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  1. 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?