恰为 6

Exactly 6

专题: Probability / 概率
难度: L3
来源: OpenQuant

题目详情

如果我掷骰子 36 次，每个数字恰好出现 6 次的概率是多少？

If I roll a dice 36 times, what is the probability that each number appears exactly 6 times?

解析

为了解决这个问题，我们可以利用多项式定理。该定理指出，对于 $n$ 试验的实验， $k$ 可能的结果和 $x_1, ..., x_k$ 期望的结果对应于每个可能的结果，可能的实验结果的数量由以下公式给出： ${n \choose x_1, .., x_k} = \frac{n!}{x_1, ..., x_k}$

我们知道骰子中出现任何面的概率是 $1/6$ ，并且我们掷骰子 36 次。因此，我们给出了如下的解决方案：

{36 \choose {6, 6, 6, 6, 6, 6}} \times (1/6)^{36} = \frac{36!}{6!^6}(\frac{1}{6})^{36} \approx 0.00025

这是一个验证我们解决方案的 python 程序：

import random
from typing import Callable
from collections import Counter

roll_dice: Callable[[None], int] = lambda: random.randrange(1, 7, 1)

def simulation() -> bool:
    # Roll the dice 36 times
    rolls = [roll_dice() for _ in range(36)]
    # Count how many times each number appeared
    counts = Counter(rolls)
    # Check that each number appeared 6 times
    return len(counts.keys()) == 6 and all([count == 6 for count in counts.values()])

iterations = 100_000
probability = sum([simulation() for _ in range(iterations)]) / iterations
print(f"The probability that each number appears exactly six times is: {probability}")

Original Explanation

To solve this question, we can leverage the multinomial theorem. This theorem states that for an experiment with $n$ trials, with $k$ possible results and $x_1, ..., x_k$ desired outcomes corresponding to each of the possible results, the number of possible experimental outcomes is given by the following formula: ${n \choose x_1, .., x_k} = \frac{n!}{x_1, ..., x_k}$

We know that the probability of any face from the dice showing up is $1/6$ and that we are rolling the dice 36 times. Therefore, we have the solution given below:

{36 \choose {6, 6, 6, 6, 6, 6}} \times (1/6)^{36} = \frac{36!}{6!^6}(\frac{1}{6})^{36} \approx 0.00025

Here is a python program validating our solution:

import random
from typing import Callable
from collections import Counter

roll_dice: Callable[[None], int] = lambda: random.randrange(1, 7, 1)

def simulation() -> bool:
    # Roll the dice 36 times
    rolls = [roll_dice() for _ in range(36)]
    # Count how many times each number appeared
    counts = Counter(rolls)
    # Check that each number appeared 6 times
    return len(counts.keys()) == 6 and all([count == 6 for count in counts.values()])

iterations = 100_000
probability = sum([simulation() for _ in range(iterations)]) / iterations
print(f"The probability that each number appears exactly six times is: {probability}")