Coin tossing problem between two players. Find probability that one player tosses heads first. Player $1$ tosses a biased coin with the probability of getting $H$ (a head) being $p$, for some $0<p<1$, and Player $2$ tosses a biased coin with the probability of getting $H$ (a head) being $q$, for some $0<q<1$. They toss their coins at the same time. The first player to get $H$ wins. If they both get $H$, the game ends in a draw. What is the probability that Player $1$ wins?
Here is what I have:
Let $P(A_k)=P(\{\text{Player $1$ wins on $k$th toss} \})=P(\{\text{Player $1$ gets $H$ on $k$th toss and Player $2$ gets $T$ on $k$th toss})=P(\{\text{Player $1$ gets $H$ on $k$th toss}\})P(\{\text{Player $2$ gets $T$ on $k$th toss}\})
=(1-p)^{k-1}(1-q)^{k-1}p.$
Then
$P(\{\text{A wins}\})=P(\cup_{k\geq 1} A_k)=\sum_{k\geq 1} P(A_k)= \sum_{k\geq 1} (1-p)^{k-1}(1-q)^{k-1}p=\frac{p}{1-p} \left(\frac{1}{1-(1-p)(1-q)}-1\right).$
Can someone tell me if I am on the right track? Thank you!
 A: $\underline{Revised\;answer}$ 
I will label player $1$ as $A$ for clarity.
I also take as per the exact wording of the question,  that both getting heads on a round is a draw , whereas both getting tails on a round extends the game, and that we want to find $ P(A\;eventually\;wins)= \Bbb P\; (say) $
$A$ can win in the $1^{st}$ round by getting H while $B$ gets $T$, and in  subsequent round(s) if and only if it extends to  subsequent round(s)


*

*$\underline{Using\; geometric\; series}$


$A$ can win in first round with $Pr = p(1-q),$ or move into next round with $Pr = (1-p)(1-q),$ and again win with $Pr = p(1-q),$ and so on.
Thus $\Bbb P = p(1-q) + (1-p)(1-q)p(1-q) + ((1-p)(1-q))^2p(1-q)\; + ...$
This a G.P. with $a = p(1-q),\; r = (1-p)(1-q)\;\;\; S(\infty) = \dfrac{a}{1-r}$
$so\; \Bbb P= \dfrac{p(1-q)}{1 - (1-p)(1-q)}$


*

*$\underline{Using\; recursion}$


$A$ wins in first round with $Pr = p(1-q)\;$ or gets back to square one with $Pr = (1-p)(1-q)$
So $\Bbb P = p(1-q) + (1-p)(1-q)\Bbb P$
which again yields, $\Bbb P = \dfrac{p(1-q)}{1 - (1-p)(1-q)}$
A: $\ P(A_w)$ = $\ P(A_H)$+$\ P(A_T,B_T,A_H)$+$\ P(A_T,B_T,A_T,B_T,A_H)$+...
$\ P(A_w)$ = $\ p$+$\ (1-p)(1-q)p$+$\ (1-p)^2(1-q)^2p$+$\ (1-p)^3(1-q)^3p$
+...
$\ P(A_w)$ = $$\sum_{k=0}^\infty p[(1-p)(1-q)]^k$$
Note that it forms a geometric sequence.
Thus $\ P(A_w)$ = $$ \frac{p}{1-(1-p)(1-q)}$$
A: To supplement user362325's answer, I will provide another method of solving this problem:
Note that, if neither player tosses heads in each of their first flips, the problem essentially reverts back to the original. Let $x_1$ be the probability that Player 1 wins. 
We know that
$$x_1 = p+(1-p)(1-q)x_1$$
where the first $p$ represents the first flip being heads and the other part is the probability that neither player flips heads, $(1-p)(1-q)$, times the original $x_1$. 
Can you take it from there?
A: Thanks for the suggestion from Carl Schildkraut.
From his expression,$$\ x_1=p+(1−p)(1−q)x_1$$
making $x_1$ as th subject,
$$\ [1-(1-p)(1-q)]x_1=p$$
$$\ x_1=\frac{p}{1-(1-p)(1-q)}$$ 
Is this numerically identical to my original answer?
