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
 A: Here is an alternative method of solution. In the following I will write $A\, \text{b.} B$ to denote the event that $A$ occurs before $B$. So you want to calculate $\textbf{P}[ 6\, \text{b.}\, 55]$.
Instead we work with $\textbf{P}[55 \, \text{b.} \, 6]$, and condition on the event $5\, \text{b.}\, 6$. So
\begin{align}
\textbf{P}[55 \, \text{b.} \, 6] & = \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6]\, \textbf{P}[5 \, \text{b.} \, 6]
\\
& = \frac12 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6].
\end{align}
To evaluate the remaining probability, we condition a second time: this time conditioning on what the roll immediately after the $5$ is.
\begin{align}
\textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6] & = \sum_{j=1}^6 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is $j$} ] \textbf{P}[ \text{next roll is $j$}] \\
& = \frac16 \sum_{j=1}^6 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is $j$} ] \\
& = \frac16 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is $6$} ]  \\
& \qquad \qquad + \frac16 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is $5$} ] \\
& \qquad \qquad + \frac46 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is neither $5$ or $6$} ] \\
&= \Big( \frac{1}{6} \times 0 \Big)+ \Big( \frac{1}{6} \times 1 \Big) +\frac46 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is neither $5$ or $6$} ] \\
& = \frac16 + \frac46 \textbf{P}[55 \, \text{b.} \, 6],
\end{align}
in the final lines we used that
$$\textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is neither $5$ or $6$} ] = \textbf{P}[55 \, \text{b.} \, 6],$$
which is true since the information that we saw a $5$ but it wasn't followed by a $5$ doesn't tell us anything about the event $55 \, \text{b.} \, 6$. So substituting this into the first equation we have
\begin{align}
\textbf{P}[55 \, \text{b.} \, 6] & = \frac12 \Big( \frac16 + \frac46 \textbf{P}[55 \, \text{b.} \, 6] \Big) \\
& = \frac{1}{12} + \frac{4}{12}\textbf{P}[55 \, \text{b.} \, 6].
\end{align}
Rearranging gives: $\textbf{P}[55 \, \text{b.} \, 6] = \frac18$, and hence 
$$\textbf{P}[6 \, \text{b.} \, 55] = \frac78.$$ 
Remark. In general this type of problem is more easily phrased in the context of a Markov chain, for which the probability you are wanting becomes a hitting probability. The proof above is essentially exactly the same as would be done using Markov processes, the exception being that the notation is simplified in that context.
A: 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} $
