偶数先于奇数

Even Before Odd

专题: Probability / 概率
难度: L4
来源: OpenQuant

题目详情

假设你掷出一个 6 面骰子，直到你看到所有 6 个面为止。在你看到所有偶数面之前，你看不到奇数面的概率是多少？

Suppose you roll a fair 6-sided die until you've seen all 6 faces. What is the probability you won't see an odd numbered face until you have seen all even numbered faces?

解析

重要的是要认识到，在这个问题中，你不应该关注掷骰子的数量，而应该关注看到面时的排序方式。也就是说，序列 $2,5,3,1,4,6$ 代表你的第一个独特次见到是 $2$ ，第二个是 $5$ ，第三个是 $3$ ，依此类推。这将是一个无效的序列，因为我们在看到所有偶数编号的面之前已经看到了奇数编号的面。

总订购量为 $6!$ 。我们可以用它作为我们的分母。

对于我们的分子，我们希望仅对第一个 $3$ 次出现的偶数进行分组，对最后一个 $3$ 的剩余奇数进行分组。

有 $3!$ 方法对奇数进行排序，以及 $3!$ 方法对偶数进行排序。

\begin{equation*} \frac{3!3!}{6!} = \frac{6\cdot6}{720} = \boxed{\frac1{20}} \end{equation*}

import random

roll_dice = lambda: random.randint(1, 6)

valid_cases = 0
num_iters = 100_000
for i in range(num_iters):

    seen_faces = set()
    while(True):

        roll = roll_dice()
        seen_faces.add(roll)

        if roll % 2 == 1 and len(seen_faces) < 4:
            break

        if len(seen_faces) == 6:
            valid_cases += 1
            break

print(valid_cases / num_iters)

Original Explanation

It's important to realize that you should not focus on the number of rolls in this question, but rather the ways to order when a face has been seen. ie) The sequence $2,5,3,1,4,6$ represents your first unique sighting being a $2$ , second being a $5$ , third being $3$ , and so on. This would be an invalid sequence as we have seen an odd numbered face before seeing all the even numbered faces.

There are $6!$ total orderings. We can use this as our denominator.

For our numerator, we want to group only even numbers for the first $3$ sightings, and the remaining odd numbers for the last $3$ .

There are $3!$ ways to order the odd numbers as well as $3!$ ways to order the even numbers.

\begin{equation*} \frac{3!3!}{6!} = \frac{6\cdot6}{720} = \boxed{\frac1{20}} \end{equation*}

import random

roll_dice = lambda: random.randint(1, 6)

valid_cases = 0
num_iters = 100_000
for i in range(num_iters):

    seen_faces = set()
    while(True):

        roll = roll_dice()
        seen_faces.add(roll)

        if roll % 2 == 1 and len(seen_faces) < 4:
            break

        if len(seen_faces) == 6:
            valid_cases += 1
            break

print(valid_cases / num_iters)