An endless series of games

First case : Omar will win the first game .

Second case : Rami will win the first game .

I will give number $1$ to the game that Omar win and number $2$ to the game Rami win .

Because we want Omar to be the winner , Rami can't win two consecutive games , so ,

In the first case : the series of winners can be like :

$1,1$

$1,2,1,1$

$1,2,1,2,1,1$

...

So we get that the probability that Omar will win in this case is :

$\sum_{n=2}^{\infty}{\left\{\frac{2}{3}\right\}}^n{\left\{\frac{1}{3}\right\}}^{n-2} = \frac{4}{7}$

In the second case we get that the series of winners can be like :

$2,1,1$

$2,1,2,1,1$

$2,1,2,1,2,1,1$

...

So we get that the probability that Omar will win in this case is :

$\sum_{n=2}^{\infty}{\left\{\frac{2}{3}\right\}}^n{\left\{\frac{1}{3}\right\}}^{n-1} = \frac{4}{21}$

At the end we see that the probability that Omar will win the series is :

$\frac{4}{7}+\frac{4}{21}=\frac{16}{21}$

Let $A$ denote the event that Omar wins, $B$ for Rami.

Probability = $AA + BAA+ ABAA + BABAA+ ABABAA+ BABABAA + ...$

$= AA((1+AB + ABAB +ABABAB...) + (B+BAB+BABAB...))$

Applying sum of infinite GP,

$=AA(\frac {1}{1-AB} + \frac {B}{1-AB})$

Putting $A=\frac {2}{3}$ and $B=\frac {1}{3}$ , we get the answer as $\dfrac {16}{21}$ .