# 牛顿-皮普斯问题

• A.6个正常的骰子独立投掷，至少出现1个6.
• B.12个正常的骰子独立投掷，至少出现2个6.
• C.18个正常的骰子独立投掷，至少出现3个6.[1]

## 概率解

${\displaystyle P(A)=1-\left({\frac {5}{6}}\right)^{6}={\frac {31031}{46656}}\approx 0.6651\,,}$
${\displaystyle P(B)=1-\sum _{x=0}^{1}{\binom {12}{x}}\left({\frac {1}{6}}\right)^{x}\left({\frac {5}{6}}\right)^{12-x}={\frac {1346704211}{2176782336}}\approx 0.6187\,,}$
${\displaystyle P(C)=1-\sum _{x=0}^{2}{\binom {18}{x}}\left({\frac {1}{6}}\right)^{x}\left({\frac {5}{6}}\right)^{18-x}={\frac {15166600495229}{25389989167104}}\approx 0.5973\,.}$

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

n变大时，P(N)会逐渐趋近于极限值1/2.

## 编程计算法

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=""))
}


[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"


## 参考文献

1. ^ Isaac Newton as a Probabilist 互联网档案馆存檔，存档日期2007-09-18., Stephen Stigler, University of Chicago
2. ^