在一次实验中，掷两个均匀的 $n$ 面骰子，每个骰子的面上分别标有数字 1，2，...，$n$（其中 $n \ge 4$）。每个骰子出现的数字定义为与桌面接触的那个面上的数字。如果两个骰子出现的数字不同，则得分记为两个数字中较大的那一个，实验结束。如果两个骰子出现相同的数字，则重新投掷这两个骰子，直到得到两个不同的数字。 一次尝试定义为投掷一次这两个骰子。

1. Calculate the probability that the first trial results in the same number on both dice.

There are $6$ outcomes where both dice show the same number: $(1, 1), (2, 2), \dots, (6, 6)$.
$\text{P}(\text{same number in 1st trial}) = \frac{6}{36} = \frac{1}{6}$

2. Calculate the probability of obtaining a score of 3 in a single trial.

For the score to be $3$, the numbers must be different and the larger number must be $3$.
The possible outcomes are $(1, 3), (2, 3), (3, 1),$ and $(3, 2)$.
$\text{P}(\text{score is 3 in 2nd trial}) = \frac{4}{36} = \frac{1}{9}$

3. Multiply the probabilities of the two independent trials to find the overall probability.

$\text{P}(\text{score of 3 with two trials}) = \text{P}(\text{same number in 1st trial}) \times \text{P}(\text{score is 3 in 2nd trial})$
$= \frac{1}{6} \times \frac{1}{9}$
$= \frac{1}{54} \text{ (Shown)}$

1. Find the probability of obtaining a score of 3 on a single trial, and the probability of a re-roll.

A score of 3 is obtained if the dice show $(1,3)$, $(3,1)$, $(2,3)$, or $(3,2)$.
$P(\text{score 3 on one trial}) = \frac{4}{36} = \frac{1}{9}$
A re-roll happens if both dice show the same number (e.g. $(1,1), (2,2), \dots, (6,6)$).
$P(\text{re-roll}) = \frac{6}{36} = \frac{1}{6}$

2. Express the total probability as an infinite geometric series representing the score being achieved on the 1st, 2nd, 3rd trial, and so on.

$P(\text{score is 3}) = P(\text{score 3 on 1st trial}) + P(\text{score 3 on 2nd trial}) + \dots$
$P(\text{score is 3}) = \frac{1}{9} + \left(\frac{1}{6}\right)\left(\frac{1}{9}\right) + \left(\frac{1}{6}\right)^2\left(\frac{1}{9}\right) + \dots$
This forms an infinite geometric series with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{6}$.

3. Evaluate the sum to infinity of the geometric series to find the overall probability.

$S_\infty = \frac{a}{1 - r}$
$P(\text{score is 3}) = \frac{\frac{1}{9}}{1 - \frac{1}{6}}$
$P(\text{score is 3}) = \frac{\frac{1}{9}}{\frac{5}{6}} = \frac{1}{9} \times \frac{6}{5} = \frac{6}{45}$
$P(\text{score is 3}) = \frac{2}{15}$

1. Calculate the probability that the score is 3 overall (from part (a)(ii)), which forms the denominator of the conditional probability.

Let $P(\text{same})$ be the probability of rolling the same number on both dice in a single trial, and $P(\text{score } 3)$ be the probability of rolling a score of 3 in a single trial.
$P(\text{same}) = \frac{6}{36} = \frac{1}{6}$
$P(\text{score } 3 \text{ in a single toss}) = P(\{ (1,3), (3,1), (2,3), (3,2) \}) = \frac{4}{36} = \frac{1}{9}$

Summing the probabilities over all possible numbers of trials to find the overall probability that the final score is 3:
$P(\text{overall score is 3}) = \frac{1}{9} + \frac{1}{6}\left(\frac{1}{9}\right) + \left(\frac{1}{6}\right)^2\left(\frac{1}{9}\right) + \dots = \frac{\frac{1}{9}}{1 - \frac{1}{6}} = \frac{2}{15}$

2. Calculate the probability of obtaining exactly four trials and a score of 3.

For exactly 4 trials, the first 3 trials must result in the same number on both dice, and the 4th trial must result in a score of 3.
$P(\text{exactly 4 trials} \cap \text{score is 3}) = \left(\frac{1}{6}\right)^3 \times \frac{1}{9} = \frac{1}{216} \times \frac{1}{9} = \frac{1}{1944}$

3. Apply the conditional probability formula to find the required probability.

$P(\text{exactly 4 trials} \mid \text{score is 3}) = \frac{P(\text{exactly 4 trials} \cap \text{score is 3})}{P(\text{overall score is 3})}$
$= \frac{\frac{1}{1944}}{\frac{2}{15}} = \frac{1}{1944} \times \frac{15}{2} = \frac{15}{3888} = \frac{5}{1296}$

1. Find the probability of obtaining a score of 3, and the probability of a tie, in a single trial.

In a single trial, to get a score of 3, the two dice must show different numbers and their maximum must be 3. The possible outcomes are $(1, 3), (3, 1), (2, 3)$, and $(3, 2)$.
$P(\text{score 3 in a trial}) = \frac{4}{n^2}$
For the experiment to not end (i.e. to proceed to the next trial), both dice must show the same number. The outcomes are $(1, 1), (2, 2), \dots, (n, n)$.
$P(\text{same number in a trial}) = \frac{n}{n^2} = \frac{1}{n}$

2. Set up the infinite geometric series for the overall probability of scoring a 3.

The overall probability of scoring a 3 is the sum of the probabilities of this event happening on the 1st trial, the 2nd trial, the 3rd trial, and so on:
$P(\text{score is } 3) = \frac{4}{n^2} + \left(\frac{1}{n}\right)\left(\frac{4}{n^2}\right) + \left(\frac{1}{n}\right)^2\left(\frac{4}{n^2}\right) + \dots$
This is an infinite geometric series with first term $a = \frac{4}{n^2}$ and common ratio $r = \frac{1}{n}$.

3. Evaluate the sum to infinity and compare it to the given expression to determine $a$ and $b$.

$S_\infty = \frac{a}{1 - r} = \frac{\frac{4}{n^2}}{1 - \frac{1}{n}}$
$S_\infty = \frac{\frac{4}{n^2}}{\frac{n-1}{n}} = \frac{4}{n^2} \times \frac{n}{n-1} = \frac{4}{n(n-1)}$
Comparing this with the given expression $\frac{a}{n(n-b)}$, we can see that $a = 4$ and $b = 1$.

1. Determine the values of $a$ and $b$ by finding the probability of scoring a 3 (from the previous part).

The experiment ends only when the two dice show different numbers. There are $n^2 - n = n(n-1)$ such pairs, and by symmetry, each is equally likely to be the final outcome.
A score of $3$ occurs for the outcome pairs $(1,3)$, $(3,1)$, $(2,3)$ and $(3,2)$, giving $4$ favourable outcomes.
Thus, the probability of a score of $3$ is $\frac{4}{n(n-1)}$.
Comparing this to $\frac{a}{n(n-b)}$, we have $a = 4$ and $b = 1$.

2. Set up an inequality representing the required probability condition.

We are given that the probability must be at least $5\%$.
$\frac{4}{n(n-1)} \ge 0.05$
$\frac{4}{n(n-1)} \ge \frac{1}{20}$

3. Solve the quadratic inequality to find the maximum integer value of $n$.

Since $n \ge 4$, we know that $n(n-1) > 0$. Multiplying both sides by $20n(n-1)$ gives:
$80 \ge n^2 - n$
$n^2 - n - 80 \le 0$
Solving the quadratic equation $n^2 - n - 80 = 0$:
$n = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-80)}}{2} = \frac{1 \pm \sqrt{321}}{2}$
The positive root evaluates to $n \approx 9.458$.
For the inequality to hold, we require $n \le 9.458$.
Since $n$ must be an integer, the greatest number of faces is $n = 9$.