Equity Research puzzles, solved step by step

Flip twice: call heads on HT, tails on TH, and flip two more times on HH or TT. HT and TH each have probability 1/3 x 2/3 = 2/9, so they are equally likely whatever the bias. A pair is kept with chance 4/9, so on average you need 9/4 pairs, which is 4.5 flips per fair result.

Why are HT and TH always equally likely?

Think of two friends who each toss the same lopsided coin once. The chance the first gets heads and the second tails is the same as the chance the first gets tails and the second heads, because the same two numbers are multiplied, just in a different order. For any bias p, HT and TH both have probability p times (1 minus p), so the order of two different outcomes is a perfectly fair coin. This is known as the von Neumann trick.

How do you get the 4.5 flips, and can you do better?

Each pair succeeds with chance 4/9. The number of pairs until the first success averages one over the success chance, 9/4, and each pair is two flips. So the fair coin costs 4.5 biased flips on average, a steep price for throwing away five pairs in every nine.

The interviewer's follow-up is efficiency. The discarded pairs still carry information: HH and TT are themselves a biased coin, and you can run the same trick on them, and on the pattern of which pairs matched. Iterating like this recovers most of the waste. The ceiling is set by how unpredictable the coin is: about 0.918 fair bits per biased flip here, or roughly 1.09 flips per fair result at best. Naming that limit is what separates a good answer from a complete one.

About 9%: duration of 6 times a 1.5 point rise in yield. For an 8-year, 7% coupon bond at par, the true fall is 8.5%, a little less than the straight-line estimate because the price curve bends. For the equity analyst, the market now charges the company about 1.5 points more for new debt, so the cost of debt in WACC should use the new yield, not the old coupon.

How does duration turn a spread move into a price move?

Think of a fixed-rate deposit you cannot break. When new deposits start paying more, yours is worth less to anyone who might buy it from you, and the longer it has left to run, the bigger the discount. Modified duration is that sensitivity in one number: the approximate percentage price change for each 1 point change in yield. A spread widening raises the bond's yield by the same amount, so 6 x 1.5 = 9%.

Why is the true fall a little smaller than 9%?

The price-yield curve is convex: it flattens as yields rise. Duration draws a straight tangent at today's yield, so for large moves it overstates price falls and understates price rises. Here the tangent says 91.0 and the bond is actually worth 91.5. For small moves the gap is negligible; at 150 basis points it is about 0.5 points, worth one sentence.

Now the equity view. The cost of debt in a WACC is what the company would pay to borrow today, so a 150 basis point widening raises it by about 1.5 points, whatever the coupon on existing bonds. With debt at 30% of capital and a 25% tax rate, that alone adds about 0.34 points to WACC. The larger message is the signal: credit investors are pricing more risk, and equity sits below the bonds, so the analyst should ask what the bond market has seen.

The probability is 0.875. The sum of two uniform numbers has a triangular density, rising from 0 to a peak at 1 and falling back to 0 at 2. The region above 1.5 is a small triangle with base 0.5 and height 0.5, area 0.125. So the chance the sum is below 1.5 is 1 minus 0.125, which is 0.875.

Why is the sum not spread evenly from 0 to 2?

Think of rolling two dice: a total of 7 turns up far more often than 2 or 12, because many pairs make 7 and only one pair makes 2. Continuous numbers behave the same way. Adding two flat distributions gives a triangle, because middle values can be reached in many ways and extreme values in very few. Combining two densities like this is called convolutionThe operation that gives the distribution of a sum of independent random quantities by sliding one density across the other and adding up the overlap..

How do you get 0.875 quickly?

Use geometry rather than integrals. Every pair of draws is a point in the unit square, all points equally likely, so probability is area. The line x + y = 1.5 cuts off a corner triangle whose legs are each 0.5 long, so its area is half of 0.5 times 0.5, which is 0.125. Everything else, 0.875, is where the sum is below 1.5. The triangle density on the right shows the same tail from the other side.

The general rule for a threshold t between 1 and 2 is 1 minus half of (2 minus t) squared; for t below 1 it is half of t squared. The research version is diversification of two independent return sources: the average of two is less likely to be extreme than either alone, which is the whole argument for combining them.

30% (130 to 91), then 25% (120 to 90), then 10% (140 to 126). Walk the series keeping the highest close so far. Each time the price falls below that running peak, track the lowest point until a new peak is set. 120 falls to 90 before 130 sets a new high; 130 falls to 91 before 140; 140 falls to 126 and has not recovered. Divide each fall by its own peak.

Why measure from the running peak rather than the start?

Picture a savings account you check every evening. If it once showed Rs 1,30,000 and now shows Rs 91,000, the loss you feel is from the best balance you ever saw, not from what you first deposited. A drawdown is the fall from the highest value reached so far to the lowest value before that high is beaten, divided by the high. The start of the series only matters until the price first rises above it.

How do you find them in one pass?

Carry two numbers as you read left to right: the running peak and the lowest price since that peak. When the price sets a new high, close the open drawdown, record it, and reset both numbers to the new high. At the end, close whatever is still open, then sort by size and return the first n. That is the programming version of the question, and it runs in a single pass over the prices.

State the edge cases before the interviewer does. An open drawdown at the end of the series counts, even though it has not recovered. Two drops inside one drawdown episode, such as 130 to 91 and a later dip to 93 before a new high, are one drawdown, not two.

6 tosses. Call S the expected tosses from the start and H the expected tosses once you have one head. From the start, one toss either gives a head (move to H) or a tail (stay at S): S = 1 + H/2 + S/2. From one head, one toss either finishes or sends you back: H = 1 + S/2. Substituting, S = 6 and H = 4.

Why is the answer not four?

Think of climbing two steps on a slippery stair where any slip sends you back to the bottom. Reaching the second step is not just two climbs; every slip costs you what you had gained. The chance of heads-heads on a single pair is one in four, but a tail after a head wipes out your progress, so the expected wait is longer than four. The clean way to count the cost of those slips is to name each position you can be in.

How do you check 6 another way?

Count sequences directly. The first HH lands on toss n in exactly F(n-1) of the 2ⁿ equally likely sequences, where F is the Fibonacci sequence: 1 way on toss 2, 1 on toss 3, 2 on toss 4, 3 on toss 5. Summing n times F(n-1) over 2ⁿ across all n gives 6.0000, the same 6. The same series shows the spread: you finish within six tosses 67% of the time, and the other third of runs can go on for a long while, which is what pulls the average up to 6.

Roughly 607 million 5G phones worth about US$ 351 billion, on stated assumptions. Assume 4 billion smartphones in use, replaced every 3 years, so about 1.33 billion are sold in 2022. Assume all premium phones, 60% of mid-range and 15% of entry phones are 5G, a blended 45.5%, which gives 607 million units. At a blended price near $579, the value is about US$ 351 billion.

Where do you start: people, phones or networks?

Think of estimating how many pairs of school shoes a town buys. You do not count shops; you count children and ask how often a pair wears out. Annual phone sales are the phones in use divided by how many years a phone lasts, so start from the installed base and the replacement cycle, then split the year's sales by price band. Say every number as an assumption: an installed base of 4 billion and a 3 year cycle are round figures for the method, to be checked against an industry tracker before they go into a note.

How do you show the interviewer you know where the estimate is weak?

Run the sensitivity on the widest step. A cycle of 2.5 years gives 728 million 5G units and 3.5 years gives 520 million, a swing larger than any plausible error in the price assumptions. Then sanity check the output: 45% of phones sold being 5G in 2022 should sit sensibly against what you know about network rollouts, and a blended price near $579 says premium phones carry most of the value. The limitation: first-time buyers are folded into the cycle, which slightly understates sales in markets still adding users.

About 6.9% a month, and a 95% one-month value at risk of about Rs 1.13 crore. Variance adds across independent days, so volatility scales with the square root of time: 1.5% x the square root of 21, which is 4.58, gives 6.87%. The one-tailed 95% point of a normal distribution is 1.645 standard deviations, so a bad month costs 1.645 x 6.87% = 11.3% of Rs 10 crore, about Rs 1.13 crore.

Why does volatility grow with the square root of time?

Think of someone taking random steps left or right along a lane. After 100 steps they are not 100 steps from where they began; many steps cancelled, and the typical distance is about 10 steps, the square root of 100. Daily returns behave the same way when one day does not predict the next. Variances add across days, so the standard deviation, which is the square root of the variance, grows with the square root of the number of days.

How do you turn a volatility into a rupee value at risk?

Value at riskA loss threshold that a position should exceed only with a stated small probability over a stated period, such as 5% over one month. answers one question: how much would a bad month cost, where bad means the worst month in twenty. Under a normal distribution 5% of outcomes fall below the mean minus 1.645 standard deviations, so the 95% one-month VaR is 1.645 x 6.87% = 11.3% of the position. On Rs 10 crore that is about Rs 1.13 crore. Assume a zero expected return for the month; over one month the drift is small against the volatility.

What does the number leave out?

Two things, and saying them separates an analyst from a calculator. The square root rule needs independent days; when bad days cluster, as they do in a sell-off, real monthly risk is higher than 6.9%. And VaR marks the edge of the tail, not what sits inside it: it says nothing about how bad the worst 5% of months can get. Real returns have fatter tails than the normal curve, so a one-in-twenty month can cost well over Rs 1.13 crore.

Pull again straight away: you survive with probability 3/4, against 2/3 after a fresh spin. The four empty chambers sit together in a row and the cylinder advances one chamber per pull. The click you heard came from one of those four, each equally likely. Only one of them, the last empty chamber before the bullets, is followed by a bullet, so the chance of dying without a spin is 1/4. A fresh spin puts you back at two bullets in six.

Why does the first click carry information?

Picture a street with four dark houses in a row, then two lit ones, laid out in a circle. You are told you are standing at a dark house and that you will walk to the next door. Three of the four dark houses have another dark house next door. The click tells you that you are somewhere in the run of four empty chambers, and because the bullets sit together, most of that run is followed by another empty chamber. A fresh spin throws that information away.

Would the answer change if the bullets were not next to each other?

Yes, and saying so shows you understand the mechanism rather than the trick. Put the bullets in chambers 1 and 4. The empty chambers are 2, 3, 5 and 6, and two of them, 3 and 6, sit in front of a bullet. With the bullets apart, pulling straight away survives only 50% of the time, so a fresh spin at 66.7% becomes the better choice. The adjacency is what makes the empty chambers cluster, and the clustering is what makes the click informative in your favour.

Why would an investing firm ask this?

Because it is a small version of the question every analyst faces after new information: does what just happened change the odds of what comes next? An outcome you have already seen is data about the state of the world, and ignoring it in favour of the unconditional average is a common and costly habit. Say that link in one line after the numbers.

About Rs 450 crore. Revenue is 5,000 students x Rs 3 lakh, Rs 150 crore a year. A 25% EBITDA margin gives Rs 37.5 crore, and at 12 times that is an enterprise value of Rs 450 crore, before subtracting any debt. The case for buying is fees paid in advance, a campus and approvals that are slow to copy, and demand that renews with a new batch every year.

How do you structure an open question like this?

The question has two halves and the interviewer wants both. It is like selling a flat: first you say what makes it worth having, the light, the location, the building, and then you give a price per square foot times the area. Give the reasons to own it in three short points, then build the price as revenue times margin times multiple, naming every input so the interviewer can push on any one of them. Leading with the structure also buys you time to do the arithmetic.

What makes the case for buying convincing?

Pick the features an investor would pay for. The strongest is that customers pay a term or a year in advance and come back every year, so the business funds its own working capital and its revenue is visible well before the year starts. Add that a campus, a faculty and regulatory approvals take years to replicate, which protects the fee level, and that demand refreshes with every new batch. Then name the risk that decides the price: how full the seats are. One caution belongs here too: in many countries, including India, degree-granting colleges are commonly run through not-for-profit trusts or societies, so confirm the legal structure first, because it decides whether an investor can own the profit at all.

Which number would you defend hardest?

The margin and the multiple, because each moves the answer one for one. At a 20% margin the value falls to Rs 360 crore, and at 10 times EBITDA it falls to Rs 375 crore; every input in the chain carries equal weight. Justify 12 times with the visibility of fee income and 25% with a campus that is nearly full. If asked for a range rather than a point, Rs 360 to 450 crore on those two sensitivities is honest.

Randomise: pay Rs 3.15 with probability about 0.159 and Rs 3.14 otherwise. Pi is 3.14159, which is 15.9% of the way from 3.14 to 3.15. Weighting the two payable amounts by those odds gives an expected payment of exactly pi. Neither side is favoured on average, which is the only sense in which an amount you cannot pay can be paid fairly.

What does fair mean when the exact amount cannot be paid?

Five children and four mangoes: nobody can get four fifths of a mango each without a knife, so a fair parent draws lots and one child misses out, with every child facing the same chance. When the exact amount is impossible, fair means fair on average: the expected payment equals what you owe. Rounding to 3.14 short-changes the restaurant every time; rounding to 3.15 overpays every time. Only a random choice between the two neighbours can hit pi exactly.

How do you find the right odds?

Call the chance of paying 3.15 by the letter p. The expected payment is 3.14 plus p times one paisa, and you want that to equal 3.14159. So p is the fraction of the gap you still owe: 0.00159 over 0.01, which is 0.1593. Say it as a balance point: the further pi sits towards 3.15, the more often you pay 3.15.

How do you actually produce odds of 0.159 with a coin?

This is the follow-up that separates people. Write p in binary: 0.0010100011 and so on. Flip a fair coin to generate your own random binary digits, one at a time, and compare each with p's digit in the same place. At the first flip that differs, you know whether your random number is below p or above it, and you pay 3.15 or 3.14 accordingly. Each flip has a one in two chance of settling it, so on average two flips decide, even though p itself is irrational. The same idea, called randomised roundingRounding a fractional quantity up or down at random, with odds set so that the expected result equals the exact fraction., is used to split odd lots of shares fairly across client accounts.