Problem 6

The Pepys–Newton Connection (1693)

Problem. A asserts that he will throw at least one six with six dice. B asserts that he will throw at least two sixes by throwing 12 dice. C asserts that he will throw at least three sixes by throwing 18 dice. Which of the three stands the best chance of carrying out his promise?

Solution. The solution involves the application of the binomial distribution. Let X be the number of sixes when n balanced dice are thrown independently, n = 1, 2, . . . Then and

For A, and

For B, and

For C, and

Thus, A is more likely than B who, in turn, is more likely than C of carrying out his promise.

# 6.1 Discussion

The above problem is of historical significance ...