equivalent states in a markov chain Two players bet on the sum of two standard six-face dice. Player A bets that the sum of 12 will occur first. Player B bets that two consecutive sum of 7s will occur first. What is the probability that A will win? Here is the same question Probability you get $12$ before two consecutive $7$s, but I do not understand the reasoning presented.
The solution I am looking at draws only four states:

*

*the starting state

*the state where the sum is 7

*the state where we get two consecutive 7-7

*the state where the sum is 12

The solution also claims that "you can use all combinations of the outcomes of the one roll and two consecutive rolls as states to construct a transition matrix and you will get the same final result"
It seems to be suggesting that to define a state we need to look at both the current sum and two consecutive sums.
1). What sequences of random variables should we be looking at to define the state here?
2). Why do we not care about states like the sum of the current rolls is 6?
 A: It might start to imagine how a game might be played.  Imagine that the rolls occur:
\begin{align}
1+5&=6 && \text{neither player cares} \\
2+5&=7 && \text{player B wins if the next roll is a $7$} \\
1+1&=2 && \text{player B doesn't end; the game resets} \\
5+1&=6 && \text{neither player cares} \\
5+6&=11 && \text{neither player cares} \\
6+6&=12 && \text{player A wins}
\end{align}
From this, it seems that there are only four states which actually matter:

*

*Player B wins if the next roll is a $7$ (which means that the last roll was also a $7$).

*Player B has won (the previous two rolls were $7$s).

*Player A has won (the previous roll was a $12$).

*Anything else (the previous roll was neither a $7$ nor a $12$).

Note that I am not thinking of the dice rolls as the states, but rather the thing which drives the transitions between states.  This is kind of a distinction without a difference, but I think that it helps to understand what is going on.
Transition probabilities work as follows:

*

*If the system is in state 1, there are three possible outcomes:  either a $7$ is rolled and the system transitions to state 2 and the game ends (this happens with probability $1/6$); a $12$ is rolled and the game ends (this happens with probability $5/6$); or anything else is rolled and the system transitions to state 4 (this happens with probability $29/36$).

*If the system is in state 2, the game is over.  It is impossible for the system to transition away from this state; that is, the system remains in state 2 with probability 1.

*If the system is in state 3, the game is over.  It is impossible for the system to transition away from this state; that is, the system remains in state 3 with probability 1.

*If the system is in state 4, then it transitions to state 1 with probability $1/6$ (if a $7$ is rolled); it transitions to state 3 with probability $1/36$ (if a $12$ is rolled); or it remains in state 4 with probability $29/36$ (if anything else is rolled).

Thus the transition matrix is given by
$$\begin{pmatrix}
0 & \frac{1}{6} & \frac{1}{36} & \frac{29}{36} \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\frac{1}{6} & 0 & \frac{1}{36} & \frac{29}{36}
\end{pmatrix}.$$
