Investment Banking puzzles, solved step by step

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

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012You flip a fair coin until you get two heads in a row. What is the expected number of flips?Expected value and gamesHardBulge bracket IBConsulting style brainteasers
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Your first instinct: how many flips on average?
Show the worked solution
Six flips on average. Track where you stand: at the start, or one head up. From the start, one flip takes you to one head or leaves you at the start; from one head, a head finishes the game and a tail sends you back. Writing the expected flips from each state as an equation and solving gives 6 from the start and 4 from one head.
Why is the obvious answer of 4 wrong?
The guess of 4 comes from the one in four chance that two given flips are both heads. Think instead of climbing a slippery two-step ladder where any slip drops you to the ground, not one rung down. A tail after a head costs you the head you already had, so progress is lost and the wait is longer than the simple odds suggest. The clean way to handle lost progress is to give each position its own equation.
From the start a head moves you to one head and a tail leaves you at the start; from one head, a head finishes and a tail sends you back, so the two equations solve to 6 expected flips from the start and 4 from one head.
How do the equations work?
Call E the expected flips from the start and E(H) the expected flips once you hold one head. Every flip costs one. From the start, half the time you move to one head and half the time you are back where you began. From one head, half the time you finish and half the time a tail sends you to the start. Each equation reads: one flip, plus the average of the waits from wherever that flip leaves you. Substitute the second into the first and E = 1.5 + 0.75E, so E = 6 and E(H) = 4. This way of setting up the problem is called a Markov chainA process where what happens next depends only on the current state, not on how you got there..
The relationship
$E = 1 + \tfrac12 E_H + \tfrac12 E \qquad E_H = 1 + \tfrac12\cdot 0 + \tfrac12 E \quad\Rightarrow\quad E = 6,\; E_H = 4$
E expected flips from the start, with no head in hand
E_H expected flips when the last flip was a head
1 the flip you are about to make
What it says in wordsFrom each position, the expected wait is one flip plus the average wait from wherever that flip lands you.
How do you check 6 a second way?
Wait for the first head, which takes 2 flips on average. Flip once more: half the time it is a head and you are done; half the time it is a tail and you start from scratch. So E = 3 + E/2, which again gives 6. Two routes to the same number is the strongest answer you can give in the room. Simulated 200,000 times, the average comes out at 6.01. Then add the twist interviewers like: waiting for heads then tails takes only 4, because a failed attempt at it, a second head, still leaves you one head up.
Where candidates lose it
Most candidates answer 4, because the chance of two heads is a quarter, and stop. The interviewer is checking whether you notice that a tail after a head resets your progress.
The second loss is trying to write a single equation for the whole game and getting tangled. Name the states first, start and one head, and write one line for each; the algebra is then two lines long.
What the interviewer asks next
What is the expected number of flips to get heads followed by tails?
What about three heads in a row?
If the coin lands heads 60% of the time, what is the expected wait for two heads in a row?
026A game doubles your stake on heads and halves it on tails, so every round has positive expected value. After ten rounds on a Rs 1 lakh stake, what is the expected wealth and what is the most likely outcome?Expected value and gamesHardBulge bracket IBConsulting style brainteasers
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After ten rounds, what is the most likely amount in hand?
Show the worked solution
Expected wealth is Rs 9.31 lakh, but the most likely outcome is Rs 1 lakh, exactly where you started. Each round multiplies the stake by 1.25 on average, and 1.25 to the tenth is 9.31. Yet the wealth you actually hold is 2 to the power of heads minus tails, and the likeliest split is five and five, which multiplies to one. The median player goes nowhere; the mean is carried by a handful of lucky runs.
Why is the average so far from the typical result?
Ten friends each put Rs 1 lakh into this game. Most finish near where they began, a few lose most of their money, and one, with a long run of heads, finishes with hundreds of lakh. Add it all up and divide by ten and the average looks wonderful; ask the friend in the middle how it went and the answer is that nothing happened. Gains and losses compound, and a doubling followed by a halving lands exactly back where you started, so the mean grows 25% a round while the median stays flat. The arithmetic average of 2 and 0.5 is 1.25; their geometric average, the square root of 2 times 0.5, is exactly 1, and compounding follows the geometric one.
The mean rises 25% a round to Rs 9.31 lakh after ten rounds while the median stays at Rs 1 lakh, because the likeliest outcome, five heads and five tails, multiplies to one and the games with eight or more heads, 5.5% of the total, supply 68% of the mean.
How do you get both numbers in two lines?
For the mean, use the fact that the expectation of a product of independent rounds is the product of the expectations: 1.25 to the tenth, 9.31. For the typical outcome, count heads. After ten flips the wealth is Rs 1 lakh times 2 to the power of heads minus tails, which is 2 to the power of 2H minus 10. Five heads is the single likeliest count, 24.6% of games, and it gives a multiplier of exactly one; it is also the median, because the outcomes sit symmetrically on either side of it. 37.7% of players finish below Rs 1 lakh and the same share above, and you can say all of that without touching the mean.
The relationship
$E[W_{10}] = \left(\tfrac12\cdot 2 + \tfrac12\cdot\tfrac12\right)^{10} = 1.25^{10} = 9.31 \qquad W_{10} = 2^{\,H-T} = 2^{\,2H-10}$
1.25 the expected multiplier of a single round, in lakh per lakh staked
H, T the number of heads and tails in the ten flips
2H - 10 heads minus tails, the net number of doublings
What it says in wordsThe mean compounds the average multiplier; the wealth you hold compounds the count of heads, and the likeliest count leaves you where you started.
Where does the Rs 9.31 lakh come from, then?
From the tail. A run of ten heads happens once in 1,024 games and turns Rs 1 lakh into Rs 1,024 lakh, which on its own adds a full Rs 1 lakh to the mean. The three best outcomes, eight or more heads, occur in 5.5% of games and contribute Rs 6.31 lakh of the Rs 9.31 lakh average, about 68% of it. Nothing is wrong with the expected value; it is the wrong statistic for a question about what will probably happen to you. The right one is the geometric meanThe growth rate that compounding actually delivers: the nth root of the product of n multipliers. It is never above the arithmetic mean. return, which here is zero.
The lesson a desk draws is about sizing. Bet only half your stake each round and the multipliers become 1.5 and 0.75, whose geometric mean is 1.0607: the typical player now grows about 6.1% a round and finishes ten rounds near Rs 1.80 lakh, even though the expected value per round has fallen from 1.25 to 1.125. Half is the fraction that maximises the typical growth rate in this game. The same arithmetic sits under volatility drag in fund returns: a strategy that gains 50% and then loses a third has a flattering average and nothing to show for it.
Where candidates lose it
Most candidates say the expected value and stop, or say the game must be good because every round is positive on average. The interviewer is waiting for you to notice that a doubling and a halving cancel exactly, so the typical outcome is no change.
The second loss is muddling the median with the mean when pressed. Say the mode and the median are both Rs 1 lakh, give the 24.6% chance of the exact five-five split, and explain that the mean lives in the tail.
What the interviewer asks next
What fraction of your stake should you risk each round to maximise your typical long-run growth, and why?
What is the probability of finishing with more than Rs 1 lakh after ten rounds?
A fund gains 50% one year and loses a third the next. What are its average and its compound returns?
038A distressed company's assets will be worth 160 or 40 next year with equal probability, and it owes 100 of debt due then. With zero interest rates and risk-neutral pricing, what are the equity and the debt worth today, and why is the equity not worthless?Expected value and gamesHardBulge bracket IBConsulting style brainteasers
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What is the equity worth today?
Show the worked solution
The equity is worth 30 and the debt 70. If assets reach 160, lenders get 100 and shareholders 60; if assets fall to 40, lenders take all 40 and shareholders get nothing, but never less than nothing. Half of 60 is 30 for equity; half of 100 plus half of 40 is 70 for debt. Together they equal the 100 the assets are worth. Limited liability makes equity a call option on the assets.
Why is the equity worth anything when assets only cover the debt?
Think of a student who borrows to start a food stall and can walk away from the loan if the stall fails. In a good year the student keeps everything above the loan; in a bad year the lender keeps the stall and the student loses nothing more. Because shareholders can lose at most what they put in but keep every rupee above the debt, equity is a call optionThe right, but not the obligation, to buy an asset at a fixed price. It pays the amount by which the asset ends above that price, or nothing. on the assets with a strike price equal to the debt. An option has value even when it sits exactly at the money, which is where this company stands today.
Debt pays the asset value up to 100 and equity pays everything above 100, so at assets of 40 debt gets 40 and equity 0, at 160 debt gets 100 and equity 60, and today equity is worth 30 and debt 70.
How do you price the two claims?
With zero interest rates and risk-neutral pricing, each claim is worth its average payoff. Equity pays 60 or 0 and is worth 30; debt pays 100 or 40 and is worth 70; the two add back to the asset value of 100. The debt trades at 70 for a promise of 100, a yield of about 42.9%, which is how a market prices distress.
The relationship
$E = \tfrac12\max(160-100,0) + \tfrac12\max(40-100,0) = 30 \qquad D = 100 - 30 = 70$
max(V - 100, 0) what shareholders get: assets above the debt, never below zero
100 today's asset value, the average of 160 and 40
What it says in wordsEquity is the average of its floored payoffs; debt is whatever is left of the assets.
What happens if the company takes more risk?
Widen the outcomes to 190 or 10, with the same average of 100. Equity now pays 90 or 0 and is worth 45; debt pays 100 or 10 and is worth 55: extra risk moves 15 of value from lenders to shareholders without the company being worth a rupee more. That is why lenders to weak companies write covenants against new risky projects, and why restructuring bankers ask who gains from each option the board is weighing.
Where candidates lose it
Candidates subtract the debt from today's assets, get zero, and call the equity worthless. That treats the equity as if it had to settle today and ignores that shareholders keep the upside while being protected from the downside.
The second slip is pricing the debt at its face value of 100. Lenders carry the bad state, so their claim is worth 70, and the market shows that as a high yield.
What the interviewer asks next
The outcomes become 190 or 10. What are equity and debt worth now?
Why might shareholders of this company vote for a risky project with a negative expected value?
How does a positive interest rate change the answer?
053You roll a fair die and are paid its face value in lakh rupees. After seeing the first roll you may roll once more and take the second result instead, whatever it turns out to be. What is the game worth, and what is your stopping rule?Expected value and gamesCoreBulge bracket IBConsulting style brainteasers
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On which first rolls do you roll again?
Show the worked solution
The game is worth Rs 4.25 lakh, and you roll again on a 1, 2 or 3. A single roll is worth 3.5 lakh on average, so the chance to roll again is worth exactly that. Keep any first roll above 3.5, which means 4, 5 or 6. Half the time you keep, averaging 5 lakh; half the time you roll again for 3.5. Half of 5 plus half of 3.5 is 4.25 lakh. The option to re-roll adds 0.75 lakh over a single roll.
Why is the rule to compare against 3.5?
Think of boarding a train with one visible seat by the door and the rest of the carriage out of sight. You walk on only if the seat you can see is worse than the average seat down the carriage, because once you walk you take whatever is there. Work backwards: the last decision has no choices left, so its value is the plain average, 3.5 lakh, and any earlier roll is kept only if it beats that. A 4, 5 or 6 beats 3.5; a 1, 2 or 3 does not; no face equals it, so the rule has no ties. The number you are comparing with is the expected valueThe average outcome if the same gamble were repeated many times: each result weighted by its probability. of rolling again.
A first roll of 1, 2 or 3 falls below the 3.5 that a second roll is worth, so you roll again, while a 4, 5 or 6 is kept, and the kept faces contribute 2.5 and the re-rolls 1.75 for a game worth 4.25 lakh.
How does the 4.25 come together?
Three faces out of six are kept and contribute (4 + 5 + 6) over 6, which is 2.5 lakh; three are re-rolled and contribute three sixths of 3.5, which is 1.75; the total is 4.25 lakh. Check the rule by moving it. Keep the 3 as well: (3 + 4 + 5 + 6) over 6 plus two sixths of 3.5 is 4.17. Re-roll the 4 too: (5 + 6) over 6 plus four sixths of 3.5 is 4.17. Both sit below 4.25, about Rs 8,333 a game, so 3.5 is the only bar that gives the full value.
The relationship
$V = \frac{4 + 5 + 6}{6} + \frac{3}{6} \times 3.5 = 2.5 + 1.75 = 4.25$
(4 + 5 + 6)/6 the kept faces, each one in six
3/6 the chance the first roll is a 1, 2 or 3 and you roll again
3.5 what the second roll is worth on average
What it says in wordsAdd what you keep, weighted by its chance, to what you get from rolling again, weighted by its chance.
What happens with a third roll, and why do interviewers ask?
Allow two re-rolls. The game from the second roll onward is now worth 4.25, so on the first roll you keep only a 5 or 6: (5 + 6) over 6 plus four sixths of 4.25 is 4.67 lakh. Every extra chance to try again raises both the value of the game and the bar for stopping, and that is the whole logic of an option: the right to swap a known result for an uncertain one is worth paying for only when the known result is poor. The same reasoning prices walking away from a signed deal or waiting for a better bid. The limit: this assumes you care only about the average. Someone who badly needs 3 lakh might keep a 3, and that is a preference, not an error.
Where candidates lose it
The common miss is a rule based on feel, such as re-rolling anything below 5 because 5 and 6 feel like wins. That throws away a 4 that beats the 3.5 you get by rolling again, and costs about Rs 8,333 of value per game.
The second loss is valuing the game at 3.5 lakh because it is still just a die. The option to re-roll is worth 0.75 lakh here, and pricing an option at zero is the error the interviewer is listening for.
What the interviewer asks next
You may re-roll twice. What is the game worth and what is the rule on the first roll?
The re-roll costs Rs 50,000. Do you still take it, and when?
How does this connect to the decision to walk away from a signed deal?
067A friend offers a bet on one roll of a fair die: you win Rs 10,000 if it shows a six and you pay Rs 2,000 if it shows anything else. Should you take it?Expected value and gamesWarm upBulge bracket IBConsulting style brainteasers
Try it first
What is the expected value of one roll?
Show the worked solution
The bet is exactly fair: its expected value is zero. A six comes up one time in six, so the win is worth 1/6 x Rs 10,000 = Rs 1,667. The other five faces each cost Rs 2,000, worth 5/6 x Rs 2,000 = Rs 1,667. They cancel. Whether to take it is then a question about risk: five rolls in six you hand over Rs 2,000, and the swing on one roll is about Rs 4,472 either way.
How do you weigh a big rare win against a small frequent loss?
Imagine playing the game six hundred times. You would expect a six about a hundred times, collecting Rs 10 lakh, and the other five hundred rolls would cost Rs 10 lakh. Expected value is each outcome multiplied by its probability, added up, and here the two sides come to Rs 1,667 each, so they cancel exactly. Say the arithmetic as fractions of 6 so the interviewer can follow: 10,000 over 6 against 5 x 2,000 over 6, and 10,000 equals 10,000.
Weighted by its one-in-six chance the Rs 10,000 win is worth Rs 1,667, and weighted by its five-in-six chance the Rs 2,000 loss is worth the same Rs 1,667, so the expected value is zero and the bet is fair.
If the maths is a tie, what decides it?
Risk appetite, and the question is really about whether you can say so. A fair bet with an uneven shape is not neutral to everyone: you lose on five rolls in six, and the spread of outcomes, about Rs 4,472 on a single roll, is large next to the stake. Someone for whom Rs 2,000 is lunch money might play for fun; someone for whom it is a week's groceries should not, because the most likely result of a single roll is a loss. Interviewers want the number, then a sentence that separates expected value from the experience of one roll.
The relationship
$E[X] = \tfrac{1}{6}(10{,}000) - \tfrac{5}{6}(2{,}000) = 1,667 - 1,667 = 0$
E[X] the expected value of one roll, in rupees
1/6 the chance of a six
5/6 the chance of any other face
What it says in wordsMultiply each payoff by its chance and add; a result of zero means the bet is fair.
How would you make the bet worth taking?
Move one number and watch the sign. Break-even on the win is 5 x Rs 2,000 = Rs 10,000, so any win above Rs 10,000 makes it positive, and at Rs 12,000 the expected value is Rs 333 a roll. Alternatively, ask to play many times: over 600 rolls the expected result is still zero, but the swing shrinks relative to the stake, which is why a casino is happy with a tiny edge and a huge number of hands. Offering that framing shows you understand why desks care about both edge and variance.
Where candidates lose it
The fast wrong answer is to take the bet because Rs 10,000 is bigger than Rs 2,000. The interviewer is checking that you weigh each payoff by its chance before you compare.
The second loss is stopping at zero. A fair bet is still a decision, and saying one sentence about the shape of the outcomes, five losses for every win, is what separates a calculator from a candidate.
What the interviewer asks next
What win on a six would make the expected value Rs 500 a roll?
Now you roll twice and win if either roll is a six. Is the bet fair?
Why might a trader take a fair bet with a tiny edge thousands of times but refuse it once?
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?
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?

E	expected flips from the start, with no head in hand
E_H	expected flips when the last flip was a head
1	the flip you are about to make

1.25	the expected multiplier of a single round, in lakh per lakh staked
H, T	the number of heads and tails in the ten flips
2H - 10	heads minus tails, the net number of doublings

max(V - 100, 0)	what shareholders get: assets above the debt, never below zero
100	today's asset value, the average of 160 and 40

(4 + 5 + 6)/6	the kept faces, each one in six
3/6	the chance the first roll is a 1, 2 or 3 and you roll again
3.5	what the second roll is worth on average

Try it first

Why is the obvious answer of 4 wrong?

How do the equations work?

How do you check 6 a second way?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the average so far from the typical result?

How do you get both numbers in two lines?

Where does the Rs 9.31 lakh come from, then?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the equity worth anything when assets only cover the debt?

How do you price the two claims?

What happens if the company takes more risk?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the rule to compare against 3.5?

How does the 4.25 come together?

What happens with a third roll, and why do interviewers ask?

Where candidates lose it

What the interviewer asks next

Try it first

How do you weigh a big rare win against a small frequent loss?

If the maths is a tie, what decides it?

How would you make the bet worth taking?

Where candidates lose it

What the interviewer asks next

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

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