TITLE: Solution using Markov chains
QUESTION [0 upvotes]: Two players A and B flip a coin. A starts; the winner is whoever gets heads first.
This problem is easily solved via geometric series yielding $P(A)=\frac{2}{3}$. I wanted to also solve this using Markov Chains, yet is unclear to me how to integrate the advantage of starting.
Any help on this matter?
(With Markov Chains I do not mean the matrix version, but the plain-old version of solving a small system of linear equations.)

REPLY [1 votes]: Just notice that if neither $A$ nor $B$ win on first try, the game starts anew.
$$
\begin{align*}
P[A\ \text{wins}]
&= P[A\ \text{wins on first try}] + P[\text{$A$ and $B$ don't win on first try}] \cdot P[A\ \text{wins}] \\
&= \frac12 + \frac14 \cdot P[A\ \text{wins}].
\end{align*}
$$
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