同一个生日

Same Birthday

专题: Probability / 概率
难度: L2
来源: OpenQuant

第 1 小问

题目详情

如果房间里有 $N \leq 365$ 人，那么其中至少 2 人生日相同的可能性有多大？

If you have $N \leq 365$ people in a room, what is the likelihood that at least 2 of them have the same birthday?

解析

为了解决这个问题，我们可以利用概率补码规则。我们必须首先计算每个人都有唯一生日的概率，然后将其减去。 $N$ 的人有不同的生日，有 $P_{N}^{365} = \frac{365}{(365-N)!}$ 排列。 $N$ 人员的生日分配总数为 $365^N$ 。因此答案是： $1 - \frac{P_{N}^{365}}{365^N}$

Original Explanation

To solve this question, we can leverage the complement rule in probability. We must first calculate the probability that everyone has a unique birthday and then subtract it from one. There are $P_{N}^{365} = \frac{365}{(365-N)!}$ permutations for $N$ people to have different birthdays. There are $365^N$ total birthday assignments for the $N$ people. Therefore the answer is: $1 - \frac{P_{N}^{365}}{365^N}$

第 2 小问

题目详情

如果房间里有 $N$ 人，那么某人与你生日相同的可能性有多大？

If you have $N$ people in a room, what is the likelihood that someone has the same birthday as you?

解析

对于这个问题，我们首先计算没有人与我们生日相同的概率，然后将其从 1 中减去。为了计算没有人与我们生日相同的概率，除了我们自己的生日之外，还有 $364^{N-1}$ 可能的生日样本。其他客人的生日有 $365^{n-1}$ 可能的方式。我们的答案是： $1 - \frac{364^{N-1}}{365^{N-1}}$

Original Explanation

For this question, we will first compute the probability that nobody has the same birthday as us and then subtract it from 1. For computing the probability that nobody has the same birthday as us, there are $364^{N-1}$ possible samples of birthdays excluding our own. There are $365^{n-1}$ possible ways for the other guests to have birthdays. Our answer is: $1 - \frac{364^{N-1}}{365^{N-1}}$