Equity Research puzzles, solved step by step

The answer is 1/6. After a first roll of 1, 3 or 5 the game ends on 2 with chance 1/3; after a 2 it cannot; after a 4 or 6 it ends on 2 with chance 1/2. Overall the game ends on 2 with chance 1/3, so a first roll of 1 carries (1/6 x 1/3) / (1/3) = 1/6. The 4 and 6 absorb the share the 2 lost.

Why does 1/5 feel right, and where does it go wrong?

Think of a cricket team told that the match was won off the last ball. That news does not just rule out one scenario; it makes the close games far more likely than the easy ones. Conditioning works the same way here. Being told the game ended on 2 is evidence, and evidence reweights every starting point by how well it explains what you saw. The 1/5 answer treats the news as if it only deleted the 2 and left the other five faces equally likely.

So ask, for each first roll, how likely it is that 2 ends up the last even number. After a 1, 3 or 5, all three evens are still missing and each is equally likely to come last, so 1/3. After a 2, the 2 is already seen and cannot come last, so 0. After a 4, only 2 and 6 are missing, and 2 comes last exactly when 6 shows first, so 1/2. The same holds after a 6.

How do you turn those chances into the answer?

Apply Bayes ruleThe chance of a cause given what you saw equals its prior chance times how likely it made the observation, divided by the total chance of the observation.. Each face starts at 1/6. Multiply by its chance of ending on 2 and add them up: three faces at 1/18, one at 0, two at 1/12, which totals 1/3. The 1 contributes 1/18 of that 1/3, which is 1/6, exactly its starting weight. The 4 and the 6 rise from 1/6 to 1/4 each, and the 2 falls to zero.

The research version of this is reading a data point. A company that beats estimates is more likely to be one that guided low, not just one that is doing well. Before you update, ask which starting stories make the thing you observed more likely, and shift weight towards them.

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.

135 over 211, about 64%. Let P(i) be the chance of reaching Rs 5 from Rs i. Each P(i) is 0.6 times P(i+1) plus 0.4 times P(i-1), with P(0) = 0 and P(5) = 1. The solution is P(i) = (1 - (2/3)^i) / (1 - (2/3)^5), and at i = 2 that is (5/9) divided by (211/243), or 135/211. A fair coin would give 40%.

How do you set the problem up without drawing every path?

Think of a cricket team chasing a small target with few wickets in hand. What matters is not the path of each ball but where the game stands now: runs needed and wickets left. The only thing that matters at any moment is how much money you hold, so write one unknown for each amount and one equation linking it to its neighbours. From Rs i you move to Rs i+1 with chance 0.6 and to Rs i-1 with chance 0.4, so P(i) = 0.6 P(i+1) + 0.4 P(i-1). The two ends are fixed: P(0) = 0 because you are broke, and P(5) = 1 because you have won.

How do you check 64% without trusting the formula?

Solve the chain by hand, starting from the bottom. Write P(1) = a. Then P(2) = (a - 0.4 x 0) / 0.6 = a/0.6 = 1.667a, and each later step follows the same rule until P(5) = 211/81 x a. Setting P(5) = 1 gives a = 81/211, or 38.4%, and P(2) = 135/211. Two routes landing on the same fraction is the check an interviewer wants to hear. It also tells you the shape: the edge matters most in the middle, where the fair line and the edge curve are furthest apart.

Say the desk version in one line. A small, repeatable edge does not make ruin impossible when you start thin: from Rs 2 you still go broke about 36 times in 100. More capital, not a bigger edge, is what pushes the ruin chance towards zero.

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.

Net income falls Rs 7.5 crore, cash rises Rs 2.5 crore, PP&E falls Rs 10 crore and equity falls Rs 7.5 crore. Pre-tax profit drops Rs 10 crore and tax drops Rs 2.5 crore, so net income is down Rs 7.5 crore. The cash flow statement adds back the Rs 10 crore of depreciation, so cash is up Rs 2.5 crore. Assets fall Rs 7.5 crore net, matching the fall in retained earnings.

Why does a non-cash expense raise cash?

A shopkeeper who can deduct the wear on her delivery van from her taxable income pays less tax, even though no money went out for the wear itself. Depreciation moves no cash, but it is deductible, so it cuts the tax bill, and the tax saved is real cash: 25% of Rs 10 crore, Rs 2.5 crore. That is the one cash effect in the whole question; everything else is accounting.

How do you say it in the room in thirty seconds?

Go statement by statement, in the order the numbers flow. Income statement: pre-tax profit down 10, tax down 2.5, net income down 7.5. Cash flow: start at minus 7.5, add back 10, cash up 2.5. Balance sheet: cash up 2.5, PP&E down 10, assets down 7.5; equity down 7.5 through retained earnings; it balances. End on the balance, because that is the check the interviewer is waiting for.

Say the assumption: the tax return uses the same depreciation as the books. If the higher charge is only in the books, cash tax does not fall, the Rs 2.5 crore goes to a deferred tax asset instead, and cash does not move at all.

With three cars the last car reaches about 1,833 miles; with n cars it reaches 1,000 x (1 + 1/2 + ... + 1/n). The cars drive together until they have burned one tankful between them, then one car refills the rest and stops. That happens after 1,000/n miles, then 1,000/(n - 1), down to the last car's 1,000. The sum has no ceiling, but it grows only like the logarithm of n.

Why does a car drop out after exactly 1,000/n miles?

Picture a group of hikers sharing water on a long desert walk. Once the group has drunk one person's worth between them, that person can pour what is left into the others' bottles, fill every one back up, and wait by the path. Carrying that hiker any further only costs water. The cars work the same way. After d miles each car has burned d miles of fuel and has room for d more. One car can refill the other n - 1 cars exactly when (n - 1) x d equals what it has left, 1,000 - d, which solves to d = 1,000/n.

Why do three cars buy 1,833 miles and not 3,000?

Follow the road left to right. Three cars run 333 miles together and burn 1,000 miles of fuel between them, one full tank. Car 3 has 667 miles of fuel left and hands 333 to each of the other two, which fills them. Two cars then run 500 miles, burning another tankful, and car 2 refills car 1. Car 1 runs its own 1,000. Every stretch burns exactly one tank, but each extra car buys a shorter stretch, because its tank has to move more cars.

What happens as n goes to infinity?

The harmonic series never converges, so enough cars can reach any distance at all. But the growth is logarithmic: ten cars reach about 2,929 miles and a hundred cars about 5,187, so every tenfold increase in cars adds only about 2,300 miles. When asked for the asymptotic answer, say both halves: unbounded, and slow. State the assumptions too: fuel moves between cars without loss, and a car that drops out is simply left behind. If the helper cars had to drive home, the answer would shrink sharply.