Derivatives Foundation puzzles, solved step by step

Puzzles
100
Traced to a firm
66
Topics
12
Hard
29
Topic
All topicsMental maths and estimation9Random walks and Markov chains7Conditional probability and Bayes7Volatility and correlation7Option pricing intuition7Expected value and optimal stopping10Market making11Option payoffs and no-arbitrage10Probability and counting11Distributions and statistics8Games and logic8Betting and sizing5
Level
AnyWarm upCoreHard
Source
AnyReported at a firmStandard
Showing 1–2 of 2 · filtered from 100Clear filters
  1. 040Five rational pirates, A the most senior down to E, split 100 gold coins. The most senior proposes a split; it passes if at least half of the pirates, his own vote included, agree, otherwise he is thrown overboard and the next most senior proposes. Each pirate wants to survive, then to get the most coins, then to see others thrown overboard. How should A split the coins to survive and keep the most?Games and logicHardOld Mission CapitalNew York · 2014

    Try it first

    How many coins can A keep?

    Show the worked solution

    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.

    Backward induction: each proposer buys the votes that would otherwise get nothingpirates leftneedsABCDEwhose vote is bought2: D, E1 of 2overboardoverboardoverboard1000nobody: own vote is half3: C, D, E2 of 3overboardoverboard9901Ethey get 0 if the proposer dies4: B, C, D, E2 of 4overboard99010Dthey get 0 if the proposer dies5: A, B, C, D, E3 of 5980101C, Ethey get 0 if the proposer diesA keeps 98, gives C and E one coin each, and B and D get nothing: C and E vote yes because B's plan gives them zeroGreen cells are the bought votes; a bought pirate is paid one coin more than the next proposal would give him
    Working up from two pirates, D keeps 100, C keeps 99 by paying E one coin, B keeps 99 by paying D one coin, and A keeps 98 by paying C and E one coin each, the two pirates who would receive nothing under B's proposal.

    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.

    The relationship
    2: (100,0)  →  3: (99,0,1)  →  4: (99,0,1,0)  →  5: (98,0,1,0,1)\text{2: }(100, 0) \;\to\; \text{3: }(99, 0, 1) \;\to\; \text{4: }(99, 0, 1, 0) \;\to\; \text{5: }(98, 0, 1, 0, 1)
    each tuplecoins to the proposer first, then to the others in order of seniority
    the 1svotes bought for one coin from pirates who would get zero in the next round
    votes neededhalf of the pirates, rounded up, including the proposer: 1 of 2, 2 of 3, 2 of 4, 3 of 5
    What it says in wordsEach proposer keeps everything except one coin for each extra vote he needs, bought from whoever the next round would leave empty-handed.

    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.

    Where candidates lose it

    The common loss is reasoning forward from fairness, offering 20 each or 50 for A, and being unable to say why those numbers rather than others. Fairness is not in the rules; survival and coins are.

    The second is paying the wrong pirates. B and D would do well without A and cannot be bought cheaply; C and E would get nothing. Candidates who pay B and D have not worked out the four-pirate round.

    What the interviewer asks next

    • The proposal now needs a strict majority of all votes. What does A propose?
    • The proposal needs a majority of the other pirates' votes, excluding the proposer. Does A survive?
    • There are 6 pirates. How does the pattern continue?
    • What if a pirate who is indifferent prefers to keep the proposer alive?

    Asked at Old Mission Capital, Prop Trading, New York, 2014 (Wall Street Oasis): There are 5 pirates and they are trying to split 100 gold coins in a rational way.

  2. 045Two players each ante Rs 1 and receive one card from an ace, a king and a queen. Player one may bet Rs 1 or check, and a check goes straight to showdown. Facing a bet, player two may call or fold. How often should player one bluff with the queen, and how often should player two call with the king?Games and logicHardOld Mission CapitalNew York · 2022

    Try it first

    How often should player one bet the queen?

    Show the worked solution

    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.

    The relationship
    King indifferent: b1+b(+2)+11+b(−2)=−1⇒b=13,Queen indifferent: 12(−2)+12(c(−2)+(1−c)(+1))=−1⇒c=13\text{King indifferent: } \tfrac{b}{1+b}(+2) + \tfrac{1}{1+b}(-2) = -1 \Rightarrow b = \tfrac13, \qquad \text{Queen indifferent: } \tfrac12(-2) + \tfrac12\big(c(-2) + (1-c)(+1)\big) = -1 \Rightarrow c = \tfrac13
    bhow often player one bets the queen
    chow often player two calls with the king
    -1the value of the alternative: folding the king, or checking the queen, each loses the ante
    1/2given the queen, player two holds the ace or the king with equal chance
    What it says in wordsEach player's frequency is chosen so that the other player's two choices are worth the same.
    Equilibrium of the ace, king, queen game: each side mixes to make the other indifferentPlayer one's cardPlayer one acts (antes 1 each, bet 1)Player two facing a betAbet alwaysit wins whenever it is calledAcall alwaysbeats every betKcheck alwaysshowdown: beats Q, loses to AKcall 1 time in 3bluff-catcher: indifferentQbet 1 in 3, check 2 in 3the bluff: indifferentQfold alwaysloses to every betPlayer one bluffs so 1 bet in 4 is a bluff: call / (pot + call) = 1 / (3 + 1)Player two calls so he folds 1 time in 3 overall: bet / (pot + bet) = 1 / (2 + 1)Value to player one at these frequencies: +1/18 of a rupee a hand; neither side gains by changing
    Player one bets the ace, checks the king and bluffs the queen one time in three, while player two calls with the ace, folds the queen and calls with the king one time in three, because those frequencies leave each player's marginal choice worth exactly the same either way.

    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.

    Where candidates lose it

    The common loss is saying player one should never bluff because the queen cannot win a showdown. Without bluffs player two simply folds his king to every bet, and player one's ace stops getting paid.

    The second is giving the bluff rate as a quarter. A quarter is the share of bets that are bluffs; because the ace is always bet, that needs the queen bet one time in three. Keep the two fractions apart and say which one you mean.

    What the interviewer asks next

    • Player two may also bet after player one checks. How does the equilibrium change?
    • The bet size doubles to Rs 2. What are the new bluffing and calling frequencies?
    • Player two never calls with the king. What is player one's best response, and what does it earn?
    • Why does player one never want to bet the king?

    Asked at Old Mission Capital, Quantitative Research, New York, 2022 (Wall Street Oasis): Asking to find the game theory optimal strategy in a simplified poker game