A and B throw a pair of die alternatively. A wins if he throws $6$ (i.e. his thrown die sum up to $6$) before B throws $7$, and B wins if he throws $7$ before A throws $6$. If first throw is done by A, find the probability that A wins the game?
Edit: Ops! I misunderstood the question in the way that A wins if he throws $6$ and immediately in the next trial, B throws $7$, and similar condition for B to win. Thanks to the answer giver. But still, how to solve it with this alternate interpretation? Thank you and sorry again!
Method I (Recursion). Three things can happen in any given round...$A$ wins, $A$ loses, the game starts over. Thus the answer, $X$, is given by the recursive formula $$X=\frac 5{36}\times 1+\frac {31}{36}\times \frac 6{36}\times 0 + \left(1-\frac 5{36}-\frac {31}{36}\times \frac 6{36}\right)X\implies \boxed {X=\frac {30}{61}}$$
Method II (Markov states) : This game has two non-final states: "$A's$ turn, $B's$ turn" Denote the probability that $A$ eventually wins given that it's $A's$ turn as $X$, and the probability that $A$ eventually wins given that it's $B's$ turn as $Y$.
Clearly: $$X=\frac 5{36}\times 1 +\frac {31}{36}\times Y$$ and
$$Y=\frac 6{36}\times 0 +\frac {30}{36}\times X$$
This is easily solved and yields our answer $$\boxed {X=\frac {30}{61}}$$
Method III: The game will end on some round with probability $1$. On any given round the winner will be $A$ with probability $\frac 5{36}$ and $B$ with probability $\frac {31}{36} \times \frac {6}{36}$. Thus the eventual winner is $A$ with probability $$\frac {\frac 5{36}}{\frac 5{36}+\frac {31}{36}\times \frac {6}{36}}=\frac {5\times 36}{5\times 36+31\times 6}=\boxed {\frac {30}{61}}$$