Newton–Pepys problem

The Newton–Pepys problem is a probability problem concerning the probability of throwing sixes from a certain number of dice.{{MathWorld|urlname=Newton-PepysProblem |title=Newton-Pepys Problem}}

In 1693 Samuel Pepys and Isaac Newton corresponded over a problem posed to Pepys by a school teacher named John Smith.Chaundy, T.W., Bullard, J.E., 1960. "John Smith’s Problem." The Mathematical Gazette 44, 253-260. The problem was:

{{quote|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.{{cite journal|doi=10.1214/088342306000000312|title=Isaac Newton as a Probabilist|journal=Statistical Science|volume=21|issue=3|pages=400|year=2006|last1=Stigler|first1=Stephen M|author-link=Stephen Stigler|arxiv=math/0701089|s2cid=17471221 }}}}

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:

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(

~~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)):Chaundy, T.W., Bullard, J.E., 1960. "John Smith’s Problem." The Mathematical Gazette 44, 253-260.

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)):{{cite journal|doi=10.1016/j.spl.2013.02.008 |first1=Damiano|last1=Varagnolo|first2=Luca |last2=Schenato|first3=Gianluigi |last3=Pillonetto |year=2013 | title=A variation of the Newton–Pepys problem and its connections to size-estimation problems | journal=Statistics & Probability Letters |volume=83 |number=5 |pages=1472–1478}}

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

References

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