Financial Analysis puzzles, solved step by step

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  1. 017A vendor offers a game: a fair coin is tossed until it shows heads. If heads comes on the first toss you get Rs 100, on the second toss Rs 200, and the prize doubles with every toss after that. The game costs Rs 1,000 to play, but the vendor has only Rs 10 lakh to pay out. What is the game really worth, and would you play?Probability and expected valueCoreEquity capital marketsConsulting-style case

    Try it first

    With the vendor's Rs 10 lakh limit, what is the game worth on average?

    Show the worked solution

    About Rs 761, so at Rs 1,000 the game is not worth playing. Each possible toss adds Rs 50 of expected value: a prize that doubles times a chance that halves. Prizes fit under the Rs 10 lakh cap for 14 tosses, which is Rs 700, and all longer runs are paid Rs 10 lakh, which adds only about Rs 61. The infinite value of the textbook game rests entirely on prizes no vendor can pay.

    Why does every toss add exactly Rs 50?

    Imagine a lottery stall where a ticket wins Rs 200 one time in four, or Rs 400 one time in eight. Each prize is worth the same on average, Rs 50, because the prize doubles exactly as the chance halves. In this game, heads first appearing on toss k has probability one over 2 to the k, and pays 100 times 2 to the k minus 1, so every possible toss contributes Rs 50 of expected value. With no limit there are infinitely many tosses, so the expected value is infinite. That is the famous St Petersburg paradox, and it is why nobody sensible pays a fortune to play.

    The relationship
    E=∑k=1∞12k min⁡(100⋅2k−1, 106)=14×50+106214≈761E = \sum_{k=1}^{\infty} \frac{1}{2^{k}}\,\min(100 \cdot 2^{k-1},\ 10^{6}) = 14 \times 50 + \frac{10^{6}}{2^{14}} \approx 761
    kthe toss on which the first head appears
    1/2^kthe chance the first head comes on toss k
    100 x 2^(k-1)the promised prize
    10^6the vendor's limit, Rs 10 lakh
    What it says in wordsFourteen tosses each add Rs 50; every longer run pays the capped Rs 10 lakh, and the chance of reaching toss 15 is one in 16,384.

    What does the Rs 10 lakh limit do to the value?

    The prize on toss 14 is Rs 100 times 2 to the 13, which is Rs 8,19,200; on toss 15 it would be Rs 16,38,400, more than the vendor holds. So only 14 tosses pay in full, worth Rs 700, and every longer run pays Rs 10 lakh, which happens one time in 16,384 and adds about Rs 61. The game is worth about Rs 761. Paying Rs 1,000 means losing about Rs 239 a game on average.

    Expected value, toss by toss: Rs 50 each until the cap binds2505007501,00001510142015+Toss on which the first head appearsValue of the game so far, RsPrice to play: Rs 1,000Capped game: Rs 761the tail adds only 61no cap: +50 a tosseach toss adds Rs 50
    Each of the first 14 tosses adds Rs 50 of expected value, reaching Rs 700, and the capped tail adds only about Rs 61, so the game is worth about Rs 761. Without the cap the value would keep rising Rs 50 a toss, but the capped game never reaches the Rs 1,000 price.

    How rich would the vendor need to be for Rs 1,000 to be fair?

    Value grows very slowly with the vendor's wealth, because each doubling of the bankroll adds just one more Rs 50 toss. A vendor holding Rs 1,000 crore would make the game worth only about Rs 1,425, and the game reaches Rs 1,000 only if the vendor can pay about Rs 2.6 crore. Say the limit too: even a fair price ignores how much risk you can stomach, since almost every game pays Rs 100 or Rs 200. That is why economists use this game to show people value money by its usefulness to them, not by its face amount.

    Where candidates lose it

    The common loss is reciting "infinite expected value" and stopping, or worse, saying you would pay any price. The interviewer added the vendor's limit precisely to see whether you notice that infinite value depends on payouts that cannot happen.

    The second loss is getting the cap wrong by assuming the tail adds a lot. Work out the last toss that pays in full, count Rs 50 per toss, and add the small capped tail; it takes thirty seconds.

    What the interviewer asks next

    • What would you pay if the vendor could pay out Rs 100 crore?
    • The vendor offers to play the game ten times in a row for Rs 7,000. Does that change your answer?
    • How does this game relate to valuing a company with a tiny chance of an enormous outcome?
  2. 087You lend to 10 borrowers. Each has a 2% chance of defaulting over the year, independently of the others. What is the chance that at least one defaults, and what is the expected number of defaults?Probability and expected valueCoreRating agenciesBank credit

    Try it first

    What is the chance that at least one of the ten defaults?

    Show the worked solution

    The chance of at least one default is 18.3%, and the expected number of defaults is 0.2. Work from the complement: each loan survives with probability 98%, so all ten survive with probability 0.98^10 = 81.7%, and at least one default is the rest. The expected number is simply 10 x 2% = 0.2, which holds whether or not the loans are independent. Independence matters for the first answer, not the second.

    Why work from the chance that nothing happens?

    Ask what the chance is that at least one of ten friends is late for dinner, each being late one time in fifty. Counting the ways someone could be late is messy: one late, two late, any combination. Counting the single way nobody is late is easy: everyone on time. 'At least one' is one minus 'none', and 'none' for independent events is just the single probabilities multiplied. Each loan survives with probability 0.98, so all ten survive with probability 0.98^10.

    To do 0.98^10 in your head, use the shortcut that (1 minus x) to the power n is close to 1 minus nx plus a correction of n(n minus 1)/2 times x squared: 1 minus 0.20 plus 45 x 0.0004 = 0.818. The exact figure is 0.8171, so the chance of at least one default is 18.29%. A second check is the Poisson shortcut, 1 minus e to the minus 0.2, which gives 18.1%.

    Ten independent 2% loans: work from the chance that none defaults81.7%016.7%11.5%20.09%3 or moreNumber of loans that defaultAt least one: 18.3%= 1 - 0.98^10 = 1 - 81.7%No default0.98^10 = 81.7%Same loans, two worldsIndependentFully linkedExpected defaults0.20.2At least one18.3%2.0%All ten defaultabout 02.0%Correlation moves risk into the tail
    With ten independent loans each carrying a 2% default chance, no default happens 81.7% of the time, so at least one default happens 18.3% of the time, while expected defaults are 0.2 whether the loans are independent or perfectly linked.

    Why is the expected number so much simpler?

    Expected values add, always. Each loan contributes 0.02 expected defaults, and ten loans contribute 0.2. The expected number of defaults is 0.2 regardless of how the loans are connected, because the expectation of a sum is the sum of the expectations even when the events are correlated. The distribution behind it is lopsided: 81.7% of the time nothing happens, 16.7% of the time exactly one loan fails, and two or more fail 1.6% of the time.

    What does correlation change?

    Take the extreme. If all ten borrowers are suppliers to one factory and fail together or survive together, there is a 2% chance that all ten default and a 98% chance that none does. Expected defaults are still 0.2. But the chance of at least one default falls to 2%, and the chance of losing the entire book jumps from practically zero to 2%. Correlation does not change the average loss; it moves probability from many small losses into rare large ones, which is the loss a lender cannot survive. That is why credit portfolio work spends its effort on concentration and correlation rather than on the average default rate, and why a portfolio of ten loans to one industry is riskier than its expected loss suggests.

    Where candidates lose it

    The fast wrong answer is 20%: ten loans times 2%. Adding probabilities only works for events that cannot happen together, and two loans can both default. The error is small here, 20% against 18.3%, but it grows quickly: with 100 such loans the same method gives 200%, which is impossible.

    The second loss is giving 18.3% and stopping. The question says independently for a reason. Saying what correlation would do, same expected defaults, fatter tail, is the part a credit interviewer is listening for.

    What the interviewer asks next

    • How many such loans do you need before at least one default is more likely than not?
    • If each loan is Rs 10 crore and recovers 40% on default, what is the expected loss on the book?
    • Two of the ten borrowers are in the same group of companies. Does the chance of at least one default go up or down?