Financial Analysis puzzles, solved step by step

A's overall rate is 9.2%, B's is 3.9%, and A is still the better underwriter in each segment. A: 2 defaults among 100 salaried plus 90 among 900 self-employed, 92 in 1,000. B: 27 plus 12, 39 in 1,000. An overall rate is a weighted average of segment rates, and the weights are the mix of the book. A's book is 90% self-employed, where everyone's rate is high, so its average is pulled up by whom it lends to, not by how well it lends.

How can a branch be better at everything and worse in total?

Two surgeons. One takes the hardest cases in the hospital and loses fewer of them than anyone else would; the other takes mostly routine cases. Count deaths per patient and the first surgeon looks worse, because her patients were sicker to begin with. An overall rate is a weighted average of segment rates, and the weights are the mix of the book, so a branch that lends mostly to the riskier segment carries a higher average even when it is better in every segment. A's rate is 0.1 x 2% + 0.9 x 10% = 9.2%; B's is 0.9 x 3% + 0.1 x 12% = 3.9%.

How do you compare the two branches fairly?

Give them the same customers. Apply each branch's segment rates to one common mix and the mix stops voting. At B's mix, 90% salaried, A's rates give 0.9 x 2% + 0.1 x 10% = 2.8% against B's actual 3.9%. At A's mix, B's rates give 0.1 x 3% + 0.9 x 12% = 11.1% against A's actual 9.2%. At a 50:50 mix, A is 6.0% and B is 7.5%. On any common mix A is lower, so A underwrites better; its higher total is the price of the book it chose to build. This reversal, where a comparison flips when groups are combined, is known as Simpson's paradox.

What should the analyst say beyond the arithmetic?

Three things. First, both statements are true at once: A underwrites better, and A's book loses more money, because capital is charged on actual losses and A chose a riskier book. Whether that was a good choice depends on the pricing: if A earns enough extra yield on self-employed loans to cover the extra 8 points of defaults, its strategy can be sound. Second, ask why the mixes differ, since location, product and targets all shape who walks in. Before judging two rates, ask whether the two populations are the same; if they are not, standardise to one mix or compare segment by segment.

Third, say the limitation. Each segment is itself a mix: of loan sizes, vintages and cities. A's salaried borrowers might be larger-ticket government employees and B's might be gig workers on salary slips, and a finer split could flip the comparison back. Standardising to the segments you can see removes the mix you know about, never the mix you do not. The honest close is that A looks the better underwriter on this cut, and you would want the segment rates by vintage before signing that view.

About 10.4% a year, forever, with the company paying out 48% of its earnings. A forward P/E equals the payout ratio divided by the cost of equity less growth, and with a fixed ROE the payout is set by growth: payout = 1 minus g/ROE. So 30 = (1 minus g/0.20) / (0.12 minus g). Multiply out: 3.6 minus 30g = 1 minus 5g, so 25g = 2.6 and g = 10.4%. Check: a dividend yield of 48% / 30 = 1.6% plus 10.4% growth is the 12% cost of equity.

Where does a multiple hide a growth forecast?

If someone offers to sell you a shop for thirty years of its current profit, they are not quoting a price; they are telling you how fast they expect the profit to grow. The multiple is the forecast. Under the Gordon growth model a share is worth next year's dividend over (r minus g), so dividing by next year's earnings gives P/E = payout / (r minus g), and a given P/E can be solved for g. The one thing people forget is that the payout is not free to choose: growth has to be paid for with retained profit.

With a fixed return on reinvested profit, growth equals ROE times the retention rate, so retention is g/ROE and payout is 1 minus g/ROE. At a 20% ROE, growing at 10% means keeping half the earnings. Put that into the multiple: 30 = (1 minus 5g) / (0.12 minus g). Cross-multiply: 3.6 minus 30g = 1 minus 5g, so 2.6 = 25g and g = 10.4%. The implied payout is 1 minus 0.104/0.20 = 48%, and the implied dividend yield is 48% of a 3.33% earnings yield, 1.6%.

Why does the ROE matter as much as the growth rate?

Because the ROE sets how much growth costs. At a 15% ROE the same 30x needs 11.1% growth, since a bigger share of earnings has to be retained to fund each point of it. At an ROE equal to the cost of equity, 12%, the multiple is 1/r = 8.3x whatever the growth rate, because every retained rupee earns exactly what shareholders could earn elsewhere. Growth adds value only when the reinvested profit earns more than the cost of equity; the multiple prices the spread between ROE and r as much as it prices g. The curve is also steep near the answer: at 10% the multiple is 25x, at 11% it is 45x, so a 30x stock is one point of growth away from either 25x or 45x.

Now say the limitation. 10.4% growth forever, only 1.6 points below the discount rate, is not a forecast anyone would defend; no company outgrows the economy indefinitely. The honest reading is that 30x prices a long period of fast growth that will fade, and a two-stage model, say 10.4% for a decade and a lower rate after, is the next thing to build. The one-line solve is still worth doing, because it converts a multiple into a sentence you can argue with: the market expects this company to compound earnings at about 10% for a very long time while earning 20% on what it retains.

Six tosses for HH on average, four for HT. For HH, a tail after a head sends you back to the start with nothing. For HT, once you hold a head, every further head keeps you one toss from finishing, so nothing is lost. Solve HH with two unknowns: from the start E0 = 1 + E0/2 + E1/2, and after a head E1 = 1 + E0/2, which gives E1 = 4 and E0 = 6. For HT, E1 = 1 + E1/2 = 2 and E0 = 4.

Why is waiting for two heads different from waiting for a head then a tail?

Suppose you will go trekking after two dry days in a row. One wet day after a dry one puts you back to zero: you need two fresh dry days. Now suppose instead you are waiting for a dry day followed by a wet one. Extra dry days cost nothing; you stay ready, and the first wet day finishes the job. A failed attempt at HH throws away the progress you had, while a failed attempt at HT keeps it, so HH pays for its failures twice: the wasted toss and the lost head. That asymmetry is the whole answer; the algebra just prices it.

Write the states. E0 is the expected tosses still needed from nothing, E1 from one head in hand. From nothing, one toss is spent and you land back at E0 with a tail or at E1 with a head: E0 = 1 + E0/2 + E1/2, so E0 = 2 + E1. For HH, from one head a head finishes and a tail drops you to E0: E1 = 1 + E0/2. Substitute: E1 = 1 + (2 + E1)/2, so E1/2 = 2, E1 = 4 and E0 = 6. For HT, from one head a tail finishes and a head keeps you at E1: E1 = 1 + E1/2, so E1 = 2 and E0 = 4.

Is there a rule for longer patterns?

Yes, and it is quick enough for the room. For each length k at which the pattern's first k symbols equal its last k symbols, add 2 to the power k. HH matches itself at k = 1 (H and H) and at k = 2 (the whole thing): 2 + 4 = 6. HT matches itself only at k = 2: 4. HHH gives 2 + 4 + 8 = 14; HTH gives 2 + 8 = 10; HTT gives just 8. Patterns that overlap with themselves take longer to appear from a fresh start, because their occurrences come in clusters, and the same long-run frequency spread into clusters means longer gaps between them. Every pattern of length n shows up once in 2 to the n positions on average; what differs is how bunched those appearances are.

The average hides a wide spread. The chance that HH has not appeared after ten tosses is 14.1%, against 1.1% for HT, so one game in seven is still running at toss ten when you wait for HH. And the figures assume a fair, independent coin: with heads at 60%, the HH wait falls to 1/p + 1/p squared, about 4.4 tosses. The finance link is any rule that needs consecutive results, such as a bonus or a covenant test that requires two good quarters in a row: one bad quarter resets the count, so the rule is harder to satisfy than its two-in-four odds suggest.