Probability and counting dice roll 
Two fair six-sided dice are rolled repeatedly until a score of 7, 9, or 11 is obtained.
You win if a score of 9 is the first of these three scores to occur.
Determine the probability that you win.

This is what I've done so far: Since probability is equally likely then
$\frac {\text{# outcome in $E$}}{\text{outcomes in $S$}}$
since order is important and the die is being replaced after each rolled then I used the formula $n^k$ so the number of outcome in $S$ is equal to $6^2=36$ where $n=6$ and $k=2$.
My problem is finding out the number of outcome in $E$.
 A: 
This is what I've done so far: Since probability is equally likely then
  $\frac {\text{# outcome in $E$}}{\text{outcomes in $S$}}$ 
  since order is important and the die is being replaced after each rolled then I used the formula $n^k$ so the number of outcome in $S$ is equal to $6^2=36$ where $n=6$ and $k=2$. 
  My problem is finding out the number of outcome in $E$.

No, $36$ is the count of all possible outcomes, but we are only interested in a restricted set of outcomes.
Let $n(7)$ count the (equally likely) outcomes of rolling two dice that result in a sum of $7$. These are $\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$, so $n(7)=6$.  Similiarly there are $n(9)$ outcomes that result in a sum of $9$, and $n(11)$ that result in a sum of $11$.  We ignore all other outcomes as the experiment is repeated until the required condition occurs (That is: one of these three events) .
The favoured event space, $E$, contains all the outcomes that result in a sum of $9$.
Likewise the conditioned space, $S$, contains all the outcomes that result in a sum of $7, 9,$ or $11$.
So you have $$\frac{|E|}{|S|} = \frac {n(9)}{n(7)+n(9)+n(11)}$$
A: For variety, here is a different approach involving a recurrence. 
Denote the probability of winning by "$P(w)$". Since there are $4$ out of $36$ ways of rolling a $9$, we win $\frac{4}{36}$ of the time on the first roll. The chance we lose on the first roll is $\frac{6}{36} + \frac{2}{36}$. So, the chance we don't win and don't lose on the first roll is $\frac{24}{36}$. 
So, the chance of winning after not-winning and not-losing on the first roll is $\frac{24}{36} \cdot P(w)$. Hence, our total chance of winning is
$P(w) = \frac{4}{36} + \frac{24}{36} \cdot P(w)$
Solving for $P(w)$, we get $\frac{1}{3}$.
