Find the probablity that a sum of 7 is rolled before a sum of 8 is rolled. Question:
In rolling a pair of a fair dice, what is the probability that a sum of $7 $is rolled before a sum of $8$ is rolled?
I've seen the answer from other's question that I can solve it like this:
Suppose A is the event that '$7$ is rolled'. $B$ is the event that '8 is rolled'.
$$P(A)=\frac{1}{6}$$
$$P(B)=\frac{5}{36}$$
According to total law of probability,
$P(A$ before $B)=P(A$ before$B|A$ happens first$)\cdot P(A$ happens first$)+P(A $before $B|B$ happens first$)\cdot P(B$ happens first$)+P(A$ before $B|A$ happens first$)\cdot P($neither A nor B happens first$)$
Suppose $P(A$ before $B)=r$
We can get an equation $$r=1\cdot\frac{1}{6}+0\cdot\frac{5}{36}+r\cdot(1-\frac{1}{6}-\frac{5}{36})$$
Therefore
$$r=\frac{6}{11}$$
But can I also solve this problem using geometric series?
Here's my solution:
Let $C$ be the event '$7$ is rolled at $n$-th trial' and $D$ be the event '$8$ is rolled after $n$-th trial'.
Hence
$P(A$ before $B)=\sum_{n=1}^{\infty}(\frac{5}{6})^{n-1}\cdot\frac{1}{6}\cdot(\frac{31}{36})^{n}=0.508$
So which of this two methods should I use to figure out the probability?
 A: Just look at the first time that a seven or an eight is rolled and neglect everything else.
For that situation you must calculate the probability that a seven shows up.
So actually to be calculated is the probability: $$\mathsf P(\text{seven rolled}\mid\text{seven rolled or eight rolled})=\frac{\frac6{36}}{\frac6{36}+\frac5{36}}=\frac6{11}$$
A: \begin{align}
P(A \text{  before } B)&= \sum_{n=1}^\infty \left(1-P(A)-P(B) \right)^{n-1}P(A)
\end{align}
That is in the first $n-1$ trials, neither $A$ nor $B$ happens but at the $n$-th toss $A$ happens. Since $B$ has not happened, $A$ happens before $B$.
A: Here's how to do it with a Markov chain.
Take states "Neither has come", "7 has come" and "8 has come", the last two being absorping states.
The transition matrix is
$$ P =\frac{1}{36}
\left[
\begin{array}*
25 & & \\
6 & 1 & \\
5 & & 1
\end{array}\right]
$$
The fundamental matrix $N$ is just $(1-\frac{25}{36})^{-1} = \frac{36}{11}$.
We get the absorbtion probabilities from $B = NR$, where $R=\frac{1}{36}[\begin{array}* 6 & 5  \end{array}]$. So it comes out
$$B=[\begin{array}* \frac{6}{11} & \frac{5}{11}  \end{array}].$$
Here the first value $\frac{6}{11}$ means the chain will be absorbed to the state "7 has come", which means that we get a $7$ before we get an $8$ (probability of absorption is $1$ so the chain will be absorped to one of the absorping states and hence we can make that intepretation).
