Debt Capital Markets puzzles, solved step by step

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

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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?
065In a trading game you are asked to make a two-way market on the sum of three fair dice. Quote a bid and an offer. The first die is then shown to be a 6. Where do you move your market, and why?Probability and expected valueHardBelvedere TradingChicago · 2022
Try it first
After the 6 is shown, where should the middle of your market be?
Show the worked solution
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.
Before any die is shown the sum centres on 10.5 with a standard deviation of 2.96; once a 6 is shown it centres on 13.0 with a standard deviation of 2.42, so the quote moves up 2.5 and narrows from 9.5 to 11.5 to about 12.2 to 13.8.
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.
The relationship
$E = 6 + 2 \times 3.5 = 13.0 \qquad \sigma = \sqrt{2 \times \tfrac{35}{12}} = 2.42 \ \text{against} \ \sqrt{3 \times \tfrac{35}{12}} = 2.96$
3.5 the expected value of one fair die
35/12 the variance of one fair die
sigma the standard deviation of the part of the sum still unknown
What it says in wordsThe expected value adds the known die to the average of the hidden ones; the uncertainty comes only from the dice still hidden.
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.
Where candidates lose it
The common loss is leaving the market where it was, or moving the middle and forgetting the width. Fair value jumps to 13.0 the moment the 6 is shown and the uncertainty is smaller, so both the level and the width should change.
The second is narrowing the width by a third because one of three dice is known. Variance falls by a third, but the standard deviation falls by only about 18%, so the market narrows by about a fifth.
What the interviewer asks next
Someone lifts your 13.8 offer three times in a row. What do you do next?
Before any die is shown, what is the probability that the sum is 13 or more?
How does your market change if the second die is also shown to be a 6?
Asked at Belvedere Trading, Equity Capital Markets, Chicago, 2022 (Wall Street Oasis): superday with two 1-1s and a group trading game
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?

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

3.5	the expected value of one fair die
35/12	the variance of one fair die
sigma	the standard deviation of the part of the sum still unknown

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

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

Where does the first quote come from?

What changes when the first die is shown?

What does the interviewer want to hear beyond the numbers?

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