Derivatives Foundation puzzles, solved step by step

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.

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.

Go first and take 1, leaving 99. Work backwards: whoever faces 11 coins loses, because any take of 1 to 10 leaves 1 to 10 for the other player to finish. The same holds for 22, 33 and every multiple of 11. From 100, taking 1 leaves 99, a multiple of 11; after that, whatever the opponent takes, you take 11 minus it, stepping down 88, 77, 66 and so on to 0, where you take the last coin.

Why work backwards from the last coin?

If you are climbing stairs with a friend and the rule is that the person who steps onto the top stair wins, you do not plan from the bottom; you ask which stair you must leave your friend on so that they cannot reach the top in one go. Games with a fixed last move are solved from the end: find the positions where the player to move loses, then find the positions from which you can push your opponent onto one of them. With 1 to 10 coins allowed, facing 1 to 10 coins is a win, you take them all. Facing 11 is a loss, because every move leaves between 1 and 10. Facing 12 to 21 is a win, since you can reduce to 11. Facing 22 is a loss again. The losing positions repeat every 11.

How do you find the period without listing every position?

The period is the largest take plus one, 11, because that is the one total a pair of moves can always be made to add up to: whatever your opponent takes between 1 and 10, you can take the balance of 11. The losing positions are the multiples of the largest take plus one, and the winning opening move is the remainder when the pile is divided by that number. 100 divided by 11 is 9 remainder 1, so take 1. If the rule allowed 1 to 7 coins, the period would be 8 and the opening would be 100 mod 8, which is 4. If the pile had been 99 to start with, you would want to go second, because the first player cannot leave a multiple of 11.

What changes if the last coin loses instead of wins?

Then you want to hand your opponent the last coin, so the position you avoid facing is 1 coin, and the losing positions shift up by one: 1, 12, 23 and so on up to 100. Facing 100 in that version you are already lost, so you would want to go second, which shows the interviewer that you re-derive the pattern rather than remember it. The method is the same in every variant: name the terminal position, step back one move at a time to find the first losing position, then find the period. The limitation of the trick is that it needs a game with perfect information and no chance; add a die that sets each turn's maximum and the clean period disappears.

A proposes 98 for himself, 0 for B, 1 for C, 0 for D and 1 for E. Work backwards. With two pirates, D keeps 100, since his own vote is half. With three, C keeps 99 and gives E 1, because E gets nothing if C dies. With four, B keeps 99 and gives D 1. With five, A needs two other votes and buys the cheapest: C and E, who get nothing under B's plan, accept one coin each. A's vote plus two makes three of five.

Why does the answer come from the end of the game, not the start?

When you negotiate a price you think about what the other side does if you walk away; what they do next determines what they will accept now. Each pirate compares the offer in front of him with what he would get under the next proposal, so to know what A must offer you first need to know what B would do, which needs C, which needs D with two pirates left. With two pirates, D proposes 100 for himself; his own vote is one of two, which is half, so it passes and E gets nothing. That is the anchor. Every earlier step is a question of who gets zero in the next round, because those are the pirates whose votes are cheapest to buy.

How does each proposer decide whose vote to buy?

He buys exactly the number of extra votes he needs, from the pirates who are cheapest, and a pirate is cheap if the next proposal gives him nothing. With three pirates, C needs one more vote; under D's plan E gets zero, so C offers E one coin and keeps 99, and E accepts because one is better than zero. With four, B needs one more; under C's plan D gets zero, so B pays D one coin. With five, A needs two more votes, and under B's plan the pirates with nothing are C and E, so A pays them one coin each and keeps 98; offering anything to B or D is wasted, because B would get 99 and D one coin without A, and neither can be bought for less. The alternation is the pattern to say out loud: each round, the pirates who were paid last time are the ones left out this time.

What assumptions is the answer resting on, and what happens when they change?

Three assumptions, and the interviewer will test at least one. Pirates are perfectly rational and know the others are. A pirate who is indifferent between two outcomes prefers the one where a rival is thrown overboard, which is why a bought pirate must be paid one coin and not zero. And the voting rule is at least half, with the proposer voting. Change the rule to a strict majority of all votes and the two-pirate case flips, D cannot pass anything and dies, so E would reject everything in the three-pirate round, and the whole ladder shifts. With a strict majority of the other pirates' votes the ladder shifts again, and a candidate who re-derives it from the two-pirate anchor is the one who gets hired. The limitation is that real negotiators are not this rational; the puzzle is a model of backward induction, not of pirates.

Player one bets the ace always, checks the king, and bluffs the queen one time in three; player two calls with the ace, folds the queen, and calls with the king one time in three. The bluff rate makes one bet in four a bluff, which is the break-even for a call of 1 into a pot of 3. The king's call rate makes player two fold one time in three overall, the break-even for a bluff of 1 into a pot of 2. Player one gains 1/18 of a rupee a hand.

Which decisions are obvious, and which one is the real question?

Start by removing the choices nobody would make. Player one always bets the ace, since it wins every showdown, and player two always calls with the ace and folds with the queen, since those cards win or lose every time; the only real decisions are player one's queen and player two's king. Player one's king should check: if it bets, player two calls only with the ace and folds the queen, so the bet loses 2 half the time and wins 1 half the time, an average of -0.5, worse than the 0 a showdown gives. That leaves two numbers to find, the queen's bluff frequency and the king's calling frequency, and each is set to make the other player's choice a matter of indifference.

How do you find the bluffing frequency?

A goalkeeper who always dives left is easy to beat; a penalty taker mixes his side so the keeper gains nothing by guessing. Poker equilibrium works the same way. Player one bluffs just often enough that player two's king gains nothing by calling over folding, and player two calls just often enough that player one's queen gains nothing by bluffing over checking. Facing a bet with the king, folding loses the ante, -1. Calling wins 2 against a bluff and loses 2 against the ace. With the ace always bet and the queen bet a fraction b of the time, a bet is a bluff with probability b / (1 + b). Setting the call's value equal to -1 gives b = 1/3, so one bet in four is a bluff.

How do you sanity check the frequencies, and who wins the game?

Use pot odds as the check. Player two calls 1 to win a pot of 3, two antes and the bet, so a call breaks even when bluffs are 1 / (3 + 1) of bets, a quarter, which is what b = 1/3 delivers. Player one bluffs 1 to win the pot of 2, so a bluff breaks even when player two folds 1 / (2 + 1) of the time; he folds the queen always and the king two times in three, which averages to one in three. Average over the six deals and player one gains 1/18 of a rupee a hand, because only he can bet; neither player can improve on that by changing his own frequency. Against a player two who never calls with the king, bluffing every queen would earn 1/6 of a rupee a hand instead of 1/18. The limitation is that equilibrium is a defence, not the most profitable play against a predictable opponent; a desk uses it as the baseline and then leans towards the other side's mistakes.

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.

Offer nothing. Every positive offer loses money on average. If you offer b and the owner accepts, you learn the value is below b, so its average is b/2, worth 1.5 x b/2 = 0.75b to you. You pay b, so each accepted deal loses 0.25b, and the deal is accepted a fraction b/100 of the time. Expected profit is minus 0.0025 b squared, negative for every b above zero. The 1.5 multiplier is not enough to overcome what acceptance tells you.

Why is the owner saying yes bad news for you?

A friend sells you their old scooter for any price you name, but only if your price beats what they privately think it is worth. If they take Rs 20,000 instantly, you have just learned the scooter is worth less than that to someone who knows it well. Acceptance is information: it tells you the true value lies below your bid, so the only companies you ever buy are the ones worth less than you paid, and the ones worth more walk away. Averaging over all possible values, as if you bought every company, is the mistake; you must average only over the values at which the owner says yes.

How do you set up the expected profit cleanly?

Condition on acceptance, then multiply by its probability. Given a bid b that is accepted, the value V is uniform on 0 to b, so E[V | accepted] = b/2, the company is worth 1.5 x b/2 = 0.75b to you, and the profit on an accepted deal is 0.75b - b = - 0.25b. Acceptance happens with probability b/100, so expected profit is - 0.25b x b/100 = - b squared over 400. At b = 50 that is minus 6.25 crore; at b = 100 it is minus 25 crore. The derivative is negative everywhere above zero, so the maximum is at b = 0.

At what multiplier does a bid start to make sense?

Replace 1.5 with a general m. Given acceptance, the company is worth m x b/2 to you against the b you pay, so the deal breaks even when m/2 = 1, that is m = 2. Unless the company is worth more than twice as much to you as to the owner, acceptance always costs you, and at exactly double every bid is a wash. That is the winner's curse in its purest form: the party with less information loses whenever the informed party decides whether to trade. The limitation is the uniform prior and the single offer; with a floor on the value, or a negotiation that reveals information, positive bids can be profitable.

Red. The back person would know his hat if he saw two blues, since only two exist; his silence rules that out. The middle person now knows the two front hats are not both blue. If he saw blue on the front person, he would know his own was red, and he would say so. His silence means the front hat is not blue. So the front person, who sees nothing, deduces red from the two silences alone.

What does a silence tell everyone else?

If a friend who can see the scoreboard says she cannot tell who is winning, you learn that the scores are close. Her not knowing is information, because you know what she would have said if they were not. Each "I don't know" rules out every world in which that person would have known, and everyone behind and in front can use it. The back person would know only in one world: two blue hats ahead, which forces his own to be red. So his silence removes that world, and the middle and front people both hear it.

How does the middle person's silence finish it?

The middle person now knows that he and the front person are not both blue. He looks at the front hat. If it is blue, he cannot be blue too, so he is red, and he would say so. He does not. The middle person's silence can only mean he sees a red hat in front, because a blue one would have told him his own colour. The front person runs the same reasoning, does not need to see anything, and says red. The full check enumerates the seven possible hat triples, removes the ones where the back person would know, then the ones where the middle person would, and every survivor has red in front.

Then say what it rests on, because that is the interview point. The puzzle needs common knowledge: everyone knows the hat counts, everyone reasons perfectly, and everyone knows the others do too. If the middle person might simply be slow, his silence carries no information and the front person learns nothing. On a trading floor the same logic runs in the other direction: a counterparty who could have traded and chose not to has told you something, and reading those non-events is part of the job.