I have the following problem:
Two men are shooting at a target. We can call them $X$ and $Y$. $X$ shoots after each hit and $Y$ after each miss. Their respective probabilities of hitting the target are $\alpha$ and $\beta$. If $n$ is the number of shots fired, what is the limiting probability of hitting the target as $n$ approaches infinity?
What I did:
I set up the transition matrix $P$ = \begin{bmatrix}\alpha&1-\alpha\\\beta&1-\beta\\ \end{bmatrix} Then I solve the system of equations $\pi = \pi P \leftrightarrow (P^T-I)\pi^T=0$ and $\pi_1+\pi_2=1$ and get $\pi = \left(\frac{\beta}{\beta+1-\alpha}, \frac{1-\alpha}{\beta+1-\alpha}\right)$.
As we want the limiting probability that the target is hit, since $\pi_1$ is the probability that $X$ shoots and $X$ only shoots after someone hits the target, I think the limiting probability should just be $\pi_1 = \frac{\beta}{\beta+1-\alpha}$.
However, the textbook has the solution as $\frac{\beta}{1-\alpha-\beta}$. I'm not sure if the textbook just has a typing error or my logic is wrong somewhere. It would be helpful if someone could corroborate my work or point out where I made a mistake.
I'm also thinking that the book might be wrong because if $\alpha$ and $\beta$ are both over $.5$ then the limiting probability will be negative, which makes no sense.
You are correct and the textbook's answer is wrong.
A quick way of getting the limiting probability is that in the long run, we switch from $X$ to $Y$ a $\pi_X (1-\alpha)$ fraction of the time, and from $Y$ to $X$ a $\pi_Y\beta$ fraction of the time. These must be equal from which we deduce that $$\pi_X : \pi_Y = \beta : 1-\alpha.$$ Normalizing, $\pi_X = \frac{\beta}{(1-\alpha) + \beta}$ and $\pi_Y = \frac{1-\alpha}{(1-\alpha) + \beta}$.
Your method, of course, works equally well.