Two fair n-sided dice, each with faces numbered 1, 2, ..., n, (where $n \ge 4$) are tossed in an experiment. The number that appears on each die is defined as the number on the face that is in contact with the table. If the numbers that appear on the 2 dice are different, the score is taken as the larger of the two numbers that appears and the experiment ends. If the same number appears on both dice, the two dice are rolled again until two different numbers are obtained. A trial is defined as a toss of two dice.

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 $n = 9$.