Debt Capital Markets puzzles, solved step by step

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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. 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.

    Two states and a reset: why two sixes in a row takes 42 rolls, not 12E0: startno six showingE1one six showingDonetwo in a rowsix, 1/6six, 1/6not a six, 5/6: back to the startnot a six, 5/6E0 = 1 + (5/6) E0 + (1/6) E1E1 = 1 + (5/6) E0 + (1/6) x 0each line: one roll, then the expected rolls still needed from wherever you landE0 = 42E1 = 36
    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
    E1=1+56E0E0=1+56E0+16(1+56E0)  ⇒  136E0=76  ⇒  E0=42E_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_0expected further rolls from the start, no six showing
    E_1expected further rolls with one six showing
    5/6, 1/6chance 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?
  2. 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(reach N from i)=1−(q/p)i1−(q/p)N1−0.818231−0.81826=11+0.81823≈64.6%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, qchance of winning and losing each trade, 0.55 and 0.45
    q/p0.45 over 0.55, about 0.818
    i, Nstarting 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.

    A small edge per trade becomes a big edge over many trades0%25%50%75%100%40%45%50%55%60%Chance of winning each tradefair coin: 50%start 3, target 6: 64.6%start 30, target 60: 99.8%
    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?
  3. 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.

    Look at two bids, then leap at the first one that beats themBid 1look, rejectBid 2look, rejectBid 3take if a recordBid 4take if a recordBid 5take if a recordset the barfirst bid above the bar wins the loanChance of picking the best bid, by bids rejected first20.0%reject 041.7%reject 143.3%reject 235.0%reject 320.0%reject 4
    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(best)=15(22+23+24)=15×2.167=0.433P(\text{best}) = \frac{1}{5}\left(\frac{2}{2} + \frac{2}{3} + \frac{2}{4}\right) = \frac{1}{5} \times 2.167 = 0.433
    1/5the 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?