Seemingly tricky dice question-probability that one event occurs before another event? The question is as follows: 

You roll two fair dice over and over.  Let $A$ be the event you see two even sums.  Let $B$ be the event you see a sum of $7$ four times.  What is the probability that event $A$ occurs before event $B$?  

I know that for mutually exclusive events with independent trials, the probability that event $E$ occurs before event $F$ is 
$$\frac{\mathbb{P}(E)}{\mathbb{P}(E) + \mathbb{P}(F)}.$$
I tried using this formula, but I ran into a problem.  I calculated 
$$\mathbb{P}(A)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4} \ \ \ \  \text{and} \ \ \ \  \mathbb{P}(B)=\left ( \frac{1}{6} \right )^4=\frac{1}{1296}.$$
However, I then realized that these probabilites are the events that two even sums occur $\textit{in a row},$ and similarly for my $\mathbb{P}(B).$
Does anyone have any suggestions on how to figure this out, or even if this formula is the one I should be using? 
Thanks in advance! 
Edit: I know there is a geometric distribution involved. 
 A: I would do a Markov chain with the states being the number of even rolls/the number of seven rolls.  Work backwards starting from the state that you have seen one even roll and three $7$s.  What is the chance that A comes first from there?
A: You have two Bernoulli processes,$S_A$ and $S_B$, and you are asked about the probability that the 2nd arrival in $S_A$ process occurs before the 4th arrival in $S_B$ process. PMF of the time of $k$-th arrival in a Bernoulli process with probability of success $p$ is
$$p_{X_k}(t)=\binom{t-1}{k-1}p^k(1-p)^{t-k}, \ t=k,k+1,\ldots$$

PS: the events "sum is even" and "sum is 7" are dependent, and it makes the question tricky indeed, and the trick is, as Dukeling explained, is to  ignore odd sums not equal to $7$; then the (conditional) probability of getting the sum of $7$ is $\frac{6}{36-12}=\frac{1}{4}$. Now the complement of the event of interest is the event that the 4th arrival in the Bernoulli process with $p=1/4$ occurs at time $4$ or $5$, and its probability is equal to
$$\binom{3}{3}p^4+\binom{4}{3}p^4(1-p)=p^4(1+4(1-p))=\frac{1}{64}$$
and the final answer is
$$1-\frac{1}{64}=\frac{63}{64}$$
