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
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  1. 015You have three six-sided dice. Red has the faces 2, 6, 7; green has 1, 5, 12; blue has 3, 4, 8, with each number on two faces. Two players each pick a die and roll; the higher number wins. Which die would you choose to play with?Games and logicCoreBelvedere TradingChicago · 2022

    Try it first

    Before working the pairs: green has the highest average face, 6 against 5 for red and blue. Does that make green the best die?

    Show the worked solution

    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.

    Red beats green, green beats blue, blue beats red, each 5 times in 9: a cycle, not a ladderRed2, 6, 7Green1, 5, 12Blue3, 4, 8red beats green 5/9green beats blue 5/9blue beats red 5/9every arrow is 5 in 9, so there is no best dieRed face against green facegreen:1512red 2redgreengreenred 6redredgreenred 7redredgreenred wins 5 of the 9 equally likely cellsthe other two pairs work the same way
    Red beats green in 5 of the 9 equally likely face pairs, green beats blue in 5 of 9 and blue beats red in 5 of 9, so the three dice form a cycle with no best die and the player who picks second always holds a 5 in 9 edge.

    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.

    The relationship
    P(R>G)=1+2+29=59,P(G>B)=0+2+39=59,P(B>R)=1+1+39=59P(R > G) = \frac{1 + 2 + 2}{9} = \frac{5}{9},\qquad P(G > B) = \frac{0 + 2 + 3}{9} = \frac{5}{9},\qquad P(B > R) = \frac{1 + 1 + 3}{9} = \frac{5}{9}
    R, G, Bthe face shown by the red, green and blue die
    9the number of equally likely face pairs, three distinct faces on each die
    5/9each die's edge over the next one around the cycle
    What it says in wordsCount the winning face pairs out of nine for each ordered pair, and the three results form a cycle.

    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.

    Where candidates lose it

    The common answer is green, because it has the biggest face and the highest average. Both facts are true and both are irrelevant: a roll pays for being higher, not for being higher by a lot, and the pairwise count is the only thing that decides it.

    The second loss is finding the cycle and still naming a die. The answer to which die is a question back: which one is the other player taking? Say that you want to choose second, and why.

    What the interviewer asks next

    • Each player rolls their die twice and the totals are compared. Does the cycle survive, and does it change direction?
    • Design a fourth die that beats all three of these more often than not, or show that none exists.
    • Where on a trading desk does moving second carry an edge, and where does it cost you?

    Asked at Belvedere Trading, Prop Trading, Chicago, 2022 (Wall Street Oasis): You have 3 dice: red has 2, 6, 7; green has 1, 5, 12; blue has 3, 4, 8. Highest number wins the game. Which one would you choose to play with?

  2. 095Three people stand in a line facing forward, wearing hats drawn from 3 red and 2 blue. The back person sees the two hats ahead, the middle person sees only the front hat, and the front person sees none. The back says "I don't know my colour", then the middle says "I don't know my colour". What colour is the front person's hat?Games and logicCoreWolverine Trading, Chicago, ILUSA · 2019

    Try it first

    What does the back person's "I don't know" rule out?

    Show the worked solution

    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.

    Each silence crosses out the worlds where that person would have knownmiddle wearsfront wearsafter the back says"I don't know"after the middle says"I don't know"redredsurvivessurvivesblueredsurvivessurvivesredbluesurvivescrossed out: he would see a bluein front and know he is redbluebluecrossed out: he would seetwo blues and know he is redBoth surviving rows have a red hat in front.The front person, seeing nothing, knows from two silences that their hat is red.
    Of the four hat pairs the back person might see on the middle and front people, his silence crosses out blue and blue, the middle person's silence then crosses out a red middle with a blue front, and both rows that survive have a red hat on the front person, which is why the front person knows the answer without seeing anything.

    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.

    The relationship
    {back silent}⇒¬(M=B∧F=B),{middle silent}⇒¬(F=B)  ⇒  F=R\{\text{back silent}\} \Rightarrow \neg(M = B \wedge F = B), \qquad \{\text{middle silent}\} \Rightarrow \neg(F = B) \;\Rightarrow\; F = R
    M, Fthe middle and front hats
    B, Rblue and red
    the arrowwhat each silence lets everyone conclude
    What it says in wordsThe first silence removes the case of two blues ahead, and the second removes a blue in front, which leaves only red.

    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.

    Where candidates lose it

    The common loss is saying the front person cannot know anything because they see nothing. That ignores that the two silences are data. The question is built to see whether you treat a non-answer as information.

    The second loss is running the logic from the front. Start with the person who has the most information, the back, ask what would have let him know, and strike that case. Then move forward one person at a time.

    What the interviewer asks next

    • Suppose the back person says "I know". What can the other two conclude?
    • With 2 red and 3 blue hats, does the same chain of silences tell the front person anything?
    • If only the back person speaks and says "I don't know", what can the middle person conclude about his own hat when he sees red in front?

    Asked at Wolverine Trading, Prop Trading, Chicago, IL, USA, 2019 (Wall Street Oasis): a brain teaser about the hat problem where 3 people go into a room with I think 3 red and 2 blue hats