The question has two parts and is:
Suppose you repeatedly roll a fair six-sided die with numbered faces 1 to 6. Determine the following
- The probablity the $n^\text{th}$ roll produces the first 6
- You roll at least one 5 before the first 6
For part 1. I determined the sample space size, $|\Omega|=6^n$, and designated the event $A$ to be where you don't roll a 6 until the $n^\text{th}$ roll, $|A|=5^{n-1}$. Hence $$P(A)=\frac{5^{n-1}}{6^n}$$.
Then on part 2. I got stuck. I designated the event $B$ to be where you roll at least one 5 before the $n^\text{th}$ roll and the event $C$ to be where you roll at least one 5 before the first 6. Then I concluded that $$C=A\cap B$$ because $A\cap B$ is the event where you both roll at least one 5 before the $n^\text{th}$ roll and a 6 on the $n^\text{th}$ roll. I already have $P(A)$. I found $P(B)$ via $1-P(B^c)$. $B^c$ is the event that you do not roll a 5 before the $n^\text{th}$ roll. So $|B^c|=5^n$ thus $$P(B^c)=\left(\frac56\right)^n\implies P(B)=1-\left(\frac56\right)^n$$ From here I no longer know what to do. I tried inclusion exclusion but I don't know $A\cup B\,$ let alone $P(A\cup B)$ and can't see how to find it.
EDIT
I have now attempted the approach of creating another event $\gamma$ where no 5 is rolled before the first 6. Then $\gamma^c$ will be the event where at least one 5 is rolled before the first 6. Then I calculate the numbers: $|\gamma|=4^{n-1}\cdot\frac16$ because we don't roll a 5 or a 6 before the $n^\text{th}$ roll and a 6 on the $n^\text{th}$ roll. Then $$P(\gamma^c)=1-P(\gamma)=1-\frac{4^{n-1}}{6^{n+1}}$$
Best Answer
Let $q_n$ be the probability that there is at least one 5 before the first 6 on the $n$-th roll.
$q_n = r_n - s_n$ where $r_n$ is the probabilty of rolling the first 6 on the $n$-th roll, and $s_n$ is the probabilty of rolling the first 6 on the $n$-th roll without rolling any 5's before that:
$$ r_n = \frac{1}{6} \left(\frac{5}{6}\right)^{n-1} $$
$$ s_n = \frac{1}{6} \left(\frac{4}{6}\right)^{n-1} $$
The total probability $q$ is an infinite sum:
$$ q = \sum_{n=1}^{\infty} q_n \;\;\;= \;\;\; \frac{1}{6} \sum_{n=1}^{\infty} \left(\frac{5}{6}\right)^{n-1} - \frac{1}{6} \sum_{n=1}^{\infty} \left(\frac{4}{6}\right)^{n-1} \;\;\;=\;\;\;\frac{1}{2} $$