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 $0.5456$.
 A: Everything other than $6$, $7$, or $8$ is irrelevant. So we can imagine that we are repeating an experiment where the probability of $6$ is $5/16$, as is the probability of $8$, while $7$ has probability $6/16$. 
Condition on the result of the first toss. Suppose it is a $6$. Then $6$'s become irrelevant, and the race is between $7$ and $8$. The first now has probability $\frac{6}{11}$ and the second has probability $\frac{5}{11}$. If we get an $8$, we win. If we get a $7$, then the probability we win is $\frac{5}{11}$. So the contribution to the probability is 
$$\frac{5}{16}\left(\frac{5}{11}+\frac{6}{11}\frac{5}{11}             \right).$$
There is the same contribution to the probability of winning from  first toss is $8$. 
Now we look at what happens if the first toss is $7$. If the next toss is $7$, we have lost. If it is $6$ or $8$, then our probability of winning is  $\frac{5}{11}$. The contribution to the probability is 
$$\frac{6}{16}\left(\frac{10}{16}\frac{5}{11}\right).$$
We conclude that the contribution from $6$ or $8$ first is $\frac{850}{(16)(121)}$, and the contribution from $7$ first is $\frac{300}{(256)(11)}$. Add. The official answer seems to be right.
A: 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 = $1 - 0.4544 = 0.5456$ 
