十层楼

我们可以通过使用指标随机变量来解决这个问题。为了快速回顾一下，指标随机变量通常用于将较大的随机变量分解为较小的二元成分。在这个问题中应用二元随机变量会将 10 层楼分解为单独的楼层。

例如，设 $\textit{S}$ 为随机变量，表示电梯停在的楼层数。问题是要求 $\textit{S}$ 或 $\mathbb{E}[S]$ 的期望。然而，我们可以将$\textit{S}$分解为10个指标随机变量的总和；我们将它们表示为 $X_1 … X_{10}$。每个 $X_i$ 都是下限 $\textit{i}$ 的指示随机变量。详细来说，如果12个人中有一个在3楼下车，那么$X_3 = 1$。如果8楼无人下车，则$X_8 = 0$。

然后这个问题简化为寻找 $\mathbb{E}[S] = \mathbb{E}[X_1 + … + X_{10}] = \mathbb{E}[X_1] + … + \mathbb{E}[X_{10}]$（通过期望线性）。我们现在需要计算 $\mathbb{E}[X_i], \forall i \in [1 … 10]$。然而，由于对称性（每个楼层与其他楼层无法区分并且具有相同的分布），我们可以简单地求解 $\mathbb{E}[X_i]$ 以获得通用 $\textit{i}$（并且 $\mathbb{E}[S]$ 之和变为 $10*\mathbb{E}[X_i]$，因为所有 $X_i$ 都是分布中的等效随机变量）。

指标随机变量的期望等于随机变量等于 1 的概率（该结论来自期望的定义）。所以问题就简化为寻找$\mathbb{P}(X_i = 1)$。口头上说，这是要求有人（至少 1 人）在 $\textit{i}$ 层下车的概率。

我们可以使用补码规则来求解这个表达式。 $\mathbb{P}(X_i = 1) = 1 - \mathbb{P}(X_i = 0)$。 $\mathbb{P}(X_i = 0)$ 是 $\textit{i}$ 层无人下车的概率。单人在 $\textit{i}$ 楼层不下车的概率为 $\frac{9}{10}$（选择除 $\textit{i}$ 楼层外的任意楼层）。因此，由于所有 12 个人以相同的可能性且彼此独立地选择楼层，因此所有 12 个人选择不在楼层 $\textit{i}$ 下车的概率为 $\frac{9}{10}^{12}$。

因此，$\mathbb{P}(X_i = 1) = 1 - \mathbb{P}(X_i = 0) = 1 - \frac{9}{10}^{12}$。向上级联，这意味着 $\mathbb{E}[X_i] = 1 - \frac{9}{10}^{12}$。这意味着 $\mathbb{E}[S] = 10 * \mathbb{E}[X_i] = 10 * (1 - \frac{9}{10}^{12}) \approx 7.17$。因此，电梯预计将停在 10 层楼中的 7 层左右，让 12 人下车。

这是一个说明此解决方案的 python 程序：

import random

def simulate_trial() -> int:
    # each of the 12 people pick a random floor between 1 and 10
    floors = [random.randint(1, 10) for _ in range(12)]

    # count the number of unique floors the elevator must stop at
    return len(set(floors))

# list of the number of floors the elevator stopped at for each trial
res = [simulate_trial() for _ in range(10_000)]

# compute the average, or expected number of floors the elevator needed to stop at over all 10,000 trials
print(sum(res) / len(res))

Original Explanation

We can solve this problem through the use of indicator random variables. For a quick recap, indicator random variables are often used to decompose a larger random variable, into smaller, binary constituents. The application of binary random variables in this problem would be breaking down the 10 floors into solitary floors.

For example, let $\textit{S}$ be the random variable indicating the number of floors the elevator stopped at. The question is asking for the expectation of $\textit{S}$, or $\mathbb{E}[S]$. However, we can break down $\textit{S}$ into the sum of 10 indicator random variables; we’ll denote them $X_1 … X_{10}$. Each $X_i$ is an indicator random variable for floor $\textit{i}$. To elaborate, if one of the 12 people gets off on floor 3, then $X_3 = 1$. If nobody gets off on floor 8, then $X_8 = 0$.

This problem then reduces to finding the $\mathbb{E}[S] = \mathbb{E}[X_1 + … + X_{10}] = \mathbb{E}[X_1] + … + \mathbb{E}[X_{10}]$ (by Linearity of Expectation). We now need to compute the $\mathbb{E}[X_i], \forall i \in [1 … 10]$. However, due to symmetry (each floor is indistinguishable from the other and has identical distribution) we can simply solve for $\mathbb{E}[X_i]$ for a generic $\textit{i}$ (and the sum $\mathbb{E}[S]$ becomes $10*\mathbb{E}[X_i]$ since all $X_i$ are equivalent random variables in distribution).

The expectation of an indicator random variable is equal to the probability of the random variable equalling 1 (this conclusion follows from the definition of expectation). So the problem reduces to finding $\mathbb{P}(X_i = 1)$. Stated verbally, this is asking for the probability that someone (at least 1) gets off at floor $\textit{i}$.

We can solve for this expression using the complement rule. $\mathbb{P}(X_i = 1) = 1 - \mathbb{P}(X_i = 0)$. $\mathbb{P}(X_i = 0)$ is the probability that nobody gets off on floor $\textit{i}$. The probability that a single person does not get off on floor $\textit{i}$ is $\frac{9}{10}$ (he/she chooses any floor except floor $\textit{i}$). Therefore, since all 12 people choose floors with equal likelihood and independently of each other, the probability that all 12 people choose not to get off at floor $\textit{i}$ is $\frac{9}{10}^{12}$.

Therefore, $\mathbb{P}(X_i = 1) = 1 - \mathbb{P}(X_i = 0) = 1 - \frac{9}{10}^{12}$. Cascading upwards, this means $\mathbb{E}[X_i] = 1 - \frac{9}{10}^{12}$. And this means $\mathbb{E}[S] = 10 * \mathbb{E}[X_i] = 10 * (1 - \frac{9}{10}^{12}) \approx 7.17$. So the elevator is expected to stop at around 7 of the 10 floors to drop off the 12 people.

Here is a python program that illustrates this solution:

import random

def simulate_trial() -> int:
    # each of the 12 people pick a random floor between 1 and 10
    floors = [random.randint(1, 10) for _ in range(12)]

    # count the number of unique floors the elevator must stop at
    return len(set(floors))

# list of the number of floors the elevator stopped at for each trial
res = [simulate_trial() for _ in range(10_000)]

# compute the average, or expected number of floors the elevator needed to stop at over all 10,000 trials
print(sum(res) / len(res))