Debt Capital Markets puzzles, solved step by step

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100
Traced to a firm
16
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13
Hard
30
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All topicsLeverage, coverage and cash flow9Mental maths and numeracy8Estimation and market sizing7Logic and brainteasers8Cost of capital and valuation riddles7Bond pricing and yield7Compounding, PIK and fees6Issuance and refinancing arithmetic8Credit spreads and default probability8Duration and convexity8Capital structure and recovery8Probability and expected value10Yield curve and forward rates6
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  1. 066Five creditor classes, ranked by seniority, must agree how to split 100 units of recovery. The most senior class still at the table proposes a split. If at least half of the classes at the table vote for it, it passes; otherwise the proposer leaves with nothing and the next most senior class proposes. Every class wants the most units and votes no when indifferent. What should the most senior class propose?Logic and brainteasersHardRestructuring

    Try it first

    How many units can the most senior class keep?

    Show the worked solution

    The senior class proposes 98 for itself, 0 for class 2, 1 for class 3, 0 for class 4 and 1 for class 5, and the plan passes. Work backwards. With two classes left, class 4 takes all 100, because its own vote is half. With three, class 3 buys class 5 with 1 unit. With four, class 2 buys class 4 with 1. With five, class 1 needs two more votes and buys classes 3 and 5, who would otherwise get nothing, with 1 unit each.

    Why start from the end of the game?

    When you must catch a train that leaves at a fixed time, you plan from the departure backwards to when you leave home. A bargaining game played in turns is solved the same way: the last stage has an obvious answer, and each earlier proposal only has to beat what each voter would get one stage later. The end here is two classes, 4 and 5. Class 4 proposes 100 for itself and votes yes, and one yes out of two is half, so the plan passes.

    Solve from the end: each proposer buys the votes that would get nothing nextthe order you solve it in2 classes left3 classes left4 classes left5 classes leftClass 1 (most senior)removedremovedremoved98Class 2removedremoved990Class 3removed9901Class 4100010Class 50101votes: 1 of 2votes: 2 of 3votes: 2 of 4votes: 3 of 5proposervote bought with 1 unitgets 0, votes no
    With two classes left class 4 takes 100; with three, class 3 keeps 99 and buys class 5; with four, class 2 keeps 99 and buys class 4; with five, class 1 keeps 98 and buys classes 3 and 5 with one unit each, because they are the classes that get nothing one stage later.

    How does each proposer buy the cheapest votes?

    With three classes left, class 3 needs two of three votes. Class 5 gets nothing if class 3 is removed, so one unit buys it: 99, 0, 1. At every stage the proposer buys the classes that would get nothing at the next stage, because one unit beats nothing. With four left, class 2 needs two of four votes and class 4 gets nothing in the three-class outcome, so the split is 99, 0, 1, 0. With five, class 1 needs three votes, its own plus classes 3 and 5, who get nothing in the four-class outcome.

    Classes leftProposerClass 1Class 2Class 3Class 4Class 5Votes needed
    2Class 4outoutout10001 of 2
    3Class 3outout99012 of 3
    4Class 2out990102 of 4
    5Class 19801013 of 5
    Units of recovery at each stage. The proposer keeps everything it does not need to spend on votes, so the final split is 98, 0, 1, 0, 1 for classes 1 to 5.

    What does this teach about a real restructuring?

    Bargaining power comes from what each party gets if the deal fails, not from its size or its rank on paper. The senior class keeps 98 because it can see what every other class would get if its plan failed, and it pays only for the votes it needs. Real restructurings do not run like this: priority rules, court oversight and class voting thresholds all change the outcome, and junior classes often hold out for a share because a fight costs everyone. Treat the puzzle as a lesson in reasoning from the fallback, not as a model of a court process.

    Where candidates lose it

    Most candidates try to be fair, or try to buy class 2 because it looks most powerful. Class 2 is the most expensive vote on the table: it becomes the proposer if class 1 is removed and would keep 99.

    The second loss is getting the three-class and four-class stages wrong by forgetting that a tie passes. State the voting rule at the start, at least half including the proposer, and check each stage against it.

    What the interviewer asks next

    • What changes if a plan needs a strict majority rather than at least half?
    • How does the answer change with seven classes?
    • If indifferent classes vote yes, what does the senior class propose?
  2. 098You have 1,000 bottles of wine and exactly one is poisoned. The poison shows its effect after 24 hours, and you have one day to find the bottle. What is the fewest testers you need, and how do you assign the bottles to them?Logic and brainteasersHardSyndicate desks

    Try it first

    How many testers do you need?

    Show the worked solution

    10 testers. Number the bottles 1 to 1,000 and write each number in binary with ten digits. Tester n drinks from every bottle whose n-th binary digit is 1. After 24 hours, the testers who fall ill spell out the poisoned bottle's number in binary. Ten testers give 1,024 possible patterns, enough for 1,000 bottles; nine give only 512.

    Why is each tester worth a binary digit?

    Think of the game twenty questions: every yes-or-no answer halves the possibilities, so twenty answers can pick out one thing in about a million. A tester is one yes-or-no answer: ill or fine. With one round of results, n testers can produce 2 to the power n different patterns, and you need at least as many patterns as bottles. 2^9 = 512 falls short of 1,000; 2^10 = 1,024 covers it. So ten is both enough and the minimum.

    Each tester is one binary digit; the ones who fall ill spell the bottleT10512T9256T8128T764T632T516T48T34T22T11Bottle 10000000001Bottle 20000000010Bottle 30000000011Bottle 5000111110100Bottle 6131001100101Bottle 10001111101000shaded = drinksTesters 10, 7, 6, 3, 1 fall ill: 512 + 64 + 32 + 4 + 1 = 61310 testers give 2^10 = 1,024 patterns, enough for 1,000 bottles; 9 give only 512
    Each bottle number written in ten binary digits tells each tester whether to drink from it, and if bottle 613 is poisoned, testers 10, 7, 6, 3, 1 fall ill, whose place values add back to 613.

    How do you read the answer back?

    Give tester n the place value 2 to the power (n minus 1): tester 1 is worth 1, tester 2 worth 2, up to tester 10 worth 512. Add the place values of the testers who fall ill and you have the bottle number. If bottle 613 is poisoned, its binary form is 1001100101, so testers 10, 7, 6, 3, 1 fall ill, and 512 + 64 + 32 + 4 + 1 = 613. Numbering 1 to 1,000 fits in ten digits because 1,000 is below 1,024, and the pattern where nobody falls ill is left spare.

    The relationship
    29=512<1,000≤1,024=2102^{9} = 512 < 1{,}000 \le 1{,}024 = 2^{10}
    2^nthe number of distinct ill-or-fine patterns n testers can produce
    What it says in wordsTen testers are the fewest whose outcomes can label every bottle.

    The desk version of the lesson: design the test so that every outcome carries information. A tester who drinks from one bottle learns about one bottle; a tester who drinks from half of them splits the problem in two.

    Where candidates lose it

    The common answers are 1,000 testers, one bottle each, or some splitting scheme that needs several rounds. The 24 hour delay allows only one round, so the testers must be designed to answer in parallel.

    The second loss is giving 10 without the proof that 9 is not enough. Say 2^9 is 512 in the same breath; it turns a remembered trick into an argument.

    What the interviewer asks next

    • You have two days instead of one, and a tester who falls ill on day one is out. How many testers do you need?
    • Exactly two bottles are poisoned. Does the binary scheme still work?
    • How many testers would a million bottles need?