Happie 是一家新奇玩意儿商店的老板，他正在为自己的商店组织一次宣传活动。利用他收购的一批极其稀有的 Lubaba 玩偶，他设计了一个游戏，玩家打开“盲盒”，试图在用于游戏的若干个完全相同的盒子中，找到隐藏在其中一个盒子里的 Lubaba 玩偶。 游戏规则如下：

• 玩家首先支付 $5 开始游戏，并打开 $n$ 个盒子中的一个。
• 如果打开的盒子是空的，玩家可以额外支付 $3 打开第二个盒子。
• 如果第二个盒子也是空的，玩家可以再次支付 $2.50 打开第三个盒子。
• 如果第三个盒子仍然是空的，玩家就输掉了游戏。未被赢取的玩偶将被捐赠出去，然后游戏重置，供下一位玩家游玩。

由于 Lubaba 玩偶被视为稀有的收藏品，可以假设在每局游戏中，玩家都会不断打开盒子，直到他们赢得玩偶，并且他们有能力支付最多 3 次允许的开盒尝试。 设 $X$ 为表示玩家在一次游戏中打开的空盒子数量的随机变量。

1. Identify the sequence of events that results in exactly two empty boxes being opened.

For $X=2$, the player opens exactly 2 empty boxes before the game ends. Since players keep opening boxes until they win the doll or reach the maximum of 3 tries, having exactly 2 empty boxes means the first two boxes were empty and the third box contained the doll.
$\text{P}(X=2) = \text{P}(\text{1st empty}) \times \text{P}(\text{2nd empty}) \times \text{P}(\text{3rd is doll})$

2. Calculate the probability using without-replacement multiplication.

There is 1 doll and $n-1$ empty boxes initially.
$\text{P}(X=2) = \left(\frac{n-1}{n}\right) \times \left(\frac{n-2}{n-1}\right) \times \left(\frac{1}{n-2}\right)$
Cancelling out the common factors $(n-1)$ and $(n-2)$:
$\text{P}(X=2) = \frac{1}{n}\ \text{ (shown)}$

1. Determine the possible values of $X$ and calculate $\text{P}(X=0)$ and $\text{P}(X=1)$.

The possible values of $X$ (number of empty boxes opened) are $0, 1, 2, 3$.
$\text{P}(X = 0) = \text{P}(\text{doll in 1st box}) = \frac{1}{n}$
$\text{P}(X = 1) = \text{P}(\text{1st empty}) \times \text{P}(\text{2nd doll}) = \frac{n-1}{n} \times \frac{1}{n-1} = \frac{1}{n}$

2. Calculate $\text{P}(X=3)$ using the property that the sum of all probabilities is 1.

Since the game ends after a maximum of 3 tries, $\sum \text{P}(X=x) = 1$.
Using $\text{P}(X=2) = \frac{1}{n}$ from part (a):
$\text{P}(X = 3) = 1 - \text{P}(X=0) - \text{P}(X=1) - \text{P}(X=2)$
$\text{P}(X = 3) = 1 - \frac{1}{n} - \frac{1}{n} - \frac{1}{n} = \frac{n-3}{n}$

3. Present the probability distribution in a table.

1. Substitute $n = 10$ into the probability distribution of $X$.

From part (b), the probability distribution of $X$ is:
$\text{P}(X = 0) = \frac{1}{n}$
$\text{P}(X = 1) = \frac{1}{n}$
$\text{P}(X = 2) = \frac{1}{n}$
$\text{P}(X = 3) = \frac{n-3}{n}$

When $n = 10$:
$\text{P}(X = 0) = \frac{1}{10}$
$\text{P}(X = 1) = \frac{1}{10}$
$\text{P}(X = 2) = \frac{1}{10}$
$\text{P}(X = 3) = \frac{7}{10}$

2. Calculate the expected value $\text{E}(X)$ and variance $\text{Var}(X)$.

$\text{E}(X) = \sum x \,\text{P}(X=x)$
$\text{E}(X) = 0\left(\frac{1}{10}\right) + 1\left(\frac{1}{10}\right) + 2\left(\frac{1}{10}\right) + 3\left(\frac{7}{10}\right) = \frac{0 + 1 + 2 + 21}{10} = 2.4$

$\text{E}(X^2) = \sum x^2 \,\text{P}(X=x)$
$\text{E}(X^2) = 0^2\left(\frac{1}{10}\right) + 1^2\left(\frac{1}{10}\right) + 2^2\left(\frac{1}{10}\right) + 3^2\left(\frac{7}{10}\right) = \frac{0 + 1 + 4 + 63}{10} = 6.8$

$\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$
$\text{Var}(X) = 6.8 - (2.4)^2 = 6.8 - 5.76 = 1.04$

3. Find the standard deviation $\sigma$ and evaluate $\text{P}(X \le \sigma)$.

$\sigma = \sqrt{1.04} \approx 1.0198$

Since $X$ takes non-negative integer values:
$\text{P}(X \le \sigma) = \text{P}(X \le 1.0198)$
$\text{P}(X \le \sigma) = \text{P}(X = 0) + \text{P}(X = 1)$
$\text{P}(X \le \sigma) = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = 0.2$

1. Determine the probabilities and the total revenue collected for each outcome.

From part (b), the probability distribution for the number of empty boxes opened ($X$) is:
$\text{P}(X=0) = \frac{1}{n}$
$\text{P}(X=1) = \frac{1}{n}$
$\text{P}(X=2) = \frac{1}{n}$
$\text{P}(X=3) = \frac{n-3}{n}$

Let $R$ be the revenue collected per game. According to the rules, the revenue for each outcome is:
If $X=0$ (finds doll in 1st box): $R = 5$
If $X=1$ (finds doll in 2nd box): $R = 5 + 3 = 8$
If $X=2$ (finds doll in 3rd box): $R = 5 + 3 + 2.50 = 10.50$
If $X=3$ (3 boxes empty, game lost): $R = 5 + 3 + 2.50 = 10.50$

2. Formulate an equation using the expected revenue and the expected cost.

For Happie to make no profit and incur no loss, the expected revenue per game must equal the cost of the doll used up per game (which is given away or donated either way).
$\text{E}(R) = \text{Cost per game}$
$5\left(\frac{1}{n}\right) + 8\left(\frac{1}{n}\right) + 10.50\left(\frac{1}{n}\right) + 10.50\left(\frac{n-3}{n}\right) = 10$

3. Solve for the number of boxes $n$.

$\frac{5 + 8 + 10.50 + 10.50(n-3)}{n} = 10$
$\frac{23.5 + 10.50n - 31.50}{n} = 10$
$\frac{10.50n - 8}{n} = 10$
$10.50n - 8 = 10n$
$0.50n = 8$
$n = 16$