牛顿-皮普斯问题

利用二项分布，三个投掷实验的概率分别为：[2]

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

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

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

该类问题的通项公式，一般的，若P(N)是投掷6n个骰子得到至少n个6的概率，则：

${\displaystyle P(N)=1-\sum _{x=0}^{n-1}{\binom {6n}{x}}\left({\frac {1}{6}}\right)^{x}\left({\frac {5}{6}}\right)^{6n-x}\,.}$

n变大时，P(N)会逐渐趋近于极限值1/2.

在R语言中，该问题可以用如下方法解：

p <- as.numeric(1/6)
s <- c(1, 2, 3)
for (i in s)
{
   x <- 0
   n <- 6*i
   for(j in 0:(i-1)) {x <- x + dbinom(j, n, p) }
   print(paste("Probability of at least ", i, " six in ", n, " fair dice: ", 1-x, sep=""))
}

结果会显示为：

[1] "Probability of at least 1 six in 6 fair dice: 0.665102023319616"
[1] "Probability of at least 2 six in 12 fair dice: 0.618667373732309"
[1] "Probability of at least 3 six in 18 fair dice: 0.597345685947723"

牛頓設想將B和C的骰子每六粒分為一組，A只可分為一組；B和C分別可分成兩組和三組，每組需要在其中一次投擲中出現6。如此可見，A的機率是最大的，因为A只需要在其中一次投擲中出現6，而B和C則分别需要重复A的過程两次和三次。