Investment Banking puzzles, solved step by step

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.

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..

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.

7 races. Race five heats of five, then race the five heat winners: the winner of race 6 is the fastest overall. Now strike out every horse that has three horses known to be faster. Only five can still be second or third: the second and third from the winner's heat, the first two from the runner-up's heat and the winner of the third-placed heat. Race them; the top two are second and third overall.

What does each race actually tell you?

Without a stopwatch, a race tells you the order only among the horses in it. Think of five classrooms each running their own sprint: you know the fastest in each room, but nothing about how one room's second best compares with another room's winner. Every horse must race at least once, so five heats are unavoidable, and they give five separate rankings that a sixth race between the heat winners stitches together. Name the heats A to E in the order their winners finished race 6, so A1 is the fastest horse of all.

Which horses can you strike out after race 6?

Any horse with three horses known to be faster cannot finish in the top three. Heats D and E go entirely, because D1 and E1 already finished behind A1, B1 and C1, and everything in those heats is slower still. In heat C only C1 survives: C2 trails C1, B1 and A1. In heat B, B1 and B2 survive, but B3 trails B1, B2 and A1. In heat A, A2 and A3 survive and A4 trails A1, A2 and A3. That leaves exactly five: A2, A3, B1, B2 and C1.

Race 7 puts those five on the track, and its first two finishers are second and third overall. A1 sits out, because it is already known to be the fastest. Could six races ever be enough? No: the five heats alone cannot name the fastest horse, and the race that does name it leaves five horses still unranked for second and third. That is the reasoning to say out loud, because it shows the 7 is a minimum rather than just a method that works.

Two thirds. You are not choosing among cards; you are looking at a face. There are three red faces you could be seeing, all equally likely: two belong to the red-red card and one to the red-white card. In two of the three cases the hidden side is red. The tempting answer of one half treats the two red-capable cards as equally likely and ignores that the red-red card shows red twice as often.

Why is one half so tempting, and where does it go wrong?

Imagine two families at a party: one with two daughters, one with a daughter and a son. If you meet one of the girls at random, she is twice as likely to come from the two-daughter family, because that family brought two girls. The red-red card is twice as likely to be the one showing red, because it has two red faces to show. Treating the two remaining cards as equally likely throws that information away, which is exactly what the question is built to catch.

How do you prove it with Bayes' rule?

Write it as a conditional probability using Bayes' ruleA formula for updating a probability after seeing evidence: the chance of the evidence given the cause, times the prior chance of the cause, divided by the overall chance of the evidence.. The chance of drawing the red-red card is one third, and if you did, you are certain to see red. The chance of seeing red at all is the share of red faces, three of six, one half. So the chance the card is red-red, given a red face, is one third times one, divided by one half: two thirds.

How do you check it without any formula?

Imagine drawing 600 times, each card about 200 times. The red-red card shows red all 200 times, the red-white card shows red about 100 times, and the white card never does. Of the 300 red sightings, 200 have red on the back. Counting outcomes in a large imagined sample is a reliable check whenever a conditional probability feels slippery. The puzzle is a version of Bertrand's box paradox, and the same reasoning sits under the Monty Hall problem.

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.

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.

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.

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.

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.

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.

Under the fixed ratio, target holders receive Rs 90 of value per share instead of Rs 100. They still get 0.5 acquirer shares, but each is now worth Rs 180. Under a fixed price they would still receive Rs 100, and the acquirer would issue 100 over 180, about 0.556 shares per target share, 11.1% more than planned. A fixed ratio shares the price risk with the target; a fixed price keeps it with the acquirer and its shareholders.

What does each structure actually fix?

Think of agreeing to swap your car for 50 grams of gold, rather than for Rs 5 lakh paid in gold. If the gold price falls before the swap, the first deal still hands you 50 grams, now worth less; the second hands you more grams so that you still get Rs 5 lakh. A fixed exchange ratioThe number of acquirer shares paid for each target share in a stock deal. fixes the number of shares the target receives; a fixed price fixes their rupee value, and the share count moves instead. Whichever number floats carries the risk of the acquirer's share price between signing and closing.

How do the two structures play out across the whole deal?

Assume the target has 10 crore shares and the acquirer 50 crore. Fixed ratio: target holders receive Rs 900 crore of value instead of Rs 1,000 crore, and own 5 crore of 55 crore shares, 9.1% of the combined company. Fixed price: they still receive Rs 1,000 crore, so the acquirer issues 5.56 crore shares, 11.1% more than planned, and the target's holders own 10.0%. The acquirer's existing shareholders give up more of the company to pay the same rupees.

Which side wants which, and how do bankers split the difference?

A target worried that the acquirer's shares will fall wants a fixed price; an acquirer worried about issuing too many shares wants a fixed ratio. A collar is the usual compromise: in one common form the ratio is fixed while the acquirer's price stays inside a band and the value is fixed outside it, so each side carries the risk only up to a point. Interviewers ask this to see whether you think about who carries risk between signing and closing, not just the headline value.

Number the bottles 1 to 1,000, write each number in 10-digit binary, and have tester k sip from every bottle whose kth digit is a 1. After 24 hours, write a 1 for each tester who is ill and a 0 for each who is not. That 10-digit binary number is the poisoned bottle. It works because 10 yes-or-no results can tell apart 2 to the 10th, 1,024 cases, which is more than 1,000.

Why does giving each tester a separate batch fail?

Picture 10 friends each tasting their own 100 bottles. One friend falls ill and you know the poison is among their 100, but not which one, and the day is over. When testers taste separate batches, each result answers one question about one batch, so 10 testers can single out at most 10 groups, not 1,000 bottles. The fix is to let testers overlap, so that every bottle is sipped by its own unique combination of testers.

How does the binary code name the bottle?

Give each tester a place value: tester 1 is worth 512, tester 2 is worth 256, and so on down to tester 10, worth 1. Bottle 357 equals 256 + 64 + 32 + 4 + 1, so testers 2, 4, 5, 8 and 10 sip from it. If exactly those five fall ill, adding their place values gives back 357, and no other bottle has the same set of tasters. Every bottle number is a different set of testers, so the pattern of illness points to one bottle only.

What is the general lesson?

Each tester is one yes-or-no question asked of every bottle at once. Ten yes-or-no answers can tell apart 1,024 possibilities, so the right design makes each answer carry one binary digit of the bottle's number. The same idea finds one faulty item among many with very few tests, and it is why 2,000 bottles need only one more tester. With two days, each tester has three outcomes, ill on day one, ill on day two, or never, and 7 testers would be enough.

About 28%. Take 1,000 companies. Twenty commit fraud and the test flags 19 of them. The other 980 are clean, and the test wrongly flags 5% of them, which is 49. So 68 companies are flagged and only 19 are frauds: 19 over 68 is 27.9%. The test is accurate, but fraud is rare, so a small error rate on the large clean group produces more false flags than true ones.

Why is the answer not 95%?

Think of a smoke alarm that sounds for 95% of real fires and stays quiet 95% of the time there is no fire. In a building where fires are rare, most alarms you hear are burnt toast, because the quiet days vastly outnumber the fires and 5% of them still ring. A test's accuracy tells you how it behaves when the truth is known; the question asks the reverse, how likely the truth is given the test, and the base rateHow common the thing is before any test is run. Here, 2% of companies commit fraud. sits between the two. With 2% frauds, a 5% error on the 98% of clean companies creates 49 false flags for every 19 true ones.

How do you work it with whole numbers?

Start with 1,000 companies and walk them down the tree. Frauds: 2% of 1,000 is 20, and 95% of those, 19, are flagged. Clean: 980, and 5% of those, 49, are wrongly flagged. 19 true flags and 49 false flags make 68 flagged companies, so the chance that a flagged company is a fraud is 19 over 68, about 28%. Working in counts rather than percentages is faster in the room and harder to get wrong, and it shows the interviewer where every number came from.

What moves the number, and what does that mean for a screen?

Two things. The base rate: if 10% of the sample committed fraud, the same test would make a flag 68% reliable. The false-flag rate: cutting it from 5% to 1% lifts the answer to 66% even at a 2% base rate. A screen for a rare event is a filter that earns a closer look, not a verdict, and the second look does the real work: run an independent test of the same accuracy on the 68 flagged companies and a double flag is 88% likely to be a fraud. That is how diligence red flags, transaction monitoring and credit early-warning lists are meant to be read. The working assumes the two tests err independently; if they look at the same ratios, the second adds less than the arithmetic suggests.

204 squares. A square of side k can start in (9 minus k) columns and (9 minus k) rows, so there are 64 squares of size 1, 49 of size 2, 36 of size 3, and so on down to a single 8 x 8. The total is the sum of the first eight square numbers, 1 + 4 + 9 + ... + 64, which is 204.

How do you count one size without listing every square?

Picture sliding a 3 x 3 photo frame across the board. Its left edge can sit on column 1, 2, 3, 4, 5 or 6; on column 7 it would hang off the side. That is 6 positions across and, by the same logic, 6 down. Count where a square's top-left corner can sit, not the squares themselves: a k x k square has (9 minus k) choices in each direction, so (9 minus k) squared positions. For k = 3 that is 36, for k = 7 it is 4, and the whole board is the single 8 x 8.

How do you add 64 + 49 + ... + 1 quickly?

Add from the top and say the running total: 64 and 49 is 113, plus 36 is 149, plus 25 is 174, plus 16 is 190, then 9, 4 and 1 take it to 204. If you know the formula for a sum of squares, use it as a check rather than a starting point. The interviewer scores the moment you see that each size contributes a square number, not the formula you remember.

Why is 1,296 the wrong answer here?

1,296 is the number of rectangles of every shape. A rectangle is fixed by choosing two of the 9 vertical grid lines and two of the 9 horizontal ones, 36 ways each, and 36 x 36 is 1,296. Squares are the rectangles whose two sides are equal, which is why the count falls from 1,296 to 204. Offering that link unprompted shows you understand the counting rather than a memorised trick.

A deferred tax liability of Rs 2.5 crore at year 1, Rs 12.5 crore at year 5, and it unwinds to zero by year 10. Book depreciation is Rs 10 crore a year; tax depreciation is Rs 20 crore for five years then nothing. In years 1 to 5 taxable profit is Rs 10 crore below book profit, so the company pays Rs 2.5 crore less tax than it charges, and the difference accrues as a liability. From year 6 the position reverses and the liability drains at Rs 2.5 crore a year.

Why is there a liability when the company has paid less tax?

Think of a shopkeeper allowed to pay this year's electricity bill next year. Cash looks better today, but the bill has not vanished; it sits as something owed. Faster tax depreciation lets the company deduct the asset's cost sooner, so it pays less tax now and more later, and the accounts record the later tax as a liability the moment the saving is taken. The income statement charges tax on book profit, Rs 2.5 crore a year more than the cash actually paid in years 1 to 5, and that extra charge is what builds the balance.

How do you get the balance without a schedule?

Compare the two values of the asset. The deferred tax liability is the tax rate times the gap between the asset's book value and its tax value, because that gap is profit the tax authority has not yet taxed. At year 1 the book value is Rs 90 crore and the tax value Rs 80 crore, a gap of 10, and 25% of 10 is Rs 2.5 crore. At year 5 the book value is Rs 50 crore and the tax value is zero, a gap of 50, so the liability is Rs 12.5 crore. At year 10 both are zero and so is the liability.

Why does a banker care about a timing difference?

Because it is cash. In years 1 to 5 the company keeps Rs 2.5 crore a year more cash than its income statement suggests, which shows up as an increase in deferred tax liabilities in operating cash flow, and in years 6 to 10 the same amount drains out. A model that uses book tax on EBIT misses both legs, and a buyer of a company with a large and growing deferred tax liability should ask whether it is still growing because the company keeps buying assets, or about to reverse. Say the limitation: the balance reverses only if the company stops adding new assets, and tax rules on depreciation differ by country and change, so confirm the current rates before building this into a model.