开心乐园餐开盒

设 $X_i$ 为指示变量，表示第 $i$ 种卡片是否至少在 6 份套餐中出现过一次：

$$X_i = \begin{cases} 1, & \text{若第 $i$ 种卡至少出现一次},\\ 0, & \text{否则}. \end{cases}$$

不同卡片的总数为 $$X = X_1 + X_2 + X_3 + X_4.$$

由期望的线性性： $$\mathbb{E}[X] = \mathbb{E}[X_1 + X_2 + X_3 + X_4] = \sum_{i=1}^{4} \mathbb{E}[X_i].$$

对每一种卡片而言，它至少出现一次的概率为 $$\mathbb{E}[X_i] = 1 - \Pr(\text{第 $i$ 种卡一次都没出现}) = 1 - \left(\frac{3}{4}\right)^6.$$

因此 $$\mathbb{E}[X] = 4 \left(1 - \left(\frac{3}{4}\right)^6\right) \approx 4 \cdot 0.822 = 3.288.$$

$$\boxed{\mathbb{E}[X] \approx 3.29}$$

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Original Explanation

Let $X_i$ be the indicator that card $i$ appears at least once among the 6 meals:

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

The total number of unique cards is $$X = X_1 + X_2 + X_3 + X_4.$$

By linearity of expectation: $$\mathbb{E}[X] = \mathbb{E}[X_1 + X_2 + X_3 + X_4] = \sum_{i=1}^{4} \mathbb{E}[X_i].$$

Each card appears with probability $$\mathbb{E}[X_i] = 1 - \Pr(\text{card $i$ never appears}) = 1 - \left(\frac{3}{4}\right)^6.$$

Hence $$\mathbb{E}[X] = 4 \left(1 - \left(\frac{3}{4}\right)^6\right) \approx 4 \cdot 0.822 = 3.288.$$

$$\boxed{\mathbb{E}[X] \approx 3.29}$$