Biased coins: Probability such that the first player throws heads Let $A$, $B$ be the two players. Each one has a coin has a probability of getting heads of $p_i$. Player $A$ always starts first. What is the probability such that $A$ wins?
Ex. The both coins land 'heads' on average 1 out of 2 times. 
The solution says $\frac{1}{2}$ as the result.
My approach was to draw a probability tree and compute the probability such that $A$ throws heads. We know, that this is geometrically distributed, but don't know how to get the result.
 A: If you drawn a tree, then you can see that either $A$ wins straight away, or $A$ flips tail and $B$ also flip tail and then $A$ gets heads, ..., or $A$ and $B$ each get $j$ tails in a row and then $A$ gets heads.
So, the probability of $A$ winning is given by summing all these probabilities:
$$
\sum_{j=0}^{+\infty} \big((1 - p_A)(1 - p_B)\big)^{j} p_A
$$
Let $c = (1 - p_A)(1 - p_B) \in [0; 1]$. The above sum can be written as:
$$
p_A \sum_{j=0}^{+\infty} c^{j} 
$$
Can you compute the result?

Edit: For future reference, I'm adding the result for the probability of $A$ winning:
$$
\frac{p_A}{1-c} = \frac{p_A}{p_A + p_B - p_A p_B}
$$

And, just for fun, I wondered which should be the probability when $p_A = p_B$ that makes the probability of $A$ winning equal to $1/2$.
$$
\frac{p_A}{2 p_A - p_A ^2} = \frac{1}{2}
$$
$$
2p_A = 2 p_A - p_A ^2
$$
$$
p_A ^2 = 0
$$
$$
p_A = 0
$$
Hence,
$$
\lim_{p_A \to 0} \frac{p_A}{2 p_A - p_A ^2} = \frac{1}{2}
$$
So, if the game goes on forever (in the limit $p_A \to 0$) then both players have $1/2$ probability of winning.
Furthermore, if $p_A = p_B$ the first player to toss the coin has the highest probability of winning. I.e., for $p_A \in [0; 1]$,
$$
\frac{p_A}{2 p_A - p_A ^2} \geq \frac{1}{2}
$$
