进行一个掷四面体骰子的游戏。骰子的四个面分别标有数字 1、2、3、4。无论骰子着地时朝下的面是什么数字，就投掷相应数量的均匀硬币。玩家的最终得分 $X$ 是硬币正面朝上的数量。

1. Establish the probability model for the number of heads based on the die outcome.

Let $D$ be the outcome on the tetrahedral die, so $\mathrm{P}(D=d) = \frac{1}{4}$ for $d = 1, 2, 3, 4$.
Given $D=d$, $d$ fair coins are tossed. The number of heads $X$ follows a Binomial distribution:
$X|D=d \sim \mathrm{B}(d, 0.5)$

2. Calculate $\mathrm{P}(X=0)$ and $\mathrm{P}(X=1)$ using total probability.

Using $\mathrm{P}(X=x) = \sum_{d=1}^4 \mathrm{P}(D=d)\mathrm{P}(X=x|D=d)$:

3. Calculate $\mathrm{P}(X=2)$, $\mathrm{P}(X=3)$ and $\mathrm{P}(X=4)$.

For $X=2$, note that $D$ must be at least $2$:

For $X=3$, $D$ must be at least $3$:

For $X=4$, $D$ must be $4$:
$\mathrm{P}(X=4) = \frac{1}{4}\left[ 0 + 0 + 0 + \binom{4}{4}\left(\frac{1}{2}\right)^4 \right] = \frac{1}{4}\left(\frac{1}{16}\right) = \frac{1}{64}$

4. Summarize the probabilities into a distribution table.

The probability distribution table for $X$ is:

1. Establish the probability distribution of $X$ as required from part (i).

Given that the die lands on $n \in \{1, 2, 3, 4\}$ with $P(N=n) = \frac{1}{4}$, the number of heads $X$ follows a binomial distribution $B(n, 0.5)$ for a given $n$.
By the law of total probability, $P(X=x) = \sum_{n=1}^{4} P(N=n) P(X=x \mid N=n)$:
$P(X=0) = \frac{1}{4} \left[ \binom{1}{0}0.5^1 + \binom{2}{0}0.5^2 + \binom{3}{0}0.5^3 + \binom{4}{0}0.5^4 \right] = \frac{15}{64}$
$P(X=1) = \frac{1}{4} \left[ \binom{1}{1}0.5^1 + \binom{2}{1}0.5^2 + \binom{3}{1}0.5^3 + \binom{4}{1}0.5^4 \right] = \frac{26}{64}$
$P(X=2) = \frac{1}{4} \left[ 0 + \binom{2}{2}0.5^2 + \binom{3}{2}0.5^3 + \binom{4}{2}0.5^4 \right] = \frac{16}{64}$
$P(X=3) = \frac{1}{4} \left[ 0 + 0 + \binom{3}{3}0.5^3 + \binom{4}{3}0.5^4 \right] = \frac{6}{64}$
$P(X=4) = \frac{1}{4} \left[ 0 + 0 + 0 + \binom{4}{4}0.5^4 \right] = \frac{1}{64}$

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

Using the expected value formula $\text{E}(X) = \sum x P(X=x)$:
$\text{E}(X) = 0\left(\frac{15}{64}\right) + 1\left(\frac{26}{64}\right) + 2\left(\frac{16}{64}\right) + 3\left(\frac{6}{64}\right) + 4\left(\frac{1}{64}\right)$
$\text{E}(X) = \frac{0 + 26 + 32 + 18 + 4}{64} = \frac{80}{64} = \frac{5}{4}$

3. Calculate the variance $\text{Var}(X)$.

First, calculate $\text{E}(X^2) = \sum x^2 P(X=x)$:
$\text{E}(X^2) = 0^2\left(\frac{15}{64}\right) + 1^2\left(\frac{26}{64}\right) + 2^2\left(\frac{16}{64}\right) + 3^2\left(\frac{6}{64}\right) + 4^2\left(\frac{1}{64}\right)$
$\text{E}(X^2) = \frac{0 + 26 + 64 + 54 + 16}{64} = \frac{160}{64} = \frac{5}{2}$
Then, use the variance formula $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$:
$\text{Var}(X) = \frac{5}{2} - \left(\frac{5}{4}\right)^2 = \frac{40}{16} - \frac{25}{16} = \frac{15}{16}$