Debt Capital Markets puzzles, solved step by step

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100
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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. 021A book holds 30 bonds whose maturity dates fall on random days of a 365-day year, independently. What is the probability that at least two of them mature on the same day?Probability and expected valueCoreFixed income asset management

    Try it first

    Gut call: roughly how likely is a shared maturity date?

    Show the worked solution

    About 70.6%. Work out the opposite: the chance that all 30 dates differ. The second bond avoids the first with 364/365, the third avoids both with 363/365, and so on down to 336/365. Multiplied together that is about 29.4%, so at least one shared date has a probability of about 70.6%. It passes 50% at just 23 bonds, because what matters is the number of pairs, not bonds.

    Why is 30 out of 365 so far off?

    In a class of 30 students, the question is not whether someone shares your birthday; it is whether any two people share one. A clash can happen between any pair, and 30 items make 435 pairs, so the chances accumulate much faster than the count of items. With 435 pairs each having a 1 in 365 chance of matching, you would expect about 1.19 matching pairs, which already tells you a clash is more likely than not.

    The chance of a shared date climbs far faster than bonds / 36525%50%75%100%1102330405060Number of bonds in the booknaive: bonds / 36523 bonds: 50.7%30 bonds: 70.6%30 bonds make 435 pairs435 / 365 = 1.19 expected clashes
    The probability of at least one shared maturity date passes 50% at 23 bonds and reaches 70.6% at 30, far above the naive 30 over 365, because 30 bonds form 435 pairs, each a chance of a clash.

    What is the clean way to calculate it?

    Count the complement. The chance of at least one shared date is one minus the chance that every date is different, and all-different is a product of shrinking fractions. Line the bonds up: the first can fall anywhere, the second must avoid one day, the third two days, and the thirtieth 29 days. Multiplying 365/365 by 364/365 and so on to 336/365 gives about 29.4%. In your head, the pairs shortcut, one minus e to the power minus 435/365, gives about 70%, close enough for the room.

    The relationship
    P(clash)=1−∏i=029365−i365≈70.6%≈1−e−(302)/365=1−e−1.19≈70%P(\text{clash}) = 1 - \prod_{i=0}^{29} \frac{365 - i}{365} \approx 70.6\% \qquad \approx 1 - e^{-\binom{30}{2}/365} = 1 - e^{-1.19} \approx 70\%
    (365 - i)/365the chance the next bond avoids the i dates already taken
    C(30,2)the number of pairs among 30 bonds, 435
    ethe base of natural logarithms, about 2.718
    What it says in wordsOne minus the chance that every bond lands on a different day; the pair count gives a quick approximation.

    Where does this show up on a desk?

    Anywhere independent events have to avoid each other. Clustered maturities are a refinancing risk: an issuer with many bonds is more likely than intuition suggests to face two large repayments close together, which is why treasurers look at the maturity profile, not the count of bonds. Say the limit: real maturity dates are not random. Issuers deliberately spread them and markets cluster issuance in certain months, so this is the benchmark for pure chance, not a description of any real book.

    Where candidates lose it

    The fast wrong answer is about 8%, 30 over 365. It counts bonds instead of pairs, and the interviewer asked this precisely because the intuition is so far off.

    The second loss is trying to add up the chances of each specific pair clashing, which double counts and gets messy. Say complement out loud, then do the product.

    What the interviewer asks next

    • How many bonds do you need for a 90% chance of a shared date?
    • What is the chance that at least one of the 30 bonds matures on one specific date, say 31 March?
    • How would you estimate the chance that two maturities fall within the same week?
  2. 069Four anchor investors are each 60% likely to put in an order for your bond, independently of one another. You need at least three of them to launch. What is the probability that you launch?Probability and expected valueCoreSyndicate desks

    Try it first

    Before you calculate: roughly what is the chance of launching?

    Show the worked solution

    About 47.5%, a little under a coin flip. Exactly three anchors can happen in 4 ways, each with probability 0.6 x 0.6 x 0.6 x 0.4, so 4 x 0.0864 = 34.6%. All four come in with probability 0.6 to the fourth, 13.0%. Adding them gives 47.5%. Each anchor is more likely than not to order, yet needing most of a small group pulls the chance of launching below half.

    How do you count the ways to get three of four?

    Four friends each say they will probably come to dinner, 60% each, and you need three to keep the booking. The chance of any one pattern, say the first three come and the fourth does not, is 0.6 x 0.6 x 0.6 x 0.4 = 8.64%, and there are four such patterns, one for each friend who stays away. So exactly three come 4 x 8.64% = 34.6% of the time. All four come 0.6 to the fourth, 13.0% of the time. The booking survives 47.5% of the time.

    Each anchor is likely, but needing three of four is a coin flip2.6%0 anchors15.4%1 anchor34.6%2 anchors34.6%3 anchors13.0%4 anchorslaunch: 34.6 + 13.0 = 47.5%each anchor 60% likely, independentlyno launch52.5%launch47.5%
    The chance that three of the four anchors order is 34.6% and that all four order is 13.0%, so the deal launches 47.5% of the time even though each anchor is 60% likely to order.
    The relationship
    P(X≥3)=(43)0.63 0.4+0.64=0.3456+0.1296=0.4752P(X \ge 3) = \binom{4}{3} 0.6^{3}\, 0.4 + 0.6^{4} = 0.3456 + 0.1296 = 0.4752
    Xthe number of anchors who order
    C(4,3)the four ways to choose which three anchors order
    0.6, 0.4each anchor's chance of ordering and of staying away
    What it says in wordsAdd the chance of exactly three anchors, counted over every pattern, to the chance of all four.

    Why is the answer below half when each anchor is likely?

    Requiring most of a small group multiplies probabilities together, and products of numbers below one shrink fast. The most likely outcomes are two or three anchors, 34.6% each, so the launch hinges on which side of that line you land. If you needed only two anchors the chance would jump to 82.1%; if you needed all four it would fall to 13.0%. The threshold matters more than any single investor.

    Say what a syndicate desk does with this. Either lower the threshold, by lining up a fifth anchor or sizing the deal so that fewer anchors are enough, or raise each anchor's probability before launch through early soundings. A fifth anchor at 60% lifts the chance of at least three from 47.5% to 68.3%; lifting each of the four to 70% gives 65.2%. The limit: anchors are rarely independent; they read the same market, so on a bad day they tend to drop out together, and the real chance is lower than this.

    Where candidates lose it

    The instinctive answer is about 60%, as if the group behaved like one anchor. Needing three of four is a joint event, and joint events are rarer than their parts.

    The second loss is forgetting the four orderings and answering 8.64% plus 13.0%, or counting only the all-four case. Say the combination count out loud: four ways to choose which anchor stays away.

    What the interviewer asks next

    • What is the chance of launching if you add a fifth anchor, also 60% likely?
    • If each anchor's chance rises to 70%, what is the chance of at least three of four?
    • Why might anchor orders be positively correlated, and what does that do to the launch probability?