Probability of a specific sequence I was preparing for my probability exam and i came across this problem. A fair dice is rolled until either 6 comes up or two $5$ comes up in a row.
For example some of the possible outcomes are 
$$52546, 6, 36, 3251541355, 55, 255 \,\, \text {etc} $$
Find the probability that the last number to come up is a $6$.
I have attempted to solve this problem this way.
Let $X$ be the number of times the die was rolled to obtain two $5$ in a row and $Y$ be the number of times the die was rolled to obtain a $6$. Intuitively i thought that there could be a  $c$ such that $c(P(Y=n)) = P(X=n)$ . Then 
$$1=\sum_{n=1}^\infty P(X=n)+P(Y=n)=\sum_{n=1}^\infty cP(Y=n)+P(Y=n)=(1+c)\sum_{n=1}^\infty P(Y=n)$$
$$\implies \sum_{n=1}^\infty P(Y=n)=\frac {1} {1+c}$$
However i am unable to find such a constant $c$. Any help / insight is appreciated.
Thanks

The absorbing chain matrix is
$$\left(
\begin{array}{ccccc}
 \text{} & o & 5 & 55 & 6 \\
 o & \frac{2}{3} & \frac{1}{6} & 0 & \frac{1}{6} \\
 5 & \frac{2}{3} & 0 & \frac{1}{6} & \frac{1}{6} \\
 55 & 0 & 0 & 1 & 0 \\
 6 & 0 & 0 & 0 & 1 \\
\end{array}
\right)$$
The states have been labeled o which represents a throw of (1,2,3,4), 5 which is the first 5 thrown, 55 which is 2 5's in a row and 6 which is the first 6 thrown.
$$ \text{Q=}\left(
\begin{array}{cc}
 \frac{2}{3} & \frac{1}{6} \\
 \frac{2}{3} & 0 \\
\end{array}
\right)$$
$$\text{R=}\left(
\begin{array}{cc}
 0 & \frac{1}{6} \\
 \frac{1}{6} & \frac{1}{6} \\
\end{array}
\right)$$
$$\text{0=}\left(
\begin{array}{cc}
 0 & 0 \\
 0 & 0 \\
\end{array}
\right)$$
Now B is the probability of ending in some absorbing state (first row) starting in some transient state (first column).
$$\text{B=}\left(
\begin{array}{ccc}
 \text{} & 55 & 6 \\
 o & \frac{1}{8} & \frac{7}{8} \\
 5 & \frac{1}{4} & \frac{3}{4} \\
\end{array}
\right)$$
When we check the element o,6 we see the answer of $ \frac{7}{8} $
