You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning? You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning? 
What I tried was 
\begin{align*}
 P(X=6) & = \frac{5}{36}\\
 P(X=8) & = \frac{5}{36}\\
 P(X=7) & = \frac{6}{36} = \frac{1}{6}
\end{align*}
I try finding the probability of getting $6$ and $8$ and not $7$
$$P(X=6)+P(X=8)+P(X \neq 7) = \frac{5}{26} + \frac{5}{36} + \frac{5}{6} = 1.11$$ 
I stick since I know that the probability cannot be greater than one. The solution key says the answer is $0.5456$.

It is easier to  compute the probability of first getting two $7's\;\;$ and take the complement
Totals other than $6,7,8$ don't matter, so we can simply take their odds in favor as $5:6:5$
Consider the three  mutually exclusive starts $(a)\;\;7-7,\;\;(b)\;\; 7-(6\;\;or\;\; 8),\;\;(c)\;\; (6\;\; or\;\; 8)-7$
In $(a),\;\; 7$ wins with $Pr = \frac{6}{16}\frac6{16}$
For $(b)\;\;and\;\;(c)$, note that for subsequent throws, it becomes a direct race between $7$ and whichever of $6$ and $8$ has not yet appeared, thus the odds in favor of $7$ becomes $6:5$
In $(b),\;\; 7\;\;$ wins with $Pr = \frac6{16}\frac{10}{16}\frac6{11}$
In $(c),\;\; 7\;\;$ wins with $Pr = \frac{10}{16}\frac6{11}\frac6{11}$
Adding up, $7$ wins with $Pr = 0.4544$
Finally, answer = $1 - 0.4544 = 0.5456$ 
