Probability of seeing one event before another when rolling two dice 
You roll two six-sided dice simultaneously and keep track of the sums.
What's the probability that we see sums $6$ and $8$ prior to seeing
the sum $7$ twice?

I know that the probability that the sum of $6, 7, 8$ are $5/36, 1/6, 5/36$, respectively. However, I don't know how to compute the probability of seeing both $6$ and $8$ prior to seeing $7$.
I thought about keeping a Markov chain with eight states: $000, 001, \ldots, 111$, where the first bit is on if we see $6$, second bit is on if we see $7$, etc.
Under my interpretation, we start in $000$ and want the probability of visiting $101$ before visiting $010$ twice. I'm not able to figure this out. Can someone please help me? Alternatively, is there an easier approach (without Markov chains?)
Thanks
 A: Suppose you roll the two dice until seeing 6 or 7 or 8. Call this the "first sum." There are three ways an outcome in the complement may occur:

*

*If the first sum you see is 7, continue rolling until seeing 6 or 7 or 8. Suppose this second sum is 7. This has probability
$$ \frac{6}{16} \cdot \frac{6}{16}, $$
because there are 6 ways to roll a sum of 7, and there are 16 ways to roll a sum of 6, 7, or 8.

*If the first sum you see is 7, continue rolling until seeing 6 or 7 or 8. Suppose this second sum is 6 or 8. Continue rolling until seeing 7 or the remaining number. Suppose this third sum is 7. This has probability
$$ \frac{6}{16} \cdot \frac{10}{16} \cdot \frac{6}{11} .$$

*If the first sum you see is 6 or 8, continue rolling until seeing 7 or the remaining number. Suppose this second sum is 7. Again continue rolling until seeing 7 or the remaining number. Suppose this third sum is 7. This has probability
$$ \frac{10}{16} \cdot \frac{6}{11} \cdot \frac{6}{11} .$$
Calculating each of these and subtracting them from 1 then gives Sil's answer, $\frac{4225}{7744}$.
