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 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.