Newton–Pepys problem

Solution

The probabilities of outcomes A, B and C are:

These results may be obtained by applying the binomial distribution (although Newton obtained them from first principles). In general, if P(N) is the probability of throwing at least n sixes with 6n dice, then:

As n grows, P(N) decreases monotonically towards an asymptotic limit of 1/2.

Example in R

The solution outlined above can be implemented in R as follows:

Newton's explanation

Although Newton correctly calculated the odds of each bet, he provided a separate intuitive explanation to Pepys. He imagined that B and C toss their dice in groups of six, and said that A was most favorable because it required a 6 in only one toss, while B and C required a 6 in each of their tosses. This explanation assumes that a group does not produce more than one 6, so it does not actually correspond to the original problem.

Generalizations

A natural generalization of the problem is to consider n non-necessarily fair dice, with p the probability that each die will select the 6 face when thrown (notice that actually the number of faces of the dice and which face should be selected are irrelevant). If r is the total number of dice selecting the 6 face, then P ( r ≥ k ; n , p ) is the probability of having at least k correct selections when throwing exactly n dice. Then the original Newton–Pepys problem can be generalized as follows:

Let ν 1 , ν 2 be natural positive numbers s.t. ν 1 ≤ ν 2 . Is then P ( r ≥ ν 1 k ; ν 1 n , p ) not smaller than P ( r ≥ ν 2 k ; ν 2 n , p ) for all n, p, k?

Notice that, with this notation, the original Newton–Pepys problem reads as: is P ( r ≥ 1 ; 6 , 1 / 6 ) ≥ P ( r ≥ 2 ; 12 , 1 / 6 ) ≥ P ( r ≥ 3 ; 18 , 1 / 6 ) ?

As noticed in Rubin and Evans (1961), there are no uniform answers to the generalized Newton–Pepys problem since answers depend on k, n and p. There are nonetheless some variations of the previous questions that admit uniform answers:

(from Chaundy and Bullard (1960)):

If k 1 , k 2 , n are positive natural numbers, and k 1 < k 2 , then P ( r ≥ k 1 ; k 1 n , 1 n ) > P ( r ≥ k 2 ; k 2 n , 1 n ) .

If k , n 1 , n 2 are positive natural numbers, and n 1 < n 2 , then P ( r ≥ k ; k n 1 , 1 n 1 ) > P ( r ≥ k ; k n 2 , 1 n 2 ) .

(from Varagnolo, Pillonetto and Schenato (2013)):

If ν 1 , ν 2 , n , k are positive natural numbers, and ν 1 ≤ ν 2 , k ≤ n , p ∈ [ 0 , 1 ] then P ( r = ν 1 k ; ν 1 n , p ) ≥ P ( r = ν 2 k ; ν 2 n , p ) .