在一个游戏中，玩家投掷一枚各个面分别标有 1 到 6 的均匀骰子。如果玩家掷出 6，他将投掷第二次，在这种情况下，他的得分是 6 与第二次掷出数字之差的绝对值。否则，他的得分就是第一次掷出的数字。设玩家的得分为 $X$。

1. Calculate the probability that the score is 1.

The score is 1 in two mutually exclusive cases: getting a 1 on the first toss, or getting a 6 on the first toss followed by a 5 on the second toss (since $|6 - 5| = 1$).
$\mathrm{P}(X=1) = \mathrm{P}(\text{1st toss is } 1) + \mathrm{P}(\text{1st toss is } 6 \text{ and 2nd toss is } 5)$
$\mathrm{P}(X=1) = \frac{1}{6} + \left(\frac{1}{6} \times \frac{1}{6}\right)$
$\mathrm{P}(X=1) = \frac{6}{36} + \frac{1}{36} = \frac{7}{36} \text{ (shown)}$

2. Determine the probabilities for all other possible values of $X$.

The possible values of $X$ are $0, 1, 2, 3, 4, 5$.
For $x \in \{2, 3, 4, 5\}$, the calculation is identical to $X=1$ (getting $x$ on the first toss, or getting 6 then $6-x$):
$\mathrm{P}(X=x) = \frac{1}{6} + \left(\frac{1}{6} \times \frac{1}{6}\right) = \frac{7}{36}$
For $X=0$, the only way is to get a 6 on the first toss and a 6 on the second toss (since $|6 - 6| = 0$):
$\mathrm{P}(X=0) = \mathrm{P}(\text{1st toss } 6) \times \mathrm{P}(\text{2nd toss } 6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

3. Tabulate the probability distribution of $X$.

1. Retrieve the probability distribution of $X$ from part (i).

Based on the rules of the game:
$\mathrm{P}(X=0) = \mathrm{P}(6 \text{ then } 6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$
For $x \in \{1, 2, 3, 4, 5\}$:
$\mathrm{P}(X=x) = \mathrm{P}(\text{first toss is } x) + \mathrm{P}(\text{first toss is } 6 \text{ and } |6 - \text{second toss}| = x)$
$\mathrm{P}(X=x) = \frac{1}{6} + \left(\frac{1}{6} \times \frac{1}{6}\right) = \frac{7}{36}$

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

$\mathrm{E}(X) = \sum x \mathrm{P}(X=x)$
$\mathrm{E}(X) = 0\left(\frac{1}{36}\right) + 1\left(\frac{7}{36}\right) + 2\left(\frac{7}{36}\right) + 3\left(\frac{7}{36}\right) + 4\left(\frac{7}{36}\right) + 5\left(\frac{7}{36}\right)$
$\mathrm{E}(X) = \frac{7}{36}(1 + 2 + 3 + 4 + 5)$
$\mathrm{E}(X) = \frac{7}{36}(15) = \frac{105}{36} = \frac{35}{12}$

1. Identify the binomial distribution and state the required probability expression.

From part (i), the probability of obtaining a score of 5 in a single game is $\mathrm{P}(X = 5) = \frac{1}{6} + \left(\frac{1}{6} \times \frac{1}{6}\right) = \frac{7}{36}$.
Let $Y$ be the number of times Mr Lim obtains a score of 5 out of 4 games.
$Y \sim \mathrm{B}\left(4, \frac{7}{36}\right)$
We need to find the probability of getting a score of 5 no more than 2 times, which is $\mathrm{P}(Y \le 2)$. This can be calculated as $1 - \mathrm{P}(Y = 3) - \mathrm{P}(Y = 4)$.

2. Calculate the probability.

$\begin{aligned} \mathrm{P}(Y \le 2) &= 1 - \binom{4}{3}\left(\frac{7}{36}\right)^3\left(\frac{29}{36}\right) - \binom{4}{4}\left(\frac{7}{36}\right)^4 \\ &= 1 - 4\left(\frac{343}{46656}\right)\left(\frac{29}{36}\right) - \frac{2401}{1679616} \\ &= 1 - \frac{39788}{1679616} - \frac{2401}{1679616} \\ &= \frac{1637427}{1679616} \\ &\approx 0.975 \text{ (to 3 s.f.)} \end{aligned}$

1. Identify the possible score combinations that give a total score of less than 2

Since the possible scores in each game are non-negative integers ($X \ge 0$), the total score $T$ in 4 games is less than 2 if and only if $T = 0$ or $T = 1$.

• $T = 0$: All 4 games result in a score of 0.
• $T = 1$: Exactly 1 game results in a score of 1, and the other 3 games result in a score of 0 (which can occur in $\binom{4}{1} = 4$ ways).

2. Calculate the probabilities for each case and find the total probability

From part (i), $\mathrm{P}(X = 0) = \frac{1}{36}$ and $\mathrm{P}(X = 1) = \frac{7}{36}$.

$\mathrm{P}(T < 2) = \mathrm{P}(T = 0) + \mathrm{P}(T = 1)$
$\mathrm{P}(T < 2) = \left(\frac{1}{36}\right)^4 + 4 \times \left(\frac{1}{36}\right)^3 \left(\frac{7}{36}\right)$
$\mathrm{P}(T < 2) = \frac{1}{1679616} + \frac{28}{1679616}$
$\mathrm{P}(T < 2) = \frac{29}{1679616}$