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橄榄球卡片

Football Cards

专题
Probability / 概率
难度
L4

题目详情

James 有 6 包橄榄球卡片。每包里只有 1 张卡。一共有 10 种不同的卡，每包中出现任意一种卡的概率都相同。求 James 在这 6 包中得到的不同橄榄球卡片种类数的期望。

James has 6 packs of football cards. Each pack only contains 1 card inside. There are 10 distinct cards he can get, each of which is equally likely to appear in any pack. Find the expected number of distinct football cards that James obtains in the 6 packs.

解析

令 XiX_i 表示 James 在 6 包卡片中至少得到过一次第 ii 种卡的指示变量：

Xi={1,if card i appears in at least one pack,0,otherwise.X_i = \begin{cases} 1, & \text{if card $i$ appears in at least one pack}, \\ 0, & \text{otherwise}. \end{cases}

则得到的不同卡片总数为 X=X1+X2+⋯+X10.X = X_1 + X_2 + \cdots + X_{10}.

由期望的线性性质： E[X]=E[X1+⋯+X10]=∑i=110E[Xi].\mathbb{E}[X] = \mathbb{E}[X_1 + \cdots + X_{10}] = \sum_{i=1}^{10} \mathbb{E}[X_i].

每种卡至少出现一次的概率为 E[Xi]=1−Pr⁡(card i never appears)=1−(910)6\mathbb{E}[X_i] = 1 - \Pr(\text{card $i$ never appears}) = 1 - \left(\frac{9}{10}\right)^6

因此 E[X]=10(1−(910)6)=10⋅(1−0.531441)≈4.686.\mathbb{E}[X] = 10 \left( 1 - \left(\frac{9}{10}\right)^6 \right) = 10 \cdot (1 - 0.531441) \approx 4.686.

E[X]≈4.69\boxed{\mathbb{E}[X] \approx 4.69}

Original Explanation

Let XiX_i be the indicator that James obtains card ii at least once in his 6 packs:

Xi={1,if card i appears in at least one pack,0,otherwise.X_i = \begin{cases} 1, & \text{if card $i$ appears in at least one pack}, \\ 0, & \text{otherwise}. \end{cases}

Then the total number of distinct cards obtained is X=X1+X2+⋯+X10.X = X_1 + X_2 + \cdots + X_{10}.

By linearity of expectation: E[X]=E[X1+⋯+X10]=∑i=110E[Xi].\mathbb{E}[X] = \mathbb{E}[X_1 + \cdots + X_{10}] = \sum_{i=1}^{10} \mathbb{E}[X_i].

Each card appears with probability E[Xi]=1−Pr⁡(card i never appears)=1−(910)6\mathbb{E}[X_i] = 1 - \Pr(\text{card $i$ never appears}) = 1 - \left(\frac{9}{10}\right)^6

Hence E[X]=10(1−(910)6)=10⋅(1−0.531441)≈4.686.\mathbb{E}[X] = 10 \left( 1 - \left(\frac{9}{10}\right)^6 \right) = 10 \cdot (1 - 0.531441) \approx 4.686.

E[X]≈4.69\boxed{\mathbb{E}[X] \approx 4.69}