Consider a game with two players A and B. Player A throws a coin with probability $p_A$ of landing heads, in which case he wins the game. If not, player B throws another coin with probability $p_B$ of landing heads, in which case B wins. If not, it is A's turn again.
Calculate the probability of both players winning the game, as a function of $p_A$ and $p_B$. If $p_A=\frac{1}{3}$, find $p_B$ so that the game is fair.
This is what I've done so far. I think I can determine the probabilities of A and B winning the game, as a function of both $p_A$ and $p_B$ and of the number of rounds $n$:
$p(W_A)={p_A}^n\left(1-p_B\right)^{n-1}$
$p(W_B)={p_B}^{n-1}\left(1-p_A\right)^{n-1}$
The way I see it, the equations are different because A gets to go first. Intuitively, B is in an unfair position unless $p_B > p_A$.
I suppose that a fair game would mean that $p(W_A)=p(W_B)$, but for what $n$? This has led me to think that there is a way to describe those probabilities without depending on the number of rounds, but I don't know how.
This is a homework question. I'm not looking for the answer, but for guidance in getting there. Let me give you some context. I am taking a graduate-level course on Stochastic Control (my research area is robotics). The course starts with a review of probability, and this particular question is part of the exercise list. The bibliography for this part of the course is Sheldon Ross' Introduction to Probability Models. Thanks for the help!