Hedge Funds puzzles, solved step by step

21. The number of rolls until the first 6 averages 6, so there are on average 5 non-six rolls plus the 6. A roll known not to be a 6 is equally likely to be 1 to 5, so it averages 3, not 3.5, giving 5 x 3 + 6 = 21. Wald's identity confirms it: 6 expected rolls times 3.5 a roll is also 21, because a stop that looks only at past rolls does not bias the total.

Why is it tempting to get 23.5?

Suppose you keep buying scratch cards until one wins. Every card before the winner is, by definition, a loser, so those cards are worth less than an average card. The stopping rule changes the rolls before the stop: each one is known not to be a 6, so it averages 3, and valuing them at 3.5 overpays by 0.5 a roll, 2.5 in all. Five non-sixes at 3.5 plus a 6 is 23.5, which counts the high side twice: once in the 3.5 and again in the final 6.

How do the two routes agree?

Route one splits the sum: the expected number of non-six rolls times their average, plus the final 6. The count of rolls is a geometric wait with success chance 1/6, so it averages 6, of which 5 are non-sixes: 5 x 3 + 6 = 21. Route two is Wald's identityFor a stopping rule that uses only rolls already seen, the expected total equals the expected number of rolls times the average roll.: the expected total is the expected number of rolls times the average roll, 6 x 3.5 = 21. The low early rolls and the high final roll balance exactly.

Why would a trading firm ask this?

Stopping rules are everywhere on a desk: exit at the first stop-loss hit, rebalance at the first breach of a band. The question checks whether you can tell when a stopping rule biases what you observe, as it does for the early rolls, and when it does not, as for the total. Say the condition too: Wald's identity needs the decision to stop to use only rolls already seen, and the expected number of rolls to be finite. A rule that could peek at the next roll would break it.

Stop on the first roll only with a 5 or 6, on the second with a 4 or more, and the game is worth 14/3, about Rs 4.67. Work backwards. The last roll is worth 3.5. With two rolls left, keep 4, 5 or 6 and reroll otherwise: worth (4 + 5 + 6)/6 + 3/6 x 3.5 = 4.25. With three rolls left, keep only what beats 4.25, a 5 or 6: worth (5 + 6)/6 + 4/6 x 4.25 = 14/3.

Why start from the last roll?

Deciding whether to take a job offer is easier if you know what your fallback is worth. Each keep-or-reroll decision compares the roll in hand with the value of the rolls still to come, so you need the value of the future first, and the only stage with no future is the last one. On the last roll you must take whatever comes, which is worth 3.5 on average. That single number lets you solve the stage before it, and so on backwards. This is {term('backward induction', 'Solving a sequence of decisions from the last one to the first, using the value of each later stage to make the earlier decision.')}, the core of dynamic programming.

How do the thresholds come out?

With two rolls left, a roll of 4, 5 or 6 beats the 3.5 you expect from rerolling, and 1, 2 or 3 does not. So the two-roll game is worth the average of the kept faces times their chance, plus the chance of rerolling times 3.5: 15/6 + 1.75 = 4.25. On the first roll the fallback is now 4.25, so a 4 is no longer good enough: only 5 or 6 is kept. That gives 11/6 plus 4/6 x 4.25, which is 1.833 plus 2.833, or 14/3.

Add the pattern. More rolls always raise the value, but by less each time: 3.5, 4.25, 4.67, then about 4.94 with four rolls. An extra option is always worth something and never worth more than what it can still improve. That is the same logic as valuing a trade you can exit early: the right to wait is priced by what the future is worth, not by the average outcome.

Both players should show heads 3/8 of the time, and the game is worth minus Rs 0.125 a round to you. Choose your mix so the opponent gains nothing by switching: 3p - 2(1 - p) = -2p + (1 - p) gives p = 3/8. The opponent's mix solves the same balance from your side, also 3/8. At those mixes you lose an eighth of a rupee a round, although the payoffs look even.

Why is a fair coin the wrong strategy?

Think of a penalty taker and a goalkeeper. If the taker always shoots left, the keeper dives left; the taker's only defence is to mix so the keeper cannot profit from guessing. A mix is right only if it leaves the other side indifferent, and a fair coin here does not: against it the opponent's tails pays you 0.5 x (-2) + 0.5 x 1 = -0.50 a round. So the opponent always shows tails, and you lose Rs 0.50 a round rather than breaking even.

How do you find the equilibrium mix?

Set up your payoff for each of your choices as the opponent's chance of heads, q, varies. Showing heads pays 3q - 2(1 - q) = 5q - 2; showing tails pays -2q + (1 - q) = 1 - 3q; they are equal at q = 3/8, where both pay -1/8. The opponent plays 3/8 heads so that nothing you do beats -1/8. By the same balance from the opponent's side, you play 3/8 heads so that nothing the opponent does pushes you below -1/8. That pair is the Nash equilibriumA pair of strategies in which neither player can do better by changing only their own choice..

Why is the game negative when the payoffs look even?

Because the balance is in the totals, not in the play. Your two winning cells need coordination the opponent will not give you, while their winning cells, the mismatches, pay the same 2 either way. The opponent can lean towards tails, starving your big Rs 3 cell, and the only price is feeding your small Rs 1 cell. Say what a desk would do with it: ask to be paid about 13 paise a round to play, or ask to swap sides.

About Rs 24.19. Round n pays Rs 2^n with probability 1/2^n, so each round adds exactly Rs 1 of expected value. That holds up to n = 23, since 2^23 is about Rs 84 lakh and 2^24 is above the Rs 1 crore cap. From round 24 on, the payout is stuck at Rs 1 crore, with total probability 1/2^23, which adds about Rs 1.19. So 23 + 1.19, about Rs 24.

Why is the uncapped game worth an infinite amount?

Each round is a doubling bet: the payout doubles while the chance of reaching it halves. Think of a raffle where a ticket twice as valuable is half as likely to win: every prize tier is worth the same to you. Round n pays 2^n with probability 1/2^n, so every round contributes exactly Rs 1 to the expected value, and there are infinitely many rounds. That is the famous paradox: few people would pay even Rs 100, yet the expected value has no bound. The resolution that matters on a desk is not psychology; it is that nobody can pay out an unlimited amount.

How does the cap change the sum?

Find the round where the cap starts to bind. 2^23 is Rs 83,88,608, under Rs 1 crore; 2^24 is Rs 1,67,77,216, over it. So rounds 1 to 23 each add Rs 1, and every round from 24 on pays the capped Rs 1 crore, which together happen with probability 1/2^23 and add 1,00,00,000 / 83,88,608, about Rs 1.19. The fair price is about Rs 24.19. Nearly all of the textbook infinity lives in outcomes the casino cannot pay.

What does a bigger casino buy you?

Very little. The value grows only with the logarithm of the cap: each doubling of the casino's bankroll adds about Rs 1. A cap of Rs 1,000 crore, a thousand times larger, lifts the fair price only to about Rs 34. That is the lesson a risk manager takes away: a payoff whose expected value rests on rare, enormous outcomes is worth what the other side can actually pay, and any estimate built on the tail should be checked against who stands behind it.