Hello I am dealing with a problem where I need to calculate the probability of a player winning a coin toss game where the two players alterantely toss a coin and the first one to toss head wins. It states that the coin does not necessairly need to be fair. I stumbled upon this proof/solution (see below) and I understand the concept behind it but I do not understand how you get from $(I)$ to $(II)$ (I assume: $p+q^2p+q^4p+... =p(1+q^2+q^4+...)=p(1+{1\over 1-q^2})$) how do I go from there? And how do I get from $(II)$ to $(III)$ could someone please elaborate? Thank you very much.
(SOLUTION)
So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_n=q^{n-1}\cdot p$ and \begin{eqnarray} P &=& P_1+P_3+P_5+... \\\\ &=&p+q^2p+q^4p+q^6p+...\quad(I)\\\\ &=& {p\over 1-q^2}\qquad\qquad\qquad\qquad\quad(II)\\\\ &=& {1\over 1+q}\qquad\qquad\qquad\qquad\quad(III) \end{eqnarray}
where $p$ is probability that head comes in one toss and $q=1−p$.
So if the coin is fair, then $p={1\over 2}=q$, so
$P= {2\over 3}$
(/SOLUTION)