Derivatives Foundation puzzles, solved step by step

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.

Expected payoffs are 10 for the call and 6.67 for the put, and neither is a price. The call pays 0, 0 and 30 across the three outcomes, averaging 10; the put pays 20, 0 and 0, averaging 6.67. A market maker charges the cost of hedging, and with zero rates call minus put must equal stock minus strike, which is 0, so prices of 10 and 6.67 would be an arbitrage against the maker. Hedge-implied odds that keep the stock worth 100 give both options the same price.

Why is the expected payoff under your odds not the price?

A shopkeeper who believes the monsoon will be good does not price umbrellas off that belief; the price is set by what the umbrellas cost to stock and what the shop next door charges. An option maker does not hold the option to expiry hoping for the payoff; they hedge it with the stock, so the price is the cost of that hedge, and the cost of the hedge depends on the stock's current price of 100, not on anyone's forecast of where it goes. Your equal odds imply the stock is expected to be worth 103.33, above today's 100; that optimism is yours to trade, not something the maker will pay you for inside an option price.

What is the one-line test that catches the mistake?

Put-call parity. Buying the call and selling the put with the same strike gives you the stock minus 100 in every outcome, which with zero rates is worth 100 - 100 = 0 today, so the call and the put must have the same price. Expected payoffs of 10 and 6.67 fail that test by 3.33, and a maker who quoted them would be lifted on the put and hit on the call until the prices met. The test needs no probabilities at all, which is the point: parity is enforced by hedging, not by views.

What odds would a maker use, and are they unique here?

Odds that make the stock worth its forward, 100 with zero rates. Any set of probabilities on 80, 100 and 130 with an average of 100 prices the two options consistently; one such set is 0.40, 0.33 and 0.27, which gives the call 0.267 x 30 = 8 and the put 0.40 x 20 = 8, equal as parity demands. The limitation is that with three outcomes and only the stock to hedge with, the set is not unique: the common level of the call and put price is pinned down by parity only up to a range, and in practice it is the market's volatility quote that chooses the point inside it.

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.

The single deck, 25.49% against 25.24%. The first card is black with probability 1/2 in either deck. The second draw is where they differ: with one deck, 26 reds remain in 51 cards, 50.98%; with two decks, 52 remain in 103, 50.49%. Multiply: 26/52 x 26/51 = 25.49% and 52/104 x 52/103 = 25.24%. Removing a black card tilts a small deck towards red more than it tilts a large one.

Why does the size of the deck matter when the mix is the same?

Take one boy out of a class of 20 with 10 boys and 10 girls, and the class is 10 girls in 19, 52.6% girls. Take one boy out of a school of 2,000 split evenly, and it is 1,000 in 1,999, 50.03%. The same removal is a bigger share of a smaller group. Drawing without replacement makes the second draw depend on the first, and the dependence is stronger the smaller the deck, which is exactly what the game rewards. You want a black card to make red more likely next, and a single deck gives you more of that lean.

What is the general pattern, and what is its limit?

With n decks shuffled together, the chance is 1/2 x 26n/(52n - 1). As n grows the second factor falls towards 1/2 and the product towards 1/4, which is the answer you would get if the draws were independent. Every finite deck beats 25%, and the smallest deck beats it by the most, because the minus one in the denominator is a larger share of a smaller count. The whole effect is small, a quarter of a percentage point between one deck and two, which is also worth saying: the interviewer wants the direction and the reason more than the third decimal.

Then show you can flip it. If the game paid on black followed by black, the single deck would be worse: 26/52 x 25/51 = 24.51% against 52/104 x 51/103 = 24.76%, and the big deck wins. Same mechanism, opposite sign. Saying that unprompted is what turns a one-line puzzle into evidence that you understand sampling without replacement rather than remembering an answer.

3.5, so pay at most Rs 3.50 for the plain game, and Rs 15.17 for the squared one, not Rs 12.25. The plain game averages 1 to 6: 21/6 = 3.5. The squared game averages 1, 4, 9, 16, 25 and 36: 91/6, about 15.17. Squaring the average roll gives only 12.25 and misses 2.92, which is the variance of the roll. A payoff that bends upward is worth more than its value at the average outcome.

What does fair price mean here?

If a friend offers to pay you whatever a die shows, and you play it a thousand times, you collect close to 3,500 rupees. Paying more than 3.50 a go loses money over the thousand games; paying less makes it. The fair price of a game is the average of what it pays, weighted by how likely each payout is, and the most a market maker would pay is a little under it, to leave room for edge. For one roll each face has weight 1/6, so the expected value is (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5. Say both numbers: fair value Rs 3.50, and a bid of perhaps Rs 3.25 if you are asked to trade it.

Why is the squared game worth more than 3.5 squared?

The trap answer squares the average roll and says 12.25. But the game does not pay the square of the average; it pays the square of whatever is rolled. A 6 pays 36, which is 23.75 above 12.25, and a 1 pays 1, only 11.25 below it. Squaring stretches the high outcomes more than it squeezes the low ones, so the average of the squares, 15.17, sits above the square of the average, 12.25. The gap has a name: E[X^2] - (E[X])^2 is the variance of the roll, 35/12, about 2.92.

Then say why a derivatives desk asks it. An option is a payoff that bends upward, like the square, so it is worth more than its payoff at the expected price, and the extra is driven by the spread of outcomes. That is the whole reason volatility has a price. The limitation: fair value assumes you are indifferent to risk and can play many times. Offered one roll of the squared game for Rs 15, a person who cannot afford to lose Rs 14 might sensibly refuse, and an interviewer will sometimes push on that.

Take the sure Rs 4, once or a thousand times. One die is worth Rs 3.50. The higher of two dice averages 161/36 = Rs 4.47, so after the Rs 1 fee it is worth Rs 3.47, the worst of the three. The sure Rs 4 beats both. Over 1,000 plays the choice only gets clearer: the games total about Rs 3,500 and Rs 3,472, each with a spread of about Rs 50, against exactly Rs 4,000.

How much is the second die really worth?

If a shop lets you take the better of two mangoes for one rupee extra, the question is how much better the better one usually is, not how good a mango can be. Two dice give you the higher face, which is often a 5 or 6, so it feels valuable. The higher of two dice is k with probability (2k - 1)/36, so its average is 161/36, about 4.47: the second die adds about Rs 0.97, a little less than the Rs 1 it costs. After the fee, option (b) is worth Rs 3.47, below the plain die at Rs 3.50 and well below the sure Rs 4.

Does playing 1,000 times change the choice?

Played once, a gambler might take (a) for the one-in-three chance of a 5 or 6, or (b) for its 30.6% chance of a 6. Both lose to Rs 4 on average and carry risk, so nobody who dislikes risk should prefer them, and nobody who is neutral to risk should either. Over 1,000 plays the spread of each game's total grows only with the square root of the plays, about Rs 54 for (a) and Rs 44 for (b), while the gap to the sure Rs 4,000 grows with the plays themselves, to Rs 500 and Rs 528. The gap is about nine standard deviations, so the sure option wins with near certainty. Repetition turns expected value into the outcome.

The interviewer usually follows with a market: what would you pay for option (b) without the fee? Fair value is Rs 4.47, so a bid around Rs 4.25 leaves edge. Then they change the numbers: if the sure amount were Rs 3.40, the plain die would become the best choice on average, and the decision for a single play would depend on how much you care about risk. The limitation is the usual one: expected value is the right yardstick only when the stakes are small relative to what you can afford to lose.