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  1. 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

    Try it first

    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.

    The rolls before the 6 are never sixes, so they average 3, not 3.5415236non-sixes: each averages (1+2+3+4+5)/5 = 3the stopThis run: 4 + 1 + 5 + 2 + 3 + 6= 21, five non-sixes at 35 x 3 + 621, correct6 rolls x 3.5 (Wald)21, correct5 x 3.5 + 623.5, wrong
    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]⋅E[X]=6×3.5=21E[S]=(E[N]−1)×3+6=5×3+6=21E[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
    Sthe total paid
    Nthe number of rolls, including the 6
    Xa 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?
  2. 081Which is more likely: at least one six in four rolls of a single die, or at least one double six in twenty-four rolls of a pair of dice?Expected value and dice gamesCoreQuant and systematic fundsProp and quant trading firms

    Try it first

    Before any arithmetic, which do you back?

    Show the worked solution

    The single six in four rolls: 51.8% against 49.1%. Use the complement for each. Four rolls with no six happen (5/6)^4 = 48.2% of the time, so at least one six is 51.8%. Twenty-four rolls of two dice with no double six happen (35/36)^24 = 50.9% of the time, so at least one is 49.1%. Only the first is better than even money.

    Why does the proportional argument give the same answer for both?

    The old gamblers' rule, in the problem usually credited to the Chevalier de Méré, went like this: a six comes up one time in six and you get four tries, so 4/6; a double six comes up one time in thirty-six and you get twenty-four tries, so 24/36, also 4/6. Think of phoning a friend four times: four tries do not give four times the chance of getting through, because once they pick up, the later calls add nothing. Adding the chance per try counts the runs with two or more hits more than once, so it overstates the chance of at least one hit, and the overstatement grows with the number of tries.

    The two bets land on opposite sides of even moneyAt least one sixin 4 rolls of one diemiss all: (5/6)^4 = 48.2%51.8%At least one double sixin 24 rolls of two dicemiss all: (35/36)^24 = 50.9%49.1%even money, 50%old rule: 66.7%40%50%60%70%axis starts at 40%At even money: bet one gains 3.5 paise a rupee, bet two loses 1.7
    At least one six in four rolls comes up 51.8% of the time and at least one double six in twenty-four rolls only 49.1%, on opposite sides of even money, while the proportional rule wrongly puts both at 66.7%.

    How does the complement settle it?

    Ask how likely it is that nothing happens. Four rolls with no six: (5/6)^4 = 625/1,296 = 48.2%, so at least one six is 51.8%. Twenty-four rolls with no double six: (35/36)^24 = 50.9%, so at least one is 49.1%. The rare event with many tries falls short, because its misses compound over six times as many rolls.

    The relationship
    1−(56)4=0.5181−(3536)24=0.4911 - \left(\tfrac{5}{6}\right)^{4} = 0.518 \qquad 1 - \left(\tfrac{35}{36}\right)^{24} = 0.491
    5/6the chance one roll of a die is not a six
    35/36the chance one roll of two dice is not a double six
    4, 24the number of tries in each bet
    What it says in wordsThe chance of at least one hit is one minus the chance that every single try misses.

    Why is the gap so small, and what is it worth as a bet?

    The two probabilities are only 2.6 points apart, which is why the question needed a careful calculation to settle and why it still tests method rather than intuition. As an even-money bet, the first earns 3.5 paise per rupee staked on average: 51.8% of winning a rupee less 48.2% of losing one. The second loses 1.7 paise per rupee. A small edge repeated many times is the whole business of a casino, and of many trading strategies. Give the second bet one more roll and it tips over: 1 - (35/36)^25 = 50.6%.

    Where candidates lose it

    The trap is the proportional argument: 4 x 1/6 = 24 x 1/36 = 2/3, so the bets are equal. It is the reasoning the question was built to catch, and any answer that makes the chance grow in a straight line with the tries fails the moment the tries pass six.

    The second loss is getting 51.8% for the first bet and assuming the second is also above half. Compute both; the point of the question is that they fall on opposite sides of 50%.

    What the interviewer asks next

    • How many rolls of two dice do you need before a double six is more likely than not?
    • You are offered even money on the second bet. What is your expected result per Rs 100 staked?
    • Why is 1 - (1 - p)^n close to 1 - e^(-np) when p is small, and what does that give for the second bet?
  3. 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

    Try it first

    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.

    Every round adds Rs 1 until the Rs 1 crore cap binds, then the tail fades outRs 10151015202324300.600.300.15cap binds after round 2323 rounds x Rs 1 = Rs 23tail adds Rs 1.19Round n in which the first tail appearsFair price: 23 + 1.19 = Rs 24.19
    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=∑n=1232n2n+107223=23+1.19=24.19E = \sum_{n=1}^{23} \frac{2^n}{2^n} + \frac{10^7}{2^{23}} = 23 + 1.19 = 24.19
    2^nthe payout if the first tail arrives on toss n
    1/2^nthe chance the first tail arrives on toss n
    10^7 / 2^23the 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?