wider or higher quote would have protected you. The limitation of the puzzle: with two interviewers trading against you, their trades carry information about nothing, because the dice are not rolled yet, so here the only reason to skew is inventory, not adverse selection.

Yes: buy at 22, because the ticket is worth 23.08. A standard deck has 12 picture cards out of 52, a 23.1% chance, so the fair value of a Rs 100 payout is 12/52 x 100 = 23.08. You buy at the offer, 22, which is below fair, and expect to make about Rs 1.08 a ticket. Selling at 20 would give away about Rs 3.08. The edge is small next to the outcome, so it pays only over many tickets.

How do you value the ticket?

If a friend offers you Rs 100 when a coin lands heads, you would pay up to Rs 50 and no more; the coin's odds set the price. A ticket that pays a fixed amount on an event is worth the payout times the probability of the event, before any talk of the quote. A standard deck has four suits, each with a jack, a queen and a king, so 12 picture cards in 52. The ticket is worth 12/52 x 100 = 3/13 x 100 = 23.08. Notice that the ace is not a picture card; counting it is the quickest way to get 16/52 and the wrong side.

Which side of the quote do you compare with?

A quote has two prices, and you can only use one of each: you buy at the offer and sell at the bid. Trade only when fair value sits outside the quote: buy when it is above the offer, sell when it is below the bid, and pass when it falls between them. Here fair is 23.08 against a 22 offer, so you buy and expect 1.08. The figure runs two other quotes for contrast: at 22 bid, 24 offered, fair is inside and there is no trade; at 24 bid, 26 offered, you sell at 24 and expect 0.92.

How much would you buy, and what could change your mind?

Each ticket pays 0 or 100, so its standard deviation is 100 x the square root of 0.231 x 0.769, about Rs 42, against an edge of Rs 1.08. The edge is real but small next to the noise, so it is worth taking in size over many independent draws and worth very little on a single ticket. Before trading, ask the two questions a desk would: is the deck standard and well shuffled, and does the person quoting know something you do not, such as a card already removed? The limitation of the clean answer is that it trusts the setup; in a game where the quoter controls the deck, a quote this generous is itself a warning.

Take 10 flips. You are betting on luck, and luck averages out as the trials pile up. With 10 flips, 6 or more heads happens 386 times in 1,024, about 37.7%. With 100 flips, 60 or more heads happens about 2.8% of the time. The share of heads settles towards 50% at the rate of one over the square root of n, so the 60% line gets harder to reach with every extra flip.

Why does the number of flips change the odds at all?

Think of a school with two cricket teams, one of eleven and one of a hundred and ten. If someone offers a prize for a team whose average height is 10 cm above the national average, the small team is the one that can win: one or two tall players move its average, while the big team's average is pinned down by sheer numbers. The same coin, flipped more often, produces a share of heads that sits ever closer to a half, so a payout that needs an unusual share wants the fewest flips you can get. This is the law of large numbers working against you, and the question is testing whether you know which side of it you are on.

How do you put a number on it without a table?

Count the small case exactly: 6 or more heads in 10 flips means adding the ways to get 6, 7, 8, 9 and 10 heads, which are 210, 120, 45, 10 and 1, a total of 386 out of 1,024, so 37.7%. For 100 flips use the normal approximation. The standard deviation of the share of heads is 0.5 over the square root of n: 15.8% for 10 flips, 5% for 100. The 60% line is 0.6 standard deviations out in the first case and 2.0 in the second, and two standard deviations in one tail is about 2.3%. The exact binomial answer is 2.8%; the approximation gets you to the right decision in one breath.

What is the interviewer listening for after the answer?

Say the general rule and then the exception. The rule: whenever a payout needs the sample to look unlike the population, choose the smallest sample. The exception: if the payout were for landing between 40% and 60%, you would want the most flips, for exactly the same reason. Read the sign of the bet before you choose the sample size: a bet on luck wants few trials and a bet on the average wants many. If the interviewer changes the wording to more than 60%, 7 or more heads in 10 is 17.2%, still far above the 100-flip figure. And if the choice is between 100 and 1,000 flips, 600 or more heads happens about 1.36e-10 of the time, which is as close to never as a desk needs.

0.09375, 1,596 and 0.1429. For 3/32, halve 1 five times to get 1/32 = 0.03125 and multiply by 3. For 38 x 42, both sit 2 away from 40, so the product is 40 squared minus 2 squared, 1,600 - 4. For 1/7, the repeating block is 142857, because 7 x 142857 = 999,999, so 1/7 = 0.142857... which rounds to 0.1429. Each one is a known anchor plus one step.

Why does the interviewer ask three small sums in a row?

A shopkeeper who totals a bill in his head is not doing long addition; he rounds to the nearest hundred and fixes the difference. Trading desks ask quick sums to see whether you reach for an anchor you already know, because that is how prices get checked in the two seconds before someone else trades. The three questions here each have a short route: fractions with a power of two below them are halvings, products of numbers either side of a round number are a difference of squares, and sevenths are one repeating block of six digits. The speed comes from knowing which route fits, and the interviewer listens to the route as much as to the number.

What is the route for each one?

For 3/32, keep halving from a half: 0.5, 0.25, 0.125, 0.0625, 0.03125. Every halving adds a digit, and 32 is five halvings, so 1/32 is 0.03125 and three of them are 0.09375. Bond traders do this all day, because US Treasury prices are quoted in 32nds. For 38 x 42, notice that both numbers sit 2 from 40, so the product is 40 squared minus 2 squared, 1,596, a trick that works for any pair placed evenly around a round number. The same trick gives 47 x 53 = 2,500 - 9 = 2,491. For 1/7, remember that 7 x 142857 = 999,999. So 1/7 is 142857 divided by 999,999, which is 0.142857 repeating, and the fifth decimal is 5, so it rounds up to 0.1429.

What does a strong candidate add after the answers?

Add the sanity check and the neighbours. A product of two numbers around a centre is always a little below the centre squared, so 1,596 must be under 1,600. Three 32nds must be just under a tenth, since 3.2 of them would make exactly 0.1. The sevenths share the same six digits in rotation, 2/7 = 0.285714 and 3/7 = 0.428571, so knowing one seventh gives you all six. The limitation is honest too: anchors cover the common cases, and for an awkward product like 37 x 46 you fall back on splitting, 37 x 46 = 37 x 50 - 37 x 4 = 1,850 - 148 = 1,702. Saying when you switch method sounds better than pretending one trick does everything.

1,000,994, then 121.5, then 9,801. 998 x 1,003 is (1,000 - 2)(1,000 + 3) = 1,000,000 + 1,000 - 6. For 2.5% of 4,860, take 10%, which is 486, and then a quarter of it, 121.5. For 99 squared, (100 - 1) squared is 10,000 - 200 + 1 = 9,801. In every case the move is the same: go to the nearest round number and fix the small error afterwards.

Why is the round number always the first move?

Buying seven items at Rs 99 each, nobody multiplies 99 by 7; you take Rs 700 and hand back Rs 7. Multiplying by a round number is free, and the correction is a small product you can hold in your head, so every mental multiplication is a round-number product plus or minus a correction. For 998 x 1,003 the round number is 1,000 for both, so the product is 1,000,000, the two cross terms are + 3 x 1,000 and - 2 x 1,000, which net to + 1,000, and the only real work is the sign of 2 x 3 at the end: a minus times a plus is a minus, so subtract 6.

How do you handle percentages that are not round?

Build the percentage out of ones you can do instantly. 2.5% is a quarter of 10%, so take a tenth of 4,860, which is 486, and quarter it: half is 243, half again is 121.5. The same habit covers 7.5% as 10% minus a quarter of 10%, 15% as 10% plus half of that, and 12.5% as an eighth. On a desk this is how you convert a basis-point move to rupees while someone is still talking: 25 basis points on Rs 4,860 crore of notional is the same 121.5, in crore.

What does the interviewer listen for beyond the answer?

Speed, but also the sanity check. 998 x 1,003 must be a little above a million because 998 x 1,003 is roughly 1,000 x 1,001, and 99 squared must end in 1 because 9 x 9 does, so a candidate who says 1,000,994 and 9,801 without pausing shows both the trick and the check. Where this breaks down is numbers that are not near anything round, such as 47 x 83; there the move is to split one factor, 47 x 80 + 47 x 3, and accept that it takes two beats rather than one.

At most Rs 100/11, about Rs 9.09. A total of 10 comes from (4, 6), (5, 5) and (6, 4), three of the 36 equally likely outcomes, so the chance is 1/12 and the odds are 11 to 1 against. A fair bet matches the odds: risk 1 to win 11, so to win Rs 100 the fair stake is 100/11. Risk Rs 10 and you lose about Rs 0.83 per game on average.

Why is the answer odds, not a probability?

A friend offers to pay you Rs 100 if your cricket team wins a match it wins one time in twelve. Paying Rs 20 to enter is a bad deal, and the way you know is by comparing what you risk to what you get: you lose Rs 20 eleven times and win Rs 100 once. A bet is fair when the stake and the prize are in the same ratio as the losing and winning chances, which is what odds express directly. One in twelve means eleven losses per win, so a fair stake is one eleventh of the prize: Rs 100/11, which is Rs 9.09. Anything above that is paying for the privilege of playing.

How do you check the stake with expected value?

Set the expected profit to zero. With stake s, you win Rs 100 with chance 1/12 and lose s with chance 11/12, so the expectation is 100/12 - 11s/12, which is zero when s = 100/11. The expected value route and the odds route must agree, and saying both in the room shows the interviewer you can move between them. Then say the direction: a lower stake is a profitable bet for you, a higher one is profitable for the house, and at Rs 10 the house edge is 8.3% of your stake, which is roughly the edge a casino takes on its better games.

What would a trader add after the number?

That the fair stake is a ceiling, not an offer. A trader asked to risk money quotes below fair, say Rs 8, and explains the gap as the edge for taking the other side, because a bet at exactly fair value makes nothing over many plays. The limitation is that 1/12 is the chance only if the dice are fair and the counting is right; if the interviewer changes the target to 7, the chance rises to 6/36 and the fair stake jumps to Rs 20, so the first thing to re-check whenever the rules change is the count of winning cells.

The call is worth Rs 0.50 and the put is worth Rs 1.00. The call pays 1 on a 5 and 2 on a 6, so its average payoff is 3/6. The put pays 3 on a 1, 2 on a 2 and 1 on a 3, so its average is 6/6. The put is worth more because the strike of 4 sits above the die's mean of 3.5, and call minus put equals 3.5 minus 4, which is minus 0.5.

Why is a die option priced by averaging the payoffs?

A school raffle with six equally likely tickets where ticket 5 pays Rs 1 and ticket 6 pays Rs 2 is worth exactly the average prize, Rs 0.50, because nothing else is uncertain and nobody can hedge a die roll. With equally likely outcomes and no hedge available, the fair price of a payoff is its expected value, so you list the payoff on each face and average. For the call with strike 4: 0, 0, 0, 0, 1, 2, which averages 0.5. For the put: 3, 2, 1, 0, 0, 0, which averages 1.0. Each payoff is floored at zero, which is what makes it an option rather than a forward.

What is the parity check, and why does it work on a die?

Add the call and subtract the put on every face. Call minus put on any single face equals the roll minus 4 exactly, because whichever side is in the money pays the gap and the other pays nothing, so the average of call minus put is the average roll minus the strike: 3.5 - 4 = - 0.5. That is put-call parity with no interest and no dividends, and it gives a one-line check: once you have the call at 0.5, the put must be 0.5 + 0.5 = 1.0. On a real option the same identity holds with the forward in place of the expected roll.

What does the interviewer ask next, and where does the analogy stop?

The next question is usually a different strike, or a market. Move the strike to 3 and the call pays 1, 2 and 3 on the top three faces, worth 1.0, while the put pays 2 and 1, worth 0.5, so the two swap values because the strike is now below the mean. Where the analogy stops is hedging: a real option is priced not by the expected payoff under your view but by the cost of replicating it with the underlying, which shifts the probabilities to the risk-neutral ones; on a die there is nothing to trade against, so the expectation under the real probabilities is the price.

89 ways. Think about the last move. Either it was a single step from step 9 or a double step from step 8, and those two cases cannot overlap, so ways(10) = ways(9) + ways(8). With ways(1) = 1 and ways(2) = 2, the sequence runs 1, 2, 3, 5, 8, 13, 21, 34, 55, 89. It is the Fibonacci rule, shifted by one place.

Why count by the last move rather than the first?

Ask how many ways there are to arrive at a railway junction and you count the lines coming in, not the stations people set out from. Every route to step n arrives from exactly one of two places, step n - 1 by a single or step n - 2 by a double, so the routes to n are the routes to those two places added together. The first move works just as well, but the last move makes the recursion read naturally from the top down, and it is the habit that generalises to harder counting problems where you condition on the final event.

How do you check 89 a second way?

Count by how many doubles you use. With k doubles and 10 - 2k singles you make 10 - k moves in total, and the number of orderings is 10 - k choose k, so the total is the sum over k from 0 to 5 of C(10 - k, k). That is 1 + 9 + 28 + 35 + 15 + 1 = 89, the same answer by a route that does not use the recursion at all. Two methods agreeing is the thing to say out loud; it also hands you the next question, since the terms tell you that four doubles and two singles is the most common shape of route.

Where does this pattern appear in trading, and where does it stop?

In anything built from steps of two sizes: the number of ways a price can move up to a level in ticks of one and two, or the number of paths in a recombining tree. The recursion is also the warm-up for dynamic programming, where the value of a position is built from the values of the positions it can reach, which is how an American option is priced on a lattice. The limitation is that Fibonacci only appears when every move is a one or a two; allow a three-step jump and the rule becomes a sum of the previous three terms, with 274 ways for ten steps.

3,136 and 5,402. Anchor 56 on 50: (50 + 6)^2 = 2,500 + 2 x 50 x 6 + 36 = 3,136. For 73 x 74, square the smaller number and add it once: 73^2 = (70 + 3)^2 = 4,900 + 420 + 9 = 5,329, then 5,329 + 73 = 5,402. Check each a second way: 56^2 = 60 x 52 + 4^2 = 3,120 + 16, and 73 x 74 = 70 x 74 + 3 x 74 = 5,180 + 222.

Why anchor on a round number instead of multiplying digit by digit?

Measuring a room, you do not count tiles one by one; you count whole rows, then the part row, then the corner. A square of side 56 is a 50 by 50 block, two 50 by 6 strips and a 6 by 6 corner. Anchoring on 50 replaces one hard multiplication with four numbers you already hold and one addition, which is the whole trick of mental arithmetic under a clock. 2,500, 300, 300 and 36 add to 3,136. The second strip is the one people drop; say both strips aloud.

How do you handle 73 x 74 when neither number is round?

Consecutive numbers are a square plus the smaller number: 73 x 74 = 73 x 73 + 73. That turns the problem into a square you can anchor, (70 + 3)^2 = 4,900 + 420 + 9 = 5,329, and one more addition, 5,402. Rewriting a product into a square and a correction works because a square has a shape you can reconstruct, and the correction is a number you already have. It also stops a common slip: people who try 70 x 74 then add 3 x 74 get it right too, but 3 x 74 = 222 is where under time pressure the 2 and the 22 get muddled.

Then check, out loud, in a different way. 56^2 is also 60 x 52 + 4^2, because (n + d)(n - d) = n^2 - d^2 with n = 56 and d = 4: 3,120 + 16 = 3,136. And 73 x 74 = 70 x 74 + 3 x 74 = 5,180 + 222 = 5,402. The interviewer times the first answer but listens for the second route; a trader who checks is worth more than one who is merely fast. The limitation: these tricks suit two-digit numbers near a round anchor; for 87 x 93 use the difference of squares around 90 instead.