Financial Analysis puzzles, solved step by step

Puzzles
100
Traced to a firm
47
Topics
13
Hard
30
Topic
All topicsAccounting flow riddles10Valuation and multiples riddles10Ratio and margin riddles8Cost of capital, leverage and rates8Compounding and time value8Mental maths8Probability and expected value9Working capital and cash riddles6Percentages and averages7Estimation and market sizing7Logic and counting brainteasers7Pricing, costing and unit economics6Data and statistics intuition6
Level
AnyWarm upCoreHard
Source
AnyReported at a firmStandard
Showing 11–12 of 12 · filtered from 100Clear filters
  1. 097A stock trades at 30 times next year's earnings. Its cost of equity is 12%, and it earns a 20% return on the profit it reinvests. What perpetual growth rate is the market pricing in?Valuation and multiples riddlesHardEquity researchBuy-side research

    Try it first

    What does the 30x multiple say about growth?

    Show the worked solution

    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%.

    The relationship
    P0E1=1−g/ROEr−g30=1−g/0.200.12−g  ⇒  g=1−30(0.12)1/0.20−30=10.4%\frac{P_0}{E_1} = \frac{1 - g/\text{ROE}}{r - g} \qquad 30 = \frac{1 - g/0.20}{0.12 - g} \;\Rightarrow\; g = \frac{1 - 30(0.12)}{1/0.20 - 30} = 10.4\%
    P0 / E1the forward P/E, price over next year's earnings
    1 minus g/ROEthe payout ratio once growth is funded from retained profit
    rthe cost of equity, 12%
    gthe perpetual growth rate the price implies
    What it says in wordsThe forward P/E is the payout ratio over the gap between cost of equity and growth, and the payout is whatever is left after funding growth at the ROE.
    A multiple is a growth forecast in disguise: where the curve meets 30x10x20x30x40x50x0%2%4%6%8%10%Perpetual growth rate gForward P/Eg = 10.4% at ROE 20%the market's forecast8%: 15x10%: 25xno growth: 1 / r = 8.3xCost of equity 12%ROE 20%30x needs g = 10.4%ROE 15%30x needs g = 11.1%ROE = 12% = rP/E stays 8.3xat any growth rate
    At a 12% cost of equity and a 20% ROE, the P/E curve starts at 8.3x with no growth, passes 15x at 8% and 25x at 10%, and crosses 30x at 10.4%, while at a 15% ROE the same 30x needs 11.1% because more of each rupee must be retained to grow.

    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.

    Where candidates lose it

    The common slip is using P/E = 1 / (r minus g), which gives 30 = 1 / (0.12 minus g) and g = 8.7%. That formula pays out every rupee and still grows, which is impossible: growth has to be funded. Put the payout in as 1 minus g/ROE and the answer moves to 10.4%.

    The second loss is reporting the number as a forecast. It is what the price implies, not what will happen, and it rests on three assumptions the interviewer wants named: a constant 20% ROE on new investment, a 12% cost of equity, and growth held forever at a rate no company sustains.

    What the interviewer asks next

    • If the return on new investment falls to 12%, what multiple is justified at any growth rate, and why?
    • The stock pays out 48% of earnings. What dividend yield does that give at 30x, and how does it reconcile with the 12% cost of equity?
    • How would you restate the implied growth as ten years of fast growth followed by 4% forever?
  2. 100You toss a fair coin until you see two heads in a row. How many tosses do you expect to need? And why is that more than the number you expect to wait for a head followed by a tail?Probability and expected valueHardConsulting-style caseKPO research support

    Try it first

    Which pattern takes longer on average to appear, HH or HT?

    Show the worked solution

    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.

    The relationship
    E0=1+12E0+12E1E1HH=1+12E0⇒E0=6E1HT=1+12E1⇒E0=4E_0 = 1 + \tfrac{1}{2}E_0 + \tfrac{1}{2}E_1 \qquad E_1^{HH} = 1 + \tfrac{1}{2}E_0 \Rightarrow E_0 = 6 \qquad E_1^{HT} = 1 + \tfrac{1}{2}E_1 \Rightarrow E_0 = 4
    E0expected tosses still needed with no useful toss in hand
    E1expected tosses still needed holding one head
    1/2the chance of a head, and of a tail, on a fair coin
    What it says in wordsEach state's expected wait is one toss plus the average of the waits in the states a head and a tail lead to.
    Two heads in a row: a tail sends you home. Head then tail: nothing is lostWaiting for H Hexpected tosses from the start: 6startE = 6one HE = 4done0 leftH, 1/2H, 1/2T, 1/2T, 1/2: all progress lostWaiting for H Texpected tosses from the start: 4startE = 4one HE = 2done0 leftH, 1/2T, 1/2T, 1/2H, 1/2: still one HOverlap rule: HH = 2 + 4 = 6 tosses, HT = 4, HHH = 14, HTH = 10, HTT = 8
    In the HH diagram the tail arrow from the one-head state runs all the way back to the start, which is why the wait is 6 tosses, while in the HT diagram a head from the one-head state only loops in place, so the wait is 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.

    Where candidates lose it

    The common wrong answer is 4 for both, from 'each pattern has probability 1/4, so wait 4 tosses'. That is the long-run frequency, not the wait from a fresh start, and it ignores that a failed HH attempt destroys the head you had. Another is 8, from doubling the single-head wait of 2 and then adding; the state equations, not intuition, settle it at 6.

    The second loss is reaching 6 and 4 with no reason. The interviewer asked why, and the one-sentence answer is that HT keeps its progress after a failure and HH does not. Offer the overlap rule afterwards: it shows you can extend the result to HHH or HTH without starting over.

    What the interviewer asks next

    • How many tosses do you expect to wait for HHH, and for HTH?
    • Two players toss one coin: one wins if HH appears first, the other if HT appears first. Who is more likely to win?
    • With a coin that shows heads 60% of the time, what is the expected wait for HH?