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 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.

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.

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.

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.