Hedge Funds puzzles, solved step by step

Puzzles: 100
Traced to a firm: 38
Topics: 14
Hard: 30

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017You roll a fair die until the first 6 appears and are paid the sum of every roll, including the final 6. What is the expected payout?Expected value and dice gamesHardQuant and systematic fundsProp and quant trading firms
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What is the expected payout?
Show the worked solution
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.
In a typical run of 4, 1, 5, 2, 3 and then 6, the five rolls before the stop are non-sixes averaging 3 and the total is 21; 5 x 3 + 6 and 6 x 3.5 both give 21, while 5 x 3.5 + 6 = 23.5 wrongly treats the early rolls as ordinary rolls.
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.
The relationship
$E[S] = E[N]\cdot E[X] = 6 \times 3.5 = 21 \qquad E[S] = (E[N]-1)\times 3 + 6 = 5\times 3 + 6 = 21$
S the total paid
N the number of rolls, including the 6
X a single roll, averaging 3.5 before any conditioning
What it says in wordsCounted either as all rolls at 3.5 or as non-sixes at 3 plus a 6, the expected payout is 21.
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.
Where candidates lose it
The slip is 5 x 3.5 + 6 = 23.5. It treats the rolls before the 6 as ordinary rolls, when the stopping rule guarantees none of them is a 6, which pulls their average down to 3.
The opposite slip is to distrust 6 x 3.5 because stopping at a 6 seems to bias it. It does not: the total is unbiased for any stopping rule that looks only at the past. Give both routes and say why they agree.
What the interviewer asks next
What is the expected payout if the final 6 is not paid?
You stop at the first 5 or 6 instead. What is the expected payout?
If you could choose to stop whenever you like, what would you pay to play?
092A casino offers the St Petersburg game: a fair coin is tossed until the first tail, and if that takes n tosses you are paid Rs 2 to the power n. The casino can pay out at most Rs 1 crore. What is a fair price to play?Expected value and dice gamesHardQuant and systematic fundsProp and quant trading firms
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Roughly what is the capped game worth?
Show the worked solution
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.
Rounds 1 to 23 each add exactly Rs 1 to the expected value, and once the Rs 1 crore cap binds the later rounds add 0.60, 0.30, 0.15 and so on, Rs 1.19 in all, so the capped game is worth about Rs 24.19.
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.
The relationship
$E = \sum_{n=1}^{23} \frac{2^n}{2^n} + \frac{10^7}{2^{23}} = 23 + 1.19 = 24.19$
2^n the payout if the first tail arrives on toss n
1/2^n the chance the first tail arrives on toss n
10^7 / 2^23 the capped Rs 1 crore times the chance of reaching round 24 or later
What it says in wordsEvery uncapped round is worth one rupee; the capped tail is worth the cap times the chance of getting that far.
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.
Where candidates lose it
Answering infinity is the trap for anyone who knows the textbook game. The interviewer added the cap to see whether you can find where it binds and redo the sum, not recite the paradox.
The other loss is dropping the tail beyond the cap and saying Rs 23, or valuing it crudely at the full Rs 1 crore times a guessed chance. The tail is a clean sum: probability 1/2^23 of receiving Rs 1 crore.
What the interviewer asks next
The cap rises to Rs 1,000 crore. What is the fair price now?
With logarithmic utility and wealth of Rs 1 lakh, roughly what would you pay for the uncapped game?
How is a book short deep out-of-the-money options like the casino in this game?

S	the total paid
N	the number of rolls, including the 6
X	a single roll, averaging 3.5 before any conditioning

2^n	the payout if the first tail arrives on toss n
1/2^n	the chance the first tail arrives on toss n
10^7 / 2^23	the capped Rs 1 crore times the chance of reaching round 24 or later

Try it first

Why is it tempting to get 23.5?

How do the two routes agree?

Why would a trading firm ask this?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the uncapped game worth an infinite amount?

How does the cap change the sum?

What does a bigger casino buy you?

Where candidates lose it

What the interviewer asks next