Here is a question from a probability textbook I am working through:
Two players bet on roll(s) of the total of two standard six-face dice. Player $A$ bets that a sum of $12$ will occur first. Player $B$ bets that two consecutive $7$'s will occur first. The players keep rolling the dice and record the sums until one player wins. What is the probability that $A$ will win?
I got the two equations$$p_0 = {1\over{36}} + {1\over6}p_1 + {{29}\over{36}}p_0, \quad p_1 = {1\over{36}} + {{29}\over{36}}p_0.$$Solving for $p_0$, I ended up with $p_0 = {5\over{13}}$. However, the answer in my book is $p_0 = {7\over{13}}$. Which is correct?
This can be modelled as an absorbing Markov chain with four states: "normal", "one $7$ appeared", "A won" and "B won". The transition matrix is $$\begin{bmatrix} 29/36&29/36&0&0\\ 1/6&0&0&0\\ 1/36&1/36&1&0\\ 0&1/6&0&1\end{bmatrix}$$ Diagonalising to take the limit of powers of this matrix, then right-multiplying by the initial state of $(1,0,0,0)^T$, yields $(0,0,7/13,6/13)^T$ – i.e. A wins with probability $\frac7{13}$ as per the provided answer.