Newton–Pepys problem

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:

"Which of the following three propositions has the greatest chance of success?


 * A. Six fair dice are tossed independently and at least one "6" appears.
 * B. Twelve fair dice are tossed independently and at least two "6"s appear.
 * C. Eighteen fair dice are tossed independently and at least three "6"s appear."

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\binom{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\binom{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:


 * $$P(n)=1-\sum_{x=0}^{n-1}\binom{6n}{x}\left(\frac{1}{6}\right)^x\left(\frac{5}{6}\right)^{6n-x}\, .$$

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:

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 $$P(r = \nu_1 k ; \nu_1 n, p) \ge P(r = \nu_2 k ; \nu_2 n, p)$$.