Newton–Pepys problem

Template:Short description The Newton–Pepys problem is a probability problem concerning the probability of throwing sixes from a certain number of dice.

In 1693 Samuel Pepys and Isaac Newton corresponded over a problem posed to Pepys by a school teacher named John Smith. The problem was:

Pepys initially thought that outcome C had the highest probability, but Newton correctly concluded that outcome A actually has the highest probability.

Solution

The probabilities of outcomes A, B and C are:

$P(A)=1-{\left(\frac{5}{6}\right)}^6=\frac{31031}{46656}\approx 0.6651,$

$P(B)=1-\sum _{x=0}^1(\genfrac{}{}{0ex}{}{12}{x}){\left(\frac{1}{6}\right)}^x{\left(\frac{5}{6}\right)}^{12-x}=\frac{1346704211}{2176782336}\approx 0.6187,$

$P(C)=1-\sum _{x=0}^2(\genfrac{}{}{0ex}{}{18}{x}){\left(\frac{1}{6}\right)}^x{\left(\frac{5}{6}\right)}^{18-x}=\frac{15166600495229}{25389989167104}\approx 0.5973.$

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:

$P(n)=1-\sum _{x=0}^{n-1}(\genfrac{}{}{0ex}{}{6n}{x}){\left(\frac{1}{6}\right)}^x{\left(\frac{5}{6}\right)}^{6n-x}.$

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:

for (s in 1:3) {          # looking for s = 1, 2 or 3 sixes
  n = 6*s                 # ... in n = 6, 12 or 18 dice
  q = pbinom(s-1, n, 1/6) # q = Prob( <s sixes in n dice )
  cat("Probability of at least", s, "six in", n, "fair dice:", 1-q, "\n")
}

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 $P(r\ge k;n,p)$ 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 ${\nu }_1,{\nu }_2$ be natural positive numbers s.t. ${\nu }_1\le {\nu }_2$. Is then $P(r\ge {\nu }_{1}k;{\nu }_{1}n,p)$ not smaller than $P(r\ge {\nu }_{2}k;{\nu }_{2}n,p)$ for all n, p, k?

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

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 $P(r\ge k_1;k_{1}n,\frac{1}{n})>P(r\ge k_2;k_{2}n,\frac{1}{n})$.

If $k,n_1,n_2$ are positive natural numbers, and $n_1<n_2$, then $P(r\ge k;k{n}_1,\frac{1}{n_1})>P(r\ge k;k{n}_2,\frac{1}{n_2})$.

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

If ${\nu }_1,{\nu }_2,n,k$ are positive natural numbers, and ${\nu }_1\le {\nu }_2,k\le n,p\in [0,1]$ then $P(r={\nu }_{1}k;{\nu }_{1}n,p)\ge P(r={\nu }_{2}k;{\nu }_{2}n,p)$.