Debt Capital Markets puzzles, solved step by step

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

Showing 1–6 of 6 · filtered from 100Clear filters

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?
023You roll a fair die repeatedly. What is the expected number of rolls to see a six, and the expected number of rolls to see two sixes in a row?Probability and expected valueHardSyndicate desks
Try it first
Expected rolls to see two sixes in a row?
Show the worked solution
6 rolls for one six, and 42 rolls for two sixes in a row. A six comes up one time in six, so the wait averages 6. For two in a row, define E0 as the expected rolls from the start and E1 from one six showing. From E1 the next roll ends it with 1/6 or sends you back with 5/6. Solving E0 = 1 + (5/6)E0 + (1/6)E1 and E1 = 1 + (5/6)E0 gives E0 = 42.
Why is the first answer 6?
If a bus comes with probability one in six each minute, on average you wait six minutes. For a repeated trial with success probability p, the expected number of tries until the first success is 1 over p. With p equal to 1/6, that is 6. Say it quickly; the interviewer is only using it to set up the second part.
Why is two in a row not 12?
Twelve assumes the progress you make is kept. In a row means a single miss after a first six erases it, so you keep paying the six-roll wait again and again, and the answer is driven by those resets. Think of climbing two steps on a slippery staircase where any slip on the second step sends you to the ground: most attempts end on step one. Two states capture this: E0 with no six showing, E1 with one six showing.
From the start a six moves you to state E1 with probability 1/6; from E1 a second six finishes, but any other roll, 5/6 of the time, sends you back to the start, so the expected rolls solve to 42 from the start and 36 from one six showing.
The relationship
$E_1 = 1 + \tfrac{5}{6}E_0 \qquad E_0 = 1 + \tfrac{5}{6}E_0 + \tfrac{1}{6}\left(1 + \tfrac{5}{6}E_0\right) \;\Rightarrow\; \tfrac{1}{36}E_0 = \tfrac{7}{6} \;\Rightarrow\; E_0 = 42$
E_0 expected further rolls from the start, no six showing
E_1 expected further rolls with one six showing
5/6, 1/6 chance of a non-six and a six on any roll
What it says in wordsEach state's expected rolls equal one roll plus the expected rolls from wherever that roll sends you.
Is there a quick check you can say out loud?
Yes: for k in a row with success probability p, the expected wait is 1/p plus 1/p squared, and so on up to 1/p to the k. For two sixes that is 6 plus 36, which is 42; for three sixes in a row it is 6 plus 36 plus 216, which is 258. The pattern shows why streaks get expensive fast. The desk version of the same idea: requiring several conditions to hold consecutively, a covenant tested on two quarters in a row, or a run of clean prints, is far rarer than requiring them separately.
Where candidates lose it
The trap answer is 12, doubling the single six, or 36, reading two in a row as a single one in 36 event. Both ignore that failure after the first six throws away progress.
The second loss is setting up E1 wrongly, sending a non-six from E1 back to E1 instead of E0. Say where each roll sends you before writing the equations.
What the interviewer asks next
What is the expected number of rolls to see a six followed immediately by a five?
How many rolls on average to see three sixes in a row?
A game pays Rs 100 when you first roll two sixes in a row and each roll costs Rs 2. Is it worth playing?
042A trader starts with 3 units of capital and stops at 0 or 6. Each trade wins or loses 1 unit. What is the probability of reaching 6 if each trade is a fair coin, and if the win probability is 55%?Probability and expected valueHardRisk managementFixed income asset management
Try it first
With a 55% edge on each trade, the chance of reaching 6 before 0 is closest to:
Show the worked solution
50% with a fair coin, and about 64.6% with a 55% win rate. With a fair coin, your capital is a fair bet, so the chance of reaching 6 from 3 is 3 over 6. With an edge, the chance is 1 over 1 plus (q over p) cubed, where q over p is 0.45 over 0.55. That gives 1 over 1.548, or 64.6%. Spread the same edge over walls ten times further away and it rises to about 99.8%.
Why is the fair-coin answer simply 3 over 6?
Picture a game where you and a friend toss a coin for Rs 1 until one of you is broke; you start with Rs 3, the friend with Rs 3. Every toss is fair, so on average nobody gains, and your expected wealth at the end must still be Rs 3. If the game ends at 0 or 6 and your expected ending wealth is 3, you must reach 6 exactly half the time. In general the fair-coin chance of reaching N from i is i over N.
How does a 55% edge change it?
The relationship
$P(\text{reach } N \text{ from } i) = \frac{1 - (q/p)^i}{1 - (q/p)^N} \qquad \frac{1 - 0.8182^3}{1 - 0.8182^6} = \frac{1}{1 + 0.8182^3} \approx 64.6\%$
p, q chance of winning and losing each trade, 0.55 and 0.45
q/p 0.45 over 0.55, about 0.818
i, N starting capital 3 and target 6
What it says in wordsThe ratio of losing to winning odds, raised to the distance from each wall, sets how strongly the edge tilts the outcome.
This is the classic gambler's ruinA random walk that stops at two walls, used to find the chance of hitting one wall before the other. set-up. A 55% edge on each trade turns into a 64.6% chance of doubling before going broke, a modest lift because the walls are only three steps away. The fair game takes 9 trades on average to finish, so the edge only gets a handful of chances to work.
Starting with 3 units and a target of 6, a 55% win rate gives a 64.6% chance of reaching the target, but with 30 units and a target of 60, still in steps of 1, the same edge gives 99.8%, because the edge has many more trades over which to work.
That is the lesson a desk wants. The same edge, bet in smaller pieces relative to capital, almost removes the risk of ruin: starting at 30 with a target of 60 and 1-unit trades, the chance of success is 99.8%. Betting a large share of capital on each trade throws the edge away, because variance gets to end the game before the edge shows. The limit is the model itself: real trades do not win or lose exactly one unit, and edges are estimated, not known.
Where candidates lose it
The common wrong answer to the second part is 55%: candidates assume the per-trade edge equals the edge on the whole game. The game is many trades long, so the edge compounds, and the answer must be higher.
The opposite error is guessing something near certainty. With walls only three steps away, variance still dominates; say {P42['prob']*100:.1f}% and then explain why position size, not the edge alone, drives the chance of ruin.
What the interviewer asks next
What is the expected number of trades before the game ends with a fair coin?
With a 45% win rate, what is the chance of reaching 6?
How does this connect to the Kelly criterion for sizing a bet?
060Pitching for a bond mandate costs Rs 50 lakh of team time. You estimate a 30% chance of winning a mandate worth Rs 4 crore in fees. Should you pitch, and what win probability makes the pitch break even?Probability and expected valueWarm upSyndicate desks
Try it first
What is the expected value of pitching, net of the cost?
Show the worked solution
Yes, pitch: the expected value is plus Rs 70 lakh, and the pitch breaks even at a 12.5% win probability. A 30% chance of Rs 4 crore is worth Rs 1.2 crore in expectation, against a certain Rs 50 lakh cost. Break-even is where the probability times Rs 4 crore equals Rs 50 lakh: 0.5 divided by 4, or 12.5%. The 30% estimate clears that by a wide margin, so even a rough probability supports pitching.
How do you compare an uncertain fee with a certain cost?
A shopkeeper deciding whether to print Rs 500 of flyers asks how much extra business they might bring, times how likely that is. Expected value multiplies each outcome by its probability and adds them up, so a 30% chance of Rs 4 crore is worth Rs 1.2 crore before costs. The cost of pitching is paid whether you win or lose, so it comes off in full: Rs 1.2 crore minus Rs 0.5 crore is plus Rs 0.7 crore.
The relationship
$EV = 0.30 \times 4.0 - 0.5 = +0.7\ \text{crore} \qquad p^{*} = \frac{0.5}{4.0} = 12.5\%$
0.30 your estimated chance of winning
4.0 the fee if you win, Rs crore
0.5 the cost of pitching, paid either way, Rs crore
p* the break-even win probability
What it says in wordsExpected fee less the certain cost; break-even is the cost divided by the prize.
The expected fee of Rs 1.2 crore less the Rs 0.5 crore cost leaves plus Rs 0.7 crore, and the expected fee line crosses the cost at a 12.5% win probability, well below the 30% estimate.
What win probability makes the pitch worth it, and how robust is the answer?
Break-even is the probability at which the expected fee just covers the cost: Rs 50 lakh over Rs 4 crore, 12.5%. The 30% estimate is more than twice that, so the decision survives a lot of error in the estimate. That is the useful part of the calculation: you rarely know a win probability precisely, but you often know whether it is comfortably above or below the break-even.
Name what the simple sum leaves out. Losing is the most likely outcome, 70% of the time, so a desk that pitches only once can easily end Rs 50 lakh down; expected value pays off across many pitches. The team's time also has an opportunity cost if it could pitch a better mandate instead, and a win may bring follow-on business that the Rs 4 crore does not capture. Each of those shifts the break-even; none reverses this answer.
Where candidates lose it
The trap is anchoring on the most likely outcome. You lose 70% of the time, so candidates say the pitch loses money. Expected value is not the most likely outcome; it is the probability-weighted average, and here that is plus Rs 70 lakh.
The second loss is forgetting to subtract the cost and answering Rs 1.2 crore, or dividing the wrong way for break-even. Say break-even as cost over prize, then check it: 12.5% of Rs 4 crore is Rs 50 lakh.
What the interviewer asks next
If you win you must share the mandate with a second bank, halving your fee. Does the answer change?
You can pitch three mandates like this but can only staff two. How do you choose?
How would you estimate the 30% win probability in the first place?
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?
097You are selling a loan and will receive five bids one at a time, in random order. You must accept or reject each bid on the spot, and a rejected bid never comes back. What rule maximises your chance of accepting the single best bid, and what is that chance?Probability and expected valueHardSyndicate desksCorporate banking
Try it first
Which rule gives the best chance of ending with the top bid?
Show the worked solution
Reject the first two bids, then accept the first bid that beats both of them; you get the best bid 43.3% of the time. Taking the first or last bid wins only 20%. The two rejected bids set a benchmark, and the rule succeeds whenever the best bid comes later and the best of the bids before it sits among the first two. Checking all 120 orderings gives 52 wins, which is 43.3%.
Why reject bids you know nothing wrong with?
House hunting in a city you do not know, you would look at a couple of flats before signing anything, simply to learn what good looks like. Sign too early and you never had a benchmark; look too long and the best one may already be gone. The rejected bids are the price of information: they set the bar that later bids must clear, and the only question is how many to spend. With five bids, spending two is the best trade.
Rejecting the first two bids and then taking the first bid that beats them picks the best of five bids 43.3% of the time, against 20% for taking the first or the last bid and 41.7% or 35.0% for rejecting one or three.
How do you get 43.3% without listing all 120 orders?
Ask where the best bid sits. If it is in the first two, you have already rejected it and lose. If it sits at position j, from 3 to 5, you take it only if no earlier bid after the first two already beat the bar, which happens when the best of the first j minus 1 bids lies in the first two. That chance is 2 out of (j minus 1), so the rule wins with probability one fifth of (2/2 + 2/3 + 2/4), which is 43.3%. Say the structure; the arithmetic takes ten seconds.
The relationship
$P(\text{best}) = \frac{1}{5}\left(\frac{2}{2} + \frac{2}{3} + \frac{2}{4}\right) = \frac{1}{5} \times 2.167 = 0.433$
1/5 the chance the best bid is in any given position
2/(j-1) the chance that, with the best bid at position j, the best earlier bid is among the two rejected
What it says in wordsAdd up, over each place the best bid could arrive, the chance the rule is still waiting when it gets there.
What is the limitation for a real loan sale?
Two things. The rule maximises the chance of the very best bid, not the expected price; a seller who cares about the average price would behave differently. And real loan sales rarely force on-the-spot decisions: a desk runs a process that collects bids together, precisely to avoid this problem. With many bids, the rule becomes the well-known look at about 37% and then leap, and the success rate falls towards about 37% too.
Where candidates lose it
The common answer is to take the first good-looking bid. It wins only 20% of the time, because a good-looking bid with no benchmark is just a random bid.
The second loss is knowing the 37% rule and applying it blindly: 37% of five is 1.85 bids, and the candidate who rounds without checking may reject one instead of two. Compute the small case directly; it takes a few lines.
What the interviewer asks next
With ten bids, how many would you reject first?
If you are paid the bid you accept rather than rewarded only for the best, does the rule change?
Rejected bidders may come back with 50% probability. How does that change your cut-off?

(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

E_0	expected further rolls from the start, no six showing
E_1	expected further rolls with one six showing
5/6, 1/6	chance of a non-six and a six on any roll

p, q	chance of winning and losing each trade, 0.55 and 0.45
q/p	0.45 over 0.55, about 0.818
i, N	starting capital 3 and target 6

0.30	your estimated chance of winning
4.0	the fee if you win, Rs crore
0.5	the cost of pitching, paid either way, Rs crore
p*	the break-even win probability

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

Why is the first answer 6?

Why is two in a row not 12?

Is there a quick check you can say out loud?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the fair-coin answer simply 3 over 6?

How does a 55% edge change it?

Where candidates lose it

What the interviewer asks next

Try it first

How do you compare an uncertain fee with a certain cost?

What win probability makes the pitch worth it, and how robust is the answer?

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 reject bids you know nothing wrong with?

How do you get 43.3% without listing all 120 orders?

What is the limitation for a real loan sale?

Where candidates lose it

What the interviewer asks next