Debt Capital Markets puzzles, solved step by step

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.

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.

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.

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.

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.

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?

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.

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.

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.

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.

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.

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.

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.

Start around 10.5, say 9.5 bid and 11.5 offered, then move to about 12.2 bid, 13.8 offered once the 6 is shown. Each fair die is worth 3.5 on average, so three are worth 10.5. After the reveal the value is 6 plus 3.5 plus 3.5, which is 13.0. The market should also narrow, because one die's uncertainty has gone: the standard deviation falls from 2.96 to 2.42, so the width shrinks by about a fifth, not a third.

Where does the first quote come from?

A shop that buys and sells used phones offers to buy below what it thinks a phone is worth and to sell above it, and earns the gap. A market maker quotes around the expected value, with a bid below and an offer above, and the width reflects how uncertain that value is. Each fair die averages 3.5, so three dice average 10.5. A quote of 9.5 bid, 11.5 offered straddles that value; the sum can still land anywhere from 3 to 18, so the market cannot be tight.

What changes when the first die is shown?

Two things, and candidates usually say only one. The expected value jumps to 6 plus 3.5 plus 3.5, which is 13.0, and the uncertainty shrinks because only two dice are still hidden. Variance adds across independent dice, 35/12 for each, so it falls from 8.75 to 5.83, a third lower; the standard deviation falls from 2.96 to 2.42, about 18% lower. Scale the width by that: 2 points becomes about 1.6, so quote roughly 12.2 bid, 13.8 offered.

What does the interviewer want to hear beyond the numbers?

Move the market the moment information arrives, because a stale quote is a free option for everyone else at the table. If you stay at 9.5 to 11.5 after a 6 shows, every player buys your offer at 11.5 against a fair value of 13.0. Then say what you would do if you suspected the other side knew more than you, for example had seen a second die: widen, or lean your quote towards the risk. The limit: the width here is a choice; a real desk sets it by competition and by how much risk it can hold, not by a formula.

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.

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.

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.

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 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.

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.

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.

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.

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.

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.