Defining the states when we roll one single die repeatedly 
We roll a single die and the game stops as soon as the sum of two successive rolls is either 5 or 7.
  We want to find the probability that the game stops at a sum of 5.

It seems like Markov chain with first-step analysis.
To find the transition matrix, I first need to define the states.
They way I define it is that if we have (1,2) then next state should be (2,x) for x=1,2,3,4,5,6.
And (1,2) is different from (2,1). 
So, there must be 36 states?
 A: I'd consider $8$ states, namely for $1\leq k\leq 6$ the states $s_k:\ $"last roll was $k\>$, but game is not yet over", and the two end states $e_5$ and $e_7$. Denote by $x(n)$ the $(1\times8)$ row vector giving the probabilities that after $n$ rolls we are in the state $s_1$, $s_2$, $\ldots\ $, $s_6$, $e_5$, $e_7$ respectively.  It follows that
$$x(1)=\left(h,h,h,h,h,h,0,0\right)\ ,$$
where I have written ${1\over6}=:h$ for typographical simplification. Let $P$ be the transition matrix. Then $p_{ik}$ denotes the probability that when in state $i$ the next roll will move us into state $k$.  One then has $x(n+1)=x(n)P$. The matrix $P$  looks as follows (note that when we are in one of the end states we stay there):
$$P=\left[\matrix{
h&h&h&0&h&0&h&h\cr
h&h&0&h&0&h&h&h\cr
h&0&h&0&h&h&h&h\cr
0&h&0&h&h&h&h&h\cr
h&0&h&h&h&h&0&h\cr
0&h&h&h&h&h&0&h\cr
0&0&0&0&0&0&1&0\cr
0&0&0&0&0&0&0&1\cr}\right]\quad.$$
Computing $x(100)=x(1)P^{99}$ leads to the conjecture that the limiting probabilities for the end states $e_5$ and $e_7$ are ${59\over153}$ and ${94\over153}$.
