Financial Analysis puzzles, solved step by step

Flip twice: heads then tails counts as heads, tails then heads counts as tails, and anything else is thrown away and flipped again. The two mixed orders each have probability 2/9, so they are equally likely whatever the bias. A pair succeeds 4/9 of the time, so each fair result takes 9/4 pairs, or 4.5 flips. Knowing the bias is exactly 1/3 lets you cut that to 2.25.

Why are heads-tails and tails-heads always equally likely?

Picture two friends flipping the same lopsided coin, one after the other. The chance the first gets heads and the second tails is the heads chance times the tails chance. The chance of the reverse is the tails chance times the heads chance. Multiplication does not care about order, so the two mixed outcomes are exactly equally likely, whatever the bias. That symmetry is the whole trick, known as the von Neumann method. Here each mixed pair has probability 1/3 times 2/3, which is 2/9.

How do you get the average of 4.5 flips?

Each pair either works or does not, independently of the last. Waiting for a success that happens with probability q takes 1/q tries on average, the same reason a die takes six rolls on average to show a six. A pair works with probability 4/9, so you need 9/4 pairs, and two flips a pair makes 4.5 flips. The method pays for its fairness with waste: 5 pairs in 9 are thrown away.

Can you do better if you know the bias exactly?

Yes, and this is usually the follow-up. With p exactly 1/3, tails-tails has probability 4/9, the same as the two mixed pairs together. Call tails-tails one side and either mixed pair the other, and only heads-heads, 1/9 of pairs, is wasted, so each fair result costs 2 times 9/8, or 2.25 flips. The von Neumann method is still the better answer when nobody tells you the bias, because it works for any p. The limit for any scheme is set by how much randomness one flip carries: about 0.92 of a fair bit here, so no method can beat roughly 1.09 flips per fair result on average.

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.

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.

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.

About 3.28 loops. Every tie cuts the number of strands by one, so there are exactly 100 ties. With k strands left, the second end you grab is one of 2k - 1 free ends, and only one of those closes a loop, so that tie adds 1 over (2k - 1) to the expected count. Summing 1 + 1/3 + 1/5 + ... + 1/199 gives 3.2843.

How do you avoid tracking the whole bowl?

Think of shaking hands at a party where each person has two hands. You do not need the seating plan to know that each handshake joins two hands. Whatever the bowl looks like, every tie reduces the count of strands by exactly one, and a strand is either an open piece or a closed loop that has left the game. So you can ignore the tangled history and ask one question per tie: does this tie close a loop, yes or no?

Pick any free end first; which one does not matter. With k strands there are 2k free ends, so 2k - 1 remain for the second pick, and exactly one of them is the other end of the same strand. That tie closes a loop with chance 1 over (2k - 1). Otherwise two strands fuse into one longer strand.

Why can you add the step expectations when the steps depend on each other?

Whether an early tie closes a loop changes nothing about how many strands remain, because every tie removes one strand either way. Expected values add even when the events are linked, so the total is simply the sum of the per-tie chances. This is linearity of expectation, and it is what makes the puzzle a two-minute answer instead of a simulation. For large n the sum is close to half the natural log of n plus about 0.98; for 100 noodles that gives 3.28, a useful sanity check.

About 15.4%. Picture 10,000 invoices. 100 are fraudulent and the screen flags 90 of them. 9,900 are clean and the screen flags 5% of them, 495. The flagged pile holds 585 invoices, of which 90 are fraud, so a flag means fraud only 90 times in 585. The rare base rate swamps the screen's accuracy.

Why is a 90% accurate screen right only 15% of the time?

Think of a smoke alarm that never misses a fire but also goes off whenever someone makes toast. In a house where fires are rare and toast is daily, almost every alarm is toast. When the thing you are hunting is rare, even a small false positive rate on the large innocent pile produces more false alarms than true hits. Here 5% of 9,900 clean invoices is 495 false flags, against only 90 true ones.

What would make the screen useful, and how would an auditor use it?

Cutting the false positive rate does far more than raising the catch rate. At a 0.5% false positive rate the flagged pile would be 90 frauds and about 49.5 clean invoices, and a flag would mean fraud 65% of the time. Raising the catch rate from 90% to 100% would only move the answer from 15.4% to about 16.8%. In practice a screen like this is a triage tool: it shrinks 10,000 invoices to 585 for a human to review, and that review is where the 495 false alarms are cleared. Say the limit too: the 1% base rate is itself an estimate, and the answer is only as good as it is.

Stop on a 5 or 6 after the first roll, on 4 or more after the second, and take whatever the third gives. The game is worth 14/3, about 4.67. Solve it from the end: a last roll is worth 3.5, so with two rolls left you keep anything above 3.5, which makes two rolls worth 4.25. With three rolls left you keep only what beats 4.25.

Why do you start from the last roll?

Think of house hunting with three viewings booked and a rule that you must take the last flat if you get that far. You cannot judge the first flat until you know what walking away from it is worth, and that depends on the viewings still to come. A stop or continue decision is only as good as your value for continuing, so you price the last stage first and carry that value backwards. The method is called backward inductionSolving a sequence of decisions from the final step back to the first, so each earlier choice is made knowing what the later ones are worth., and it is how an option to wait is valued in finance too.

On the third roll there is no choice: you get the face, and a fair die averages (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5. That 3.5 is the price of walking away from the second roll. So on the second roll you keep a 4, 5 or 6, each of which beats 3.5, and re-roll a 1, 2 or 3. Half the time you keep an average of 5; half the time you collect 3.5. Two rolls are worth 0.5 x 5 + 0.5 x 3.5 = 4.25.

What changes when you hold three rolls?

The bar goes up. With three rolls in hand, walking away from the first roll is worth 4.25, so a 4 is no longer good enough: only a 5 or a 6 beats it. Two faces in six you keep, averaging 5.5; four faces in six you roll on and collect 4.25. That is (2/6) x 5.5 + (4/6) x 4.25 = 1.83 + 2.83 = 4.67.

What do you add to show you see the pattern?

Two observations. First, the bar rises with the number of chances left. A candidate who applies one rule, keep 4 or more, on every roll gets 4.625 instead of 4.667: a small loss that shows the continuation value was never priced. Second, each extra roll is worth less than the one before: the second roll adds 0.75, the third only 0.42, and a fourth would add 0.28. An extra option is worth less when the options you already hold are good. The limit to say out loud: this strategy maximises the average, which is right for a player who plays many times; someone playing once who needs at least 4 would play differently.

About 67%. If the price is the probability-weighted average of the two outcomes, p x 500 + (1 - p) x 350 = 450, so p = (450 - 350) / (500 - 350) = 100 / 150 = 2/3. The price sits two thirds of the way from the failure value to the offer. The answer is only as good as the Rs 350 failure estimate, which nobody can observe directly.

Why does a price between two outcomes reveal a probability?

Picture a resale ticket for a cricket match that may be rained off. If the match is played the ticket is worth Rs 1,000; if it is washed out you get a Rs 400 refund. If tickets change hands at Rs 800, buyers are betting on play two times in three. When a price can end at one of two known values, where it sits between them is the market's probability, read off by distance from the bad outcome. A merger target is the same ticket: it ends at the offer price or falls back to where it would trade alone.

What changes once you allow for time?

Deals take months to close, and an arbitrageur who ties up Rs 450 wants paying for the wait. Say closing is six months away and the required return is 8% a year, 4% for the half year. Then the expected payoff must be 450 x 1.04 = Rs 468, and p = (468 - 350) / 150 = 78.7%. Ignoring time value understates the implied probability, because part of the gap to the offer is simply the return for waiting.

What would you check before trusting the number?

The failure value first, because it is an estimate and the answer swings on it. If the shares would fall only to Rs 380, say because the market has risen since the bid, the implied probability drops to 58.3%. Next the shape of the bet: Rs 50 to gain against Rs 100 to lose, so an arbitrage desk needs real confidence in the regulatory approvals, the buyer's financing and the shareholder vote. The limit to say aloud is that a probability read from prices also carries a premium for bearing deal risk, so it is not a pure forecast of completion.

Play: one roll pays 30 x 3.5 = Rs 105 on average against a Rs 100 fee, an edge of Rs 5. With the re-roll, keep a 3 or better and re-roll a 1 or 2; the game is worth Rs 118.33, an edge of Rs 18.33. A fresh roll is worth 105 less the Rs 20 fee, 85, so re-roll only a result paying less than 85: 30 and 60 qualify, 90 does not.

Why is the first game worth playing?

A fair die averages 3.5, so the average payout is 30 x 3.5 = Rs 105, more than the Rs 100 you pay. On expected value the game pays Rs 5 a play, so a player who can repeat it should play. Half the time you lose money, since a 1, 2 or 3 pays 30, 60 or 90, and the other half you win 120, 150 or 180; the wins are larger than the losses, which is where the edge sits.

How does the re-roll change the rule?

Think of returning a shirt. You keep it if what you have is worth more than what the shop will hand you after the restocking fee. Re-roll only when the roll in hand is worth less than a fresh roll net of its fee, which is 105 - 20 = Rs 85. A 1 pays 30 and a 2 pays 60, both below 85: re-roll. A 3 pays 90, above 85: keep it, even though 3 is below the average face. The fee is what makes a 3 worth keeping.

What does the option itself cost and earn?

The re-roll lifts the game from 105 to 118.33, so the option is worth Rs 13.33 on average, and it also cuts the chance of losing money from a half to a third, because a bad first roll gets a second chance. Re-rolling a 3 as well would give 117.50, worse by 0.83; a free re-roll would make re-rolling the 3 right and lift the game to 127.5. The threshold moves with the fee: below a fee of Rs 15 the 3 is worth re-rolling, above it the 3 is kept. The limit to say: the worst path, a 1 or 2 followed by a 1, costs Rs 90, and a player who cannot afford that should not value the game by its average.

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%.

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.

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.