一家新奇玩意儿商店的老板Happie正在为他的商店组织一场宣传活动。利用他购得的一批超稀有Lubaba玩偶，他设计了一个游戏，玩家通过打开“神秘盒子”来尝试寻找隐藏在众多完全相同的游戏盒子之中的一个Lubaba玩偶。 游戏规则如下：

• 玩家首付 $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。
• If the second box is empty, the player can pay $2.50 again to open a third box.
• 如果第三个盒子依然是空的，玩家输掉游戏。未被赢走的玩偶将被捐赠，游戏将重置以供下一位玩家游玩。

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赢得玩偶，并且他们有能力支付最多允许的 3 次开盒尝试。 设随机变量 $X$ 表示一个玩家在一局游戏中打开的空盒子数量。

1. Apply the multiplication principle for dependent events to find the probability of opening 2 empty boxes followed by the box with the doll.

$\begin{aligned} P(X=2) &= P(\text{1st box empty}) \times P(\text{2nd box empty}) \times P(\text{3rd box has doll}) \\ &= \frac{n-1}{n} \times \frac{n-2}{n-1} \times \frac{1}{n-2} \\ &= \frac{1}{n} \end{aligned}$

1. Identify the possible values of $X$ and calculate $P(X=0)$ and $P(X=1)$.

Since a player can open a maximum of 3 boxes, the possible numbers of empty boxes opened ($X$) are $0, 1, 2,$ and $3$.

If $X=0$, the first box contains the doll:
$P(X=0) = \frac{1}{n}$

If $X=1$, the first box is empty and the second box contains the doll:
$P(X=1) = \frac{n-1}{n} \times \frac{1}{n-1} = \frac{1}{n}$

2. Determine $P(X=3)$ and present the probability distribution table.

Since the sum of all probabilities in a probability distribution must equal $1$, and we are given $P(X=2) = \frac{1}{n}$ from part (a):

$P(X=3) = 1 - P(X=0) - P(X=1) - P(X=2)$
$P(X=3) = 1 - \frac{1}{n} - \frac{1}{n} - \frac{1}{n} = \frac{n-3}{n}$

(Alternatively, calculating directly: $P(X=3) = \frac{n-1}{n} \times \frac{n-2}{n-1} \times \frac{n-3}{n-2} = \frac{n-3}{n}$)

The probability distribution of $X$ is:

1. Evaluate the probability distribution of $X$ for $n=10$.

From part (b), the probability distribution for $X$ is:
$P(X=0) = \frac{1}{n} = \frac{1}{10} = 0.1$
$P(X=1) = \frac{1}{n} = \frac{1}{10} = 0.1$
$P(X=2) = \frac{1}{n} = \frac{1}{10} = 0.1$
$P(X=3) = \frac{n-3}{n} = \frac{7}{10} = 0.7$

2. Calculate the expectation, variance, and standard deviation of $X$.

$E(X) = 0(0.1) + 1(0.1) + 2(0.1) + 3(0.7) = 2.4$
$E(X^2) = 0^2(0.1) + 1^2(0.1) + 2^2(0.1) + 3^2(0.7) = 6.8$
$\text{Var}(X) = E(X^2) - [E(X)]^2 = 6.8 - 2.4^2 = 1.04$
$\sigma = \sqrt{1.04} \approx 1.0198$

3. Evaluate $P(X \le \sigma)$.

Since $X$ only takes discrete integer values,
$P(X \le 1.0198) = P(X=0) + P(X=1)$
$= 0.1 + 0.1 = 0.2$

1. Determine the total revenue collected from a player for each possible value of $X$.

Let $R$ be the revenue collected per game.
If $X=0$, the player wins on the first box and pays $\char36{}5$.
If $X=1$, the player wins on the second box and pays $\char36{}5 + \char36{}3 = \char36{}8$.
If $X=2$, the player wins on the third box and pays $\char36{}5 + \char36{}3 + \char36{}2.50 = \char36{}10.50$.
If $X=3$, the player loses after opening $3$ empty boxes and pays $\char36{}5 + \char36{}3 + \char36{}2.50 = \char36{}10.50$.

2. Set up the equation for the expected revenue.

From the probability distribution of $X$ found in (b), $P(X=0)=\frac{1}{n}$, $P(X=1)=\frac{1}{n}$, $P(X=2)=\frac{1}{n}$, and $P(X=3)=\frac{n-3}{n}$.
The expected revenue is:
$\text{E}(R) = 5\left(\frac{1}{n}\right) + 8\left(\frac{1}{n}\right) + 10.5\left(\frac{1}{n}\right) + 10.5\left(\frac{n-3}{n}\right)$
$\text{E}(R) = \frac{5 + 8 + 10.5 + 10.5(n-3)}{n}$
$\text{E}(R) = \frac{23.5 + 10.5n - 31.5}{n} = \frac{10.5n - 8}{n}$

3. Equate the expected revenue to the cost of the doll and solve for $n$.

Since one doll is either won or donated away in every game, the expected cost per game is always $\char36{}10$.
For Happie to make no profit and incur no loss, the expected revenue must equal the cost per game ($\text{E}(R) = 10$).
$\frac{10.5n - 8}{n} = 10$
$10.5n - 8 = 10n$
$0.5n = 8$
$n = 16$