三个骰子的最小值

3 Die Minimum

专题: Statistics / 统计
难度: L9
来源: OpenQuant

题目详情

假设你掷出三个 100 面骰子。最低滚动的预期值是多少？

Suppose you roll three fair 100-sided die. What is the expected value of the lowest roll?

解析

我们将三个骰子中最小值的值称为 $X$ 。使用生存函数求期望：

E[X] = \sum_{k=1}^{100} P(X \geq k)

$X$ 至少为 $k$ 的概率是

P(X\ge k)=\frac{(100-k+1)^3}{100^3}.

因此

E[X] = \frac{100^3}{100^3} + \frac{99^3}{100^3} + \frac{98^3}{100^3} + \dots + \frac{1^3}{100^3} = \frac{1}{100^3}\left(100^3+99^3+98^3+\dots+1^3\right).

利用立方和公式

1^3+2^3+\cdots+100^3=(1+2+\cdots+100)^2,

且

\sum_{k=1}^{100}k=100\cdot\frac{1+100}{2}=5050,

所以

E[X]=\frac{5050^2}{100^3}=\boxed{25.5025}.

import random

roll_dice = lambda: random.randint(1, 100)

mins = []
num_iters = 1_000_000
for i in range(num_iters):

    d1 = roll_dice()
    d2 = roll_dice()
    d3 = roll_dice()

    mins.append(min(d1, d2, d3))

print(sum(mins) / num_iters)

Original Explanation

Let's call the value of the minimum of the three die $X$ . Using expectation by summation of survival we get:

E[X] = \sum_{k=1}^{100}P(X \geq k)

The probability that $X$ is at least $k$ is $\frac{(100-k+1)^3}{100^3}$ so,

E[X] = \frac{100^3}{100^3} + \frac{99^3}{100^3} + \frac{98^3}{100^3} + \dots + \frac{1^3}{100^3}

= \frac1{100^3}({100^3} + {99^3} + {98^3} + \dots + {1^3})

The sum of cubes can also be written from $1$ to $n$ can be rewritten to the square of the sum from $1$ to $n$ thus,

E[X] = \frac{(1 + 2 + 3 + \dots+ 100)^2}{100^3}

\Longrightarrow \sum_{k=1}^{100}k = 100\cdot\frac{1+100}{2} = 5050

\Longrightarrow E[X] = \frac{5050^2}{100^3} = \boxed{25.5025}

import random

roll_dice = lambda: random.randint(1, 100)

mins = []
num_iters = 1_000_000
for i in range(num_iters):

    d1 = roll_dice()
    d2 = roll_dice()
    d3 = roll_dice()

    mins.append(min(d1, d2, d3))

print(sum(mins) / num_iters)