Derivatives Foundation puzzles, solved step by step
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020One glass holds 100 ml of wine and another holds 100 ml of water. You take a spoonful of wine, tip it into the water and stir. Then you take a spoonful of the mixture and tip it back into the wine glass. Is there now more wine in the water glass, or more water in the wine glass?Prop trading firms
Try it first
Decide before any arithmetic: after the two spoonfuls,
Show the worked solution
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 Each glass ends with exactly 100 ml, so every millilitre of wine that left the wine glass and did not come back has been replaced by a millilitre of water: the two foreign amounts must be equal.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.
With a 10 ml spoon, the wine glass goes from 100 ml of wine to 90 ml and then back to 100 ml holding 9.09 ml of water, while the water glass goes to 110 ml and back to 100 ml holding 9.09 ml of wine, so the two foreign amounts are equal. 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.
The relationship10 the spoon, in millilitres 100/110 the share of water in the stirred water glass after the first transfer 10/110 the share of wine in that glass What it says in wordsThe water carried into the wine glass equals the wine left behind in the water glass, both 9.09 ml for a 10 ml spoon.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.
Stage Wine glass Water glass Start 100 wine 100 water After spoon 1 90 wine 100 water + 10 wine After spoon 2 90.91 wine + 9.09 water 90.91 water + 9.09 wine Tracking a 10 ml spoon through both transfers leaves each glass at 100 ml with 9.09 ml of the other liquid, which is what the conservation argument predicted without any arithmetic. Where candidates lose it
The common answer is more wine in the water, because the first spoon was undiluted. It anchors on one transfer and forgets that the second spoon also took some of that wine back.
The second loss is reaching the right answer by long arithmetic and then failing the follow-up, such as an unstirred glass or several transfers, because there was no argument underneath. Give the volume argument first and use the numbers only as a check.
What the interviewer asks next
- The return spoon is 5 ml instead of 10 ml. Which glass now holds more of the other liquid, and by how much?
- You repeat the two-spoon swap many times. What do both glasses converge to?
- Where on a trading desk does checking a total first save you from tracking every transfer?
035There are 100 coins on the table. Players take turns removing 1 to 10 coins, and whoever takes the last coin wins. Do you want to go first, and what is your first move?Quant trading
Try it first
Go first or second, and what is the opening?
Show the worked solution
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.
Every multiple of 11 from 0 to 99 is a losing position for the player who must move, so the first player takes 1 to leave 99 and then answers every take of t with 11 minus t, stepping down through 88, 77 and 66 until the last coin. 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.
The relationship11 the largest allowed take plus one, the amount you can always complete in a pair of moves 100 mod 11 the remainder when 100 is divided by 11; take exactly this many What it says in wordsTake the remainder on your first move, then keep each pair of moves summing to 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.
Where candidates lose it
The common loss is taking 10, because a bigger move feels like a stronger start. It leaves 90, which is not a multiple of 11, and a prepared opponent takes 2 to leave 88 and wins from there.
The second is knowing the answer and not the reason. Say why 11 is the period: any take of 1 to 10 can be completed to 11. Without that sentence the interviewer will change the numbers and watch you stall.
What the interviewer asks next
- Players may take 1 to 7 coins instead. Do you go first, and what is the opening?
- The player who takes the last coin loses. Do you go first?
- There are two piles, 100 and 60, and you may take from either pile. Who wins?
- Each turn a die sets the maximum take. Is there still a strategy, and what is it?
061You and I each show heads or tails at the same time. You win Rs 3 if we both show heads, Rs 1 if we both show tails, and you lose Rs 2 if we show different faces. What mix should you play, and is the game worth playing?Quant tradingProp trading firms
Try it first
Two wins and two losses in the table. Is this game good for you?
Show the worked solution
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.
Your expected payoff is 5p - 2 if I show heads and 1 - 3p if I show tails, and since I will always pick the lower line, the best you can do is the crossing at p = 3/8, where both lines give minus 1/8, so the game is worth minus Rs 0.125 to you per round. 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.
The relationshipp your probability of showing heads 5p - 2 your expected payoff when I show heads 1 - 3p your expected payoff when I show tails V the value of the game to you per round What it says in wordsEqualising your payoff across my two replies gives a three-eighths mix and a value of minus one eighth of a rupee per round.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.
Where candidates lose it
The common answer is that the game is fair or favourable, from summing the four cells. The sum of a payoff table says nothing when the opponent chooses the column. Set up the two lines and find where they cross.
The second loss is solving for the right p and then saying the game is fine because 3 and 1 are bigger than 2. State the value, minus 1/8, and say you need a fee of at least that to play.
What the interviewer asks next
- What is my optimal mix, and what does it earn me?
- Change the heads-heads payoff to Rs 4. Does the game become worth playing?
- I am known to show heads 60% of the time regardless. What should you do now?
- Why do both players end up with the same 3/8 here, and is that a coincidence?
066A company is worth a uniformly random amount between Rs 0 and Rs 100 crore to its owner, who knows the exact value. It is worth 1.5 times that amount to you. You make one take-it-or-leave-it offer, and the owner accepts if your offer exceeds the value. What should you offer?Hedge fundsMarket making
Try it first
The company is worth 50% more to you than to the owner. What do you bid?
Show the worked solution
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.
Expected profit from an offer b is minus a quarter of b times the acceptance chance b/100, a parabola that is zero at b = 0 and falls to minus 25 crore at b = 100, so no positive offer earns anything and the best move is not to bid; an offer of 60, for instance, buys a company worth 45 to you on average and loses 9 in expectation. 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.
The relationshipb your offer in Rs crore b/100 the chance the owner accepts, since the value is uniform on 0 to 100 1.5 x b/2 what the company is worth to you on average once you know the value is below b What it says in wordsThe expected profit from any offer is a negative multiple of the offer squared, so the best offer is zero.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.
Where candidates lose it
Candidates bid around 50 or 75 by averaging over the whole range of values, forgetting that they only buy when the owner accepts. The interviewer wants you to say that acceptance is information before you touch any arithmetic.
The second loss is getting zero and not being able to say what would change it. Give the general condition: the multiplier must exceed two for any positive bid to pay.
What the interviewer asks next
- At what multiplier does a positive bid first break even, and what is the best bid at a multiplier of 3?
- The value is uniform on Rs 50 to Rs 100 crore instead. Does a positive bid make sense now?
- How is this the same problem as a market maker being hit only when they are wrong?
- You get two offers rather than one, and the owner rejects the first. Does that change anything?