Debt Capital Markets puzzles, solved step by step

Puzzles: 100
Traced to a firm: 16
Topics: 13
Hard: 30

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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 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
$P(\text{clash}) = 1 - \prod_{i=0}^{29} \frac{365 - i}{365} \approx 70.6\% \qquad \approx 1 - e^{-\binom{30}{2}/365} = 1 - e^{-1.19} \approx 70\%$
(365 - i)/365 the chance the next bond avoids the i dates already taken
C(30,2) the number of pairs among 30 bonds, 435
e the 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?
048A plane has 100 seats and 100 passengers with assigned seats. The first passenger is drunk and sits in a random seat; every later passenger takes their own seat if it is free, otherwise a random free seat. What is the probability the last passenger gets their own seat, and the probability for the Nth passenger?Probability and expected valueCoreBelvedere TradingChicago · 2022
Try it first
The chance that passenger 100 gets their own seat is:
Show the worked solution
The last passenger gets their own seat with probability exactly 1/2. Every random choice picks seat 1, seat 100, or another seat that just passes the problem on. Seat 1 and seat 100 are always equally likely, and whichever goes first decides it. For passenger k, the same argument over the seats that still matter gives (n minus k plus 1) over (n minus k plus 2): 99 in 100 for passenger 2, falling to 1/2 for the last.
Which seats actually matter to the last passenger?
Think of a game of musical chairs where every displaced person grabs a random empty chair. It looks like a mess, but most grabs only move the mess along to someone else. For the last passenger, only two seats matter: seat 1, the drunk's own, and seat 100, their own. If a displaced passenger takes seat 1, the chain stops and everyone after sits correctly. If someone takes seat 100, the last passenger loses. Any other seat hands the same situation to a later passenger.
Every random pick lands on seat 1, which ends the chain and saves the last passenger, on seat 100, which dooms them, or on another seat, which just passes the choice along, and since seats 1 and 100 are always equally likely the last passenger's chance is exactly one half.
How do you get the answer for any passenger?
Apply the same logic to passenger k. For them, the deciding seats are seat 1 and the seats of passengers k to n, which are still unclaimed by their owners when the chain reaches them. Passenger k loses only if their own seat is picked before seat 1, and among the n minus k plus 2 seats that matter, only one of them is theirs. The chance of the chain ending well for them is therefore n minus k plus 1 over n minus k plus 2.
The relationship
$P(k \text{ gets own seat}) = \frac{n - k + 1}{n - k + 2} \qquad k = 100: \frac{1}{2} \qquad k = 2: \frac{99}{100}$
n number of seats and passengers, 100
k the passenger's boarding position, from 2 to n
What it says in wordsPassenger k is safe unless their own seat is the one picked first among the seats that still matter to them.
Check the ends of the formula, which is what an interviewer will do. Passenger 2 loses only if the drunk sits in seat 2, a 1 in 100 chance, so 99 in 100 is right. Passenger 99 has a 2 in 3 chance, and passenger 100 has 1 in 2. The answer does not depend on the size of the plane for the last passenger: with 10 seats or 1,000 it is still one half, which is the fact worth saying out loud.
Where candidates lose it
Candidates try to track the chain of displaced passengers and drown in cases. The interviewer is waiting to see if you spot that only two seats matter; say that first and the answer follows in one line.
The second loss is answering 1/2 for the last passenger and then guessing 1/2 for everyone. The reported question asked for the Nth passenger, so have the general formula and its two sanity checks ready.
What the interviewer asks next
What is the expected number of passengers who end up in the wrong seat?
What if the first two passengers are both drunk?
Why is the answer for the last passenger independent of the number of seats?
Asked at Belvedere Trading, Equity Capital Markets, Chicago, 2022 (Wall Street Oasis): Drunk passenger on a plane, what's the probability the Nth passenger gets his assigned seat
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.
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
$P(X \ge 3) = \binom{4}{3} 0.6^{3}\, 0.4 + 0.6^{4} = 0.3456 + 0.1296 = 0.4752$
X the number of anchors who order
C(4,3) the four ways to choose which three anchors order
0.6, 0.4 each 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?
077You have two bowls, 60 white balls and 40 black balls. Split all 100 balls between the bowls any way you like. One bowl is then chosen at random and one ball drawn from it. How do you maximise the chance of drawing white, and what is that chance?Probability and expected valueCoreDeutsche BankMumbai · 2024
Try it first
Pick the best chance you think a clever split can reach.
Show the worked solution
Put one white ball alone in bowl A and the other 99 balls, 59 white and 40 black, in bowl B. Bowl A gives white every time and bowl B gives white 59 times in 99. Each bowl is chosen half the time, so the chance is 0.5 x 1 + 0.5 x 59/99, about 79.8%, against 60% for an even split.
Why does a bowl with one ball count as much as a bowl with 99?
Picture two teams tossing a coin to bat first: the coin does not care that one side has eleven players and the other has one. The bowls are chosen the same way. The coin picks a bowl, not a ball, so a bowl holding a single white ball gets the same half of the draws as a bowl holding ninety nine. That makes one lone white ball the cheapest way to buy certainty for half of all outcomes, and it costs bowl B only one white ball out of sixty.
Bowl A with one white ball gives white every time; bowl B with 59 white and 40 black gives white 59.6% of the time, so the average is 79.8%, well above the 60% an even split gives.
How do you show that nothing beats it?
Bowl A cannot do better than 100%, and one white ball is the least that gets it there. Every further white ball moved into bowl A is wasted there and missed in bowl B, and every black ball moved into bowl A drags it below 100%. So bowl A is one white ball and bowl B takes what is left. Checking all 2,499 possible splits by computer agrees: the best is 79.80%, and only the one-ball split reaches it.
The relationship
$P(\text{white}) = \tfrac{1}{2}\cdot 1 + \tfrac{1}{2}\cdot\frac{W-1}{W+B-1} = \tfrac{1}{2} + \tfrac{1}{2}\cdot\frac{59}{99} \approx 0.798$
W white balls, 60
B black balls, 40
1/2 the chance each bowl is chosen
(W-1)/(W+B-1) the white share in bowl B once one white ball is set aside
What it says in wordsHalf the time you get the certain bowl, half the time you get everything else.
The formula also shows how the answer moves. With 50 white and 50 black, the version a candidate reported from an interview, it gives 74.7%. More white in the pool lifts bowl B and the total; bowl A is already at its ceiling. Here the split adds about 20 percentage points because one ball is turned into half of the probability.
Where candidates lose it
Most people answer 60% because the pool is 60% white and they assume a split cannot change the pool. It cannot change the pool, but it changes the weights, because the bowl is chosen before the ball.
The second loss is stopping at the number. Give the one-line reason no other split does better, then generalise: one half plus one half of (W minus 1) over (W plus B minus 1).
What the interviewer asks next
What if the bowl is chosen with probability proportional to how many balls it holds?
With three bowls and the same 100 balls, what is the best split and the best chance?
You are paid Rs 100 for a white ball and nothing for black. What is the most you would pay to play once?
Asked at Deutsche Bank, Equity Capital Markets, Mumbai, 2024 (Wall Street Oasis): After the distribution, one bowl will be selected at random, and then one ball will be randomly drawn from that bowl
1004% of issuers in a sector default within a year. An early warning model flags 75% of the issuers that will default and 10% of those that will not. An issuer is flagged. What is the probability that it defaults?Probability and expected valueCoreBelvedere TradingChicago · 2022
Try it first
Pick the closest before you calculate.
Show the worked solution
About 23.8%. Picture 1,000 issuers: 40 will default and 960 will not. The model flags 75% of the 40, which is 30, and 10% of the 960, which is 96. Of the 126 flagged issuers, 30 default, so the chance is 30 / 126 = 23.8%. The flag multiplies the risk about six times, but most flags are still false alarms because defaults are rare.
Why is the answer not 75%?
A smoke alarm that goes off for every real fire and also for one in ten batches of toast will ring mostly for toast, because toast is far more common than fire. The early warning model is the same. The chance of default given a flag depends on how common defaults are to begin with, and when the base rate is low, false flags from the large healthy group swamp the true flags from the small defaulting group. 75% answers a different question: how often a defaulter gets flagged.
Of 1,000 issuers, the 40 that will default produce 30 flags and the 960 that will not produce 96 false flags, so only 30 of the 126 flagged issuers default, a probability of 23.8%.
The relationship
$P(D \mid F) = \frac{P(F \mid D)P(D)}{P(F \mid D)P(D) + P(F \mid \bar D)P(\bar D)} = \frac{0.75 \times 0.04}{0.75 \times 0.04 + 0.10 \times 0.96} = \frac{30}{126} = 23.8\%$
P(D) the base rate of default, 4%
P(F | D) the chance a defaulter is flagged, 75%
P(F | not D) the chance a healthy issuer is flagged, 10%
What it says in wordsOf all the flags, the share that come from real defaulters is the chance a flag means default.
How should a credit desk use a 23.8% flag?
As a reason to look harder, not a verdict. A flag lifts the default probability from 4% to 23.8%, roughly six times, which is worth a review but not a sale on its own. A second, independent signal compounds the evidence: if a separate test with the same accuracy also flagged the issuer, the probability would rise to about 70%. The honest limitation is that two warning signals about the same company are rarely independent, so the real lift from a second flag is usually smaller.
Where candidates lose it
The common error is answering 75%, swapping the probability of a flag given default for the probability of default given a flag. It is the most frequent mistake in Bayes questions, and interviewers set it up deliberately.
The second loss is doing the formula silently and producing 23.8% with no picture. Say the 1,000 issuers out loud, 30 true flags and 96 false ones, and the interviewer can follow every step.
What the interviewer asks next
What false alarm rate would the model need for a flag to mean a 50% chance of default?
The sector's default rate doubles to 8%. What does a flag mean now?
Two independent models both flag the issuer. What is the probability of default?
Asked at Belvedere Trading, Capital Markets, Chicago, 2022 (Wall Street Oasis): The technical portion of the interview consisted of probability questions including one questions relating to Bayes' theorem

(365 - i)/365	the chance the next bond avoids the i dates already taken
C(30,2)	the number of pairs among 30 bonds, 435
e	the base of natural logarithms, about 2.718

n	number of seats and passengers, 100
k	the passenger's boarding position, from 2 to n

X	the number of anchors who order
C(4,3)	the four ways to choose which three anchors order
0.6, 0.4	each anchor's chance of ordering and of staying away

W	white balls, 60
B	black balls, 40
1/2	the chance each bowl is chosen
(W-1)/(W+B-1)	the white share in bowl B once one white ball is set aside

Try it first

Why is 30 out of 365 so far off?

What is the clean way to calculate it?

Where does this show up on a desk?

Where candidates lose it

What the interviewer asks next

Try it first

Which seats actually matter to the last passenger?

How do you get the answer for any passenger?

Where candidates lose it

What the interviewer asks next

Try it first

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

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

Where candidates lose it

What the interviewer asks next

Try it first

Why does a bowl with one ball count as much as a bowl with 99?

How do you show that nothing beats it?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the answer not 75%?

How should a credit desk use a 23.8% flag?

Where candidates lose it

What the interviewer asks next