Probability of winning a head on a coin The problem I am asking is generated from this problem:
Carla and Dave each toss a coin twice. The one who tosses the greater number of heads wins a prize. Suppose that Dave has a fair coin $[P_d(H)=.5]$, while Carla has a coin for which the probability of heasd on a single toss is $.4$ $[P_c(H)=.4$.
Here is the question I am asking:

In the experiment of that problem, Dave tosses a fair coin twice while Carla also tosses a coin twice. Fro Carla's coin, the probability of head on a single toss is $0.4$. What is the probability that Dave will win the prize provided that the experiment is repeated whenever a tie occurs.

I know the probability for $P(C_1)=P_c(HT)+P_c(TH)=2(.4)(.6)=.48$ and $P(D_2)=P_d(HH)=(.5)^2=.25$
I know that we can probably use the sum of Geometric series:
$$
\sum\limits_{n=1}^\infty (ar)^{x-1}= \frac{a}{1 - r}
$$
Can someone please help me to solve this problem? I am not sure about this.
 A: Outline: Let $p$ be the probability that Dave (ultimately) wins. There are two ways this can happen: (i) Dave wins in he first round or (ii) There is a first-round tie, but Dave ultimately wins.
Let $\delta$ be the probability that Dave wins in the first round. I am sure you can compute $\delta$. I believe it is $0.39$. 
Let $\tau$ be the probability of a first-round tie. You can compute $\tau$. 
We have then
$$p=\delta+\tau p.$$
Solve this linear equation for $p$.
Remark: You can also calculate $p$ via an infinite geometric series. We have
$$p=\delta+\delta\tau+\delta\tau^2+\cdots=\frac{\delta}{1-\tau}.$$
A: The easy way is to recognize that you can ignore the event of a tie.  If Dave's chance of winning on one turn is $d$, Carla's chance of winning on one turn is $c$, then Dave's chance of winning overall is $\frac d{c+d}$ and Carla's is $\frac c{c+d}$  You can show this by summing the geometric series as you suggest.  The chance of nobody winning on a given turn is $1-c-d$, so Dave's chance of winning is $d + (1-c-d)d+ (1-c-d)^2d+\dots$ where the factors of $(1-c-d)$ come from turns nobody won.  Now sum the series and you will get $\frac d{c+d}$
