Investment Banking puzzles, solved step by step

Puzzles: 100
Traced to a firm: 36
Topics: 12
Hard: 29

Showing 11–17 of 17 · filtered from 100Clear filters

080You roll a fair six-sided die repeatedly. On average, how many rolls will it take before every face from 1 to 6 has appeared at least once?Expected value and gamesHardBulge bracket IBConsulting style brainteasers
Try it first
Gut call before you work it: how many rolls on average?
Show the worked solution
About 14.7 rolls. Break the hunt into stages. The first roll always gives a new face. After that a new face comes up with chance 5/6, then 4/6, and so on down to 1/6. The average wait for an event of chance p is 1/p, so the stages take 1, 1.2, 1.5, 2, 3 and 6 rolls, which add to 6 x (1 + 1/2 + ... + 1/6) = 14.7.
Why does breaking it into stages make it easy?
Collecting cricket cards from cereal packets feels quick at first: almost every packet has a new card. The last card is the painful one, because nearly every packet repeats something you already own. The die is the same. Each stage is a simple waiting game with a fixed chance of success, and the average wait for a chance p is 1/p rolls. With k faces already found, a new one turns up with chance (6 minus k)/6, so that stage takes 6/(6 minus k) rolls on average.
Why is the wait 1/p? If a new face comes up one roll in three, you expect to wait three rolls, the same way a bus that comes one minute in ten keeps you waiting about ten. That one fact, applied six times, is the whole puzzle.
The six stages take 1, 1.2, 1.5, 2, 3 and 6 rolls on average, adding to 14.7, and the final face alone accounts for 6 rolls because only one roll in six finds it.
How do you add it up quickly in the room?
The relationship
$E[N] = \frac{6}{6} + \frac{6}{5} + \frac{6}{4} + \frac{6}{3} + \frac{6}{2} + \frac{6}{1} = 6 \sum_{k=1}^{6} \frac{1}{k} \approx 14.7$
E[N] the expected number of rolls to see all six faces
6/(6 minus k) the average wait for a new face once k faces are found
What it says in wordsAdd the average wait for each new face; the waits grow because new faces get rarer.
Pair the fractions to keep the arithmetic out loud: 1 plus 6 is 7, 1.2 plus 3 is 4.2, 1.5 plus 2 is 3.5, total 14.7. The last face costs six rolls on its own, more than the first four stages put together. That shape, cheap early progress and an expensive tail, is worth one sentence because it shows you see the structure, not just the sum.
One limitation: 14.7 is an average. The spread is wide, and plenty of runs take more than 20 rolls, so if the interviewer asks for a number you would bet on for a single run, say the average is not a likely outcome.
Where candidates lose it
The instinct is six, or a vague 10, because candidates picture one roll per face. Saying a number without the stage structure gives the interviewer nothing to follow, and an unexplained guess counts as a miss even if it is close.
The second loss is getting the stages right and fumbling the sum. Pair the terms that make round numbers and say 14.7, not fourteen point something.
What the interviewer asks next
How many rolls on average to see all faces of a 20-sided die?
What is the expected number of rolls until you see a 6?
Roughly how many rolls would you need to be 90% sure of seeing every face?
083A stock trades at Rs 100 and pays no dividend. A one-year European call struck at Rs 100 costs Rs 10, and the one-year interest rate is 5%. What should a put with the same strike and expiry cost?Rates, risk and optionsHardDebt capital marketsSales and trading
Try it first
Pick the put price before you work it.
Show the worked solution
About Rs 5.24. A call plus cash worth the strike's present value pays the larger of the share price and 100 at expiry, and so does a put plus the share. Identical payoffs must cost the same today: 10 + 100/1.05 = P + 100. The present value of 100 is 95.24, so the put is 5.24. No pricing model or volatility estimate is needed.
Why can you price the put without knowing volatility?
If two combo meals contain exactly the same food, a restaurant cannot charge different prices for long: everyone would buy the cheaper one. Options work the same way. Two portfolios that pay the same in every outcome must cost the same today, so if you know the price of one, the other follows without any view on where the share goes. Volatility matters for what a call is worth, but the call's price already contains it. The put inherits it through the parity relationship.
Build the two portfolios. A: buy the call and put aside 95.24 at 5%, which grows to exactly 100. B: buy the put and buy one share. If the share ends at 130, A exercises the call for 30 and adds its 100 cash, B holds a share worth 130 and lets the put expire: both have 130. If the share ends at 70, A lets the call expire and holds 100 cash, B exercises the put to sell the share for 100: both have 100.
A call plus cash of 95.24 and a put plus one share both pay the larger of the share price and 100 at expiry, so their prices today must match, which pins the put at 5.24.
What does the formula say, and what if the put traded at Rs 7?
The relationship
$C + \frac{K}{1+r} = P + S \quad\Rightarrow\quad P = 10 + \frac{100}{1.05} - 100 \approx 5.24$
C call price, Rs 10
P put price, the unknown
S share price today, Rs 100
K/(1+r) the strike discounted one year at 5%, 95.24
What it says in wordsA call plus the discounted strike is worth the same as a put plus the share.
If the put traded at 7, portfolio B would cost 107 against 105.24 for A, so you would sell B and buy A, pocketing 1.76 today with payoffs that cancel at expiry. Selling B means writing the put and short selling the share. Traders arbitraging that gap is what holds parity in place, which is why quoted put and call prices sit close to it in practice.
Say the assumptions: European options, so no early exercise; no dividend before expiry; and annual compounding. With continuous compounding the strike's present value is 100 x e to the minus 0.05, about 95.12, and the put comes out near 5.12. Give the convention, not just the number.
Where candidates lose it
The instinctive answer is Rs 10, on the idea that at-the-money calls and puts must cost the same. They do not, because the call holder defers paying the strike for a year, and that deferral is worth something when rates are positive.
The second loss is remembering parity but flipping a sign and answering 4.76. Build the two portfolios out loud before writing the formula and the signs take care of themselves.
What the interviewer asks next
If the stock paid a Rs 2 dividend before expiry, how would the put price change?
Why does parity hold exactly for European options but only as a bound for American ones?
What position replicates a long put using only the call, the share and borrowing?
084Four people must cross a narrow bridge at night with one torch. They take 1, 2, 5 and 10 minutes to cross. At most two can be on the bridge at once, a pair walks at the slower person's pace, and the torch must be carried on every crossing. What is the fastest total time?Logic and brainteasersHardConsulting style brainteasersSales and trading
Try it first
What is the fastest time you can get everyone across?
Show the worked solution
17 minutes. The 1 and 2 cross (2), the 1 returns (1), the 5 and 10 cross together (10), the 2 returns (2), and the 1 and 2 cross again (2). The obvious plan, the fastest person escorting each of the others, takes 19. The saving comes from pairing the two slowest so the 5 minute walk overlaps the 10 minute one.
Why does the escort plan feel right but lose two minutes?
Picture sharing taxis to the airport. Putting the two people with the latest flights in the same cab costs nothing extra for either; putting each in a cab with someone in a hurry makes every ride as slow as its slowest passenger. A pair always moves at the slower pace, so the 5 minute walker is free if he travels with the 10, and costs a full five minutes if he travels with anyone else. The escort plan pays for the 10, the 5 and the 2 separately.
Escorting everyone with the fastest walker takes 19 minutes because the 10 and the 5 each pay for their own crossing, while sending the 10 and the 5 together takes 17, since the 5 minute walk is absorbed inside the 10.
How do you know 17 cannot be beaten?
Count what every plan must pay. Five crossings are needed: three trips over and two back with the torch. The 10 must cross at least once, costing 10. If the 5 crosses separately from the 10, that is another 5 on top. Pairing them leaves two forward trips and two returns to do with the 1 and the 2, and the cheapest way to do that is 2 + 1 + 2 + 2. The trade is paying the 2 minute walker an extra return in order to save the 5 minute walker's crossing, and 2 is less than 5 minus 1.
The relationship
$\text{pair the slowest if } 2b < a + c: \quad 2 \times 2 = 4 < 1 + 5 = 6$
a the fastest walker, 1 minute
b the second fastest, 2 minutes
c the second slowest, 5 minutes
What it says in wordsSending the two slowest together wins when two trips by the second fastest cost less than the fastest walker plus the second slowest.
The rule is worth stating because it is not always pairing. If the walkers took 1, 4, 5 and 10 minutes, twice 4 is 8, more than 1 plus 5, so escorting wins: 21 minutes against 23 for pairing. Showing the interviewer you know when your trick stops working is better than the trick.
Where candidates lose it
Most candidates find 19 with the escort plan, test one or two variations and declare it optimal, because using the fastest walker for every return looks obviously efficient. The interviewer is waiting to see whether you notice the slow walkers can share a crossing.
The second loss is reaching 17 by trial and error and being unable to say why. Have the one-line reason ready: the 5 hides inside the 10, at the price of one extra 2 minute return.
What the interviewer asks next
What if the times are 1, 4, 5 and 10?
Add a fifth person who takes 12 minutes. What is the fastest time now?
Where does the same logic of batching slow jobs together show up on a deal team?
087A DCF sets its terminal value at 10x final-year EBITDA. Free cash flow runs at half of EBITDA and the WACC is 10%. What perpetual growth rate is that exit multiple quietly assuming?Valuation riddlesHardElite boutique IBBulge bracket IB
Try it first
Roughly what growth rate does 10x EBITDA imply here?
Show the worked solution
About 4.8% a year, forever. Write the terminal value both ways. The multiple says 10 x EBITDA. The Gordon formula says 0.5 x EBITDA x (1 + g) / (0.10 - g). Setting them equal, EBITDA cancels: 1.0 - 10g = 0.5 + 0.5g, so g = 0.5 / 10.5 = 4.76%. Every exit multiple implies a growth rate, and the check is whether that rate is believable for this business.
Why does a multiple contain a growth rate at all?
If a buyer wants a 5% return and pays 25 times this year's rent for a flat, they are counting on the rent rising: with flat rent, a 5% return supports a price of only 20 times. A multiple is a shorthand price for a stream of future cash flows, so every multiple implies some combination of growth and discount rate. In a DCF the exit multiple sits beside an explicit WACC, which means the growth assumption can be solved for exactly.
Set the two terminal values equal. The multiple gives 10 x EBITDA. The perpetuity growth method gives next year's free cash flow over (r minus g), and next year's cash flow is 0.5 x EBITDA x (1 + g). EBITDA cancels from both sides, leaving one equation in g, which solves to 4.76%.
With free cash flow at half of EBITDA and a 10% WACC, a 10x exit multiple implies perpetual growth of 4.8%, and implied growth rises with the multiple while bending toward the 10% WACC, which it can never reach.
The relationship
$m = \frac{f(1+g)}{r-g} \;\Rightarrow\; g = \frac{mr - f}{m + f} = \frac{1.0 - 0.5}{10 + 0.5} \approx 4.76\%$
m exit multiple of EBITDA, 10x
f free cash flow as a share of EBITDA, 0.5
r WACC, 10%
g implied perpetual growth
What it says in wordsRearranging the perpetuity formula turns any exit multiple into the growth rate it assumes.
How do you judge whether 4.8% is believable?
Perpetual growth means growth forever, so it should not exceed the long-run nominal growth of the economy whose currency the cash flows are in, or the business eventually becomes larger than that economy. Whether 4.8% passes depends on that currency's inflation and real growth, so state the benchmark you are using rather than a number from memory. The useful habit is running the check both ways: an exit multiple that implies an unbelievable growth rate is a sign the multiple came from a peer set that does not match this business.
Say the convention too. If you apply the multiple to final-year EBITDA but take next year's cash flow without the (1 + g) uplift, the answer becomes 10% minus 0.5/10, which is 5.0%. A small difference, but naming the convention shows you know where it comes from.
Where candidates lose it
The common answer is that an exit multiple has no growth assumption, which is exactly backwards. Candidates who have only used the multiple method never think to translate it, and the interviewer is testing that translation.
The second loss is solving the algebra but not commenting on the result. Say the number, then say what you would compare it against and why.
What the interviewer asks next
What exit multiple implies zero growth here?
If free cash flow conversion rises to 70% of EBITDA, what growth does 10x imply?
Which method would you lead with in a fairness opinion, and why show both?
092You may bet any fraction of your bankroll, again and again, on a coin that lands in your favour 60% of the time and pays even money. What fraction maximises the long-run growth of your bankroll, and what happens if you bet twice that fraction?Expected value and gamesHardBulge bracket IBConsulting style brainteasers
Try it first
Which stake grows your bankroll fastest over many bets?
Show the worked solution
Bet 20% each time; bet 40% and long-run growth falls to about zero, slightly negative. With even money, the growth-maximising fraction is the win probability minus the loss probability, 0.6 - 0.4 = 0.2. That earns about 2.0% compound growth a bet. At 40%, three wins and two losses leave you with 0.988 of what you started, a small loss, even though every bet still has a positive expected value.
If every bet has positive expected value, why not bet more?
A shopkeeper with a healthy margin who stakes the whole shop on each delivery will, sooner or later, lose one delivery and the shop with it. The margin was real; the sizing ruined it. When you reinvest the results, what compounds is the typical outcome, not the average one, and big stakes make one loss cost more than one win earns. At a 40% stake a win takes you to 1.4 and a loss to 0.6; one of each leaves 0.84, a 16% loss on a pair that broke even in count.
So test a typical stretch: in five bets you expect three wins and two losses. Betting 20%, 1.2 cubed times 0.8 squared is 1.106, about 11% richer. Betting 40%, 1.4 cubed times 0.6 squared is 0.988, slightly poorer. Same coin, same edge, and only the stake changed.
Long-run growth per bet peaks at about 2.0% when you stake 20% of the bankroll and falls to slightly below zero at 40%, so doubling the right stake throws away all of the growth that a genuine 60% edge provides.
Where does 20% come from?
The relationship
$g(f) = p \ln(1+f) + q \ln(1-f), \qquad g'(f) = 0 \Rightarrow f^* = p - q = 0.6 - 0.4 = 0.2$
f the fraction of bankroll bet each time
p, q the chances of winning and losing, 0.6 and 0.4
g(f) the average log growth per bet, what compounds over time
What it says in wordsMaximise the average of the log of wealth, and with even-money odds the best stake equals your edge.
This is the Kelly criterion, named after John Kelly, who published it in 1956. Over 100 bets the typical bankroll at a 20% stake is about 7.5 times the start; at 40% it is about 0.78 times. Over-betting a real edge can destroy growth as surely as having no edge. In practice many people bet half the Kelly fraction: at 10% growth is 1.50% a bet, about three quarters of the maximum, with far smaller swings, which matters because real edges are estimated, not known.
Where candidates lose it
The first trap is answering 100% because expected value rises with the stake. It is a single-bet answer to a repeated-bet question, and it ends in ruin the first time you lose.
The second trap is answering 60%, confusing the win probability with the stake. Say the rule out loud, edge over odds, and show the three-wins-two-losses check so the interviewer sees why 40% fails.
What the interviewer asks next
What is the Kelly stake if the coin pays 2 to 1 and wins 40% of the time?
Why do practitioners bet half Kelly or less?
How does this sizing logic carry over to a portfolio manager's position limits?
095A sponsor bought a company at 8x EBITDA of Rs 100 crore, funded with Rs 480 crore of debt. It exits at 9x EBITDA of Rs 130 crore with Rs 300 crore of debt still outstanding. What is the MOIC, and how much of the equity gain came from EBITDA growth, multiple expansion and debt paydown?Deal mathsHardElite boutique IBPrivate equity
Try it first
Which source contributed least to the equity gain?
Show the worked solution
MOIC is 2.72x: equity grows from Rs 320 crore to Rs 870 crore. Of the Rs 550 crore gain, EBITDA growth contributes 240 (30 x 8x), multiple expansion 130 (one turn on 130), and debt paydown 180 (480 - 300). The three add exactly to 550. Over five years that would be an IRR of about 22%, though the question gives no holding period.
How do you get entry and exit equity quickly?
Equity is what is left of the business's value after the debt. Entry: 8 x 100 is an enterprise value of 800, less 480 of debt, so the sponsor put in 320. Exit: 9 x 130 is 1,170, less 300 of debt, leaving 870. MOIC is exit equity over entry equity, 870 / 320 = 2.72x. Say both enterprise values out loud before subtracting debt, so the interviewer can follow each step.
Why split the gain into three sources?
Think of a house you bought, renovated and sold. Your profit came partly from adding a room, partly from the neighbourhood getting more popular, and partly from paying down the mortgage. Only the first is a skill you can repeat on the next house. Attribution shows which part of the return came from the sponsor's own work and which came from the market, which is what investors in the fund want to know. EBITDA growth and debt paydown come from running the business; multiple expansion mostly comes from market conditions at exit.
Entry equity of Rs 320 crore grows by 240 from EBITDA growth, 130 from the multiple rising one turn and 180 from debt repaid, reaching Rs 870 crore, so more than three quarters of the gain came from operations and deleveraging rather than the multiple.
The relationship
$\Delta E = \underbrace{(130-100)\times 8}_{240} + \underbrace{(9-8)\times 130}_{130} + \underbrace{(480-300)}_{180} = 550$
(130 - 100) x 8 EBITDA growth valued at the entry multiple
(9 - 8) x 130 multiple expansion applied to exit EBITDA
480 - 300 debt repaid from the business's cash flow
What it says in wordsThe equity gain is growth at the old multiple, plus the new multiple's lift on the new EBITDA, plus debt repaid.
Say the convention. The growth and multiple pieces share a cross term: 30 of extra EBITDA times one extra turn, which is 30. Here it sits inside multiple expansion. Value growth at the exit multiple instead and growth becomes 270 while multiple expansion falls to 100. The total does not change, but the story does, so state which order you used.
Where candidates lose it
The common error is forgetting debt paydown, because it is not a change in value of the business. Candidates find 240 and 130, get 370, and cannot reconcile to the 550 gain.
The second loss is quoting an IRR without a holding period. MOIC needs no time; IRR does. Ask, or state the assumption, before giving a percentage.
What the interviewer asks next
What is the IRR if the holding period is three years instead of five?
What happens to the MOIC if the exit multiple falls back to 8x?
Which of the three sources would a fund's investors value most, and why?
098Two associates agree to meet at a coffee shop, each arriving at a random time between 12:00 and 1:00, independently. Each will wait 15 minutes for the other and then leave. What is the probability that they meet?ProbabilityHardBulge bracket IBSales and trading
Try it first
Pick before you draw anything.
Show the worked solution
7/16, about 43.8%. Put one arrival time on each axis of a square from 12:00 to 1:00. They meet when the two times are within 15 minutes, a band along the diagonal. They miss in two corner triangles with legs of 45 minutes, each covering (3/4)^2 / 2 = 9/32 of the square. So the chance of meeting is 1 - 9/16 = 7/16.
Why turn a timing problem into a square?
If you and a friend each pick a random seat in a cinema, every pair of choices is equally likely, so the chance of sitting together is the share of seat pairs that are neighbours. Arrival times work the same way, except they are continuous. When two independent times are uniformly likely, every point in the square of arrival pairs is equally likely, so a probability is simply an area. The question becomes: what fraction of the square has the two times within 15 minutes?
Plotting both arrival times on a square, the associates meet inside the diagonal band where their times are within 15 minutes, which covers 7/16 of the square, while the two grey corners where one arrives more than 15 minutes ahead cover 9/16.
Why is the corner easier to measure than the band?
The band has an awkward shape, but its complement is two right triangles. In the lower right corner, associate 1 arrives more than 15 minutes after associate 2, who has already left. That triangle runs from 12:15 to 1:00 along each side, 45 minutes, or 3/4 of the hour. Each triangle covers half of (3/4) squared, which is 9/32, so the two together cover 9/16 and the band covers the remaining 7/16.
The relationship
$P(\text{meet}) = 1 - (1 - w)^2 = 1 - \Big(\tfrac{3}{4}\Big)^2 = \tfrac{7}{16} \approx 43.75\%$
w the waiting time as a share of the hour, 15/60 = 1/4
(1 - w)^2 the two miss triangles together
What it says in wordsThe chance of meeting is one minus the area of the two corners where one person arrives too late.
The formula gives quick answers to the follow-ups. If each waits 30 minutes, w is a half and they meet with probability 1 - 1/4 = 75%. The limitation is the assumption of uniform arrivals: real people arrive close to the agreed time, which raises the chance of meeting, so the answer is for the model, not for actual associates.
Where candidates lose it
The common wrong answer is 25%, fifteen minutes out of sixty. It forgets that either person can arrive first and that the window shrinks near the ends of the hour.
The second loss is trying to integrate by cases out loud and getting lost. Draw the square, shade the corners, and the arithmetic is two lines.
What the interviewer asks next
How long must each wait for the chance of meeting to be 50%?
What if one associate waits 15 minutes and the other waits 30?
What if they must arrive between 12:00 and 2:00 instead?

E[N]	the expected number of rolls to see all six faces
6/(6 minus k)	the average wait for a new face once k faces are found

C	call price, Rs 10
P	put price, the unknown
S	share price today, Rs 100
K/(1+r)	the strike discounted one year at 5%, 95.24

a	the fastest walker, 1 minute
b	the second fastest, 2 minutes
c	the second slowest, 5 minutes

m	exit multiple of EBITDA, 10x
f	free cash flow as a share of EBITDA, 0.5
r	WACC, 10%
g	implied perpetual growth

Try it first

Why does breaking it into stages make it easy?

How do you add it up quickly in the room?

Where candidates lose it

What the interviewer asks next

Try it first

Why can you price the put without knowing volatility?

What does the formula say, and what if the put traded at Rs 7?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the escort plan feel right but lose two minutes?

How do you know 17 cannot be beaten?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a multiple contain a growth rate at all?

How do you judge whether 4.8% is believable?

Where candidates lose it

What the interviewer asks next

Try it first

If every bet has positive expected value, why not bet more?

Where does 20% come from?

Where candidates lose it

What the interviewer asks next

Try it first

How do you get entry and exit equity quickly?

Why split the gain into three sources?

Where candidates lose it

What the interviewer asks next

Try it first

Why turn a timing problem into a square?

Why is the corner easier to measure than the band?

Where candidates lose it

What the interviewer asks next