What's the probability that Abe will win the dice game? Abe and Bill are playing a game.  A die is rolled each turn.
If the die lands 1 or 2, then Abe wins.
If the die lands 3, 4, or 5, then Bill wins.
If the die lands 6, another turn occurs.
What's the probability that Abe will win the game?
I think that the probability is $\frac{2}{5}$ just by counting the number of ways for Abe to win. I'm not sure how to formalize this though in terms of a geometric distribution.

I agree with you that $\dfrac{2}{5}$ is obvious. End of story.  But if you really want to sum a series, abbreviate by $A$ the event "$1$ or $2$" and by $S$ the event "$6$." Then Abe can win in various ways. These are $A$ (wins immediately), $SA$ (get a $6$, then win), $SSA$, $SSSA$, and so on. 
These have probabilities $\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{1}{6}\cdot\dfrac{2}{6}$, and so on. So we want to sum the series
$$a+ar+ar^2+ar^3+\cdots,$$
where $a=\dfrac{2}{6}$ and $r=\dfrac{1}{6}$.
By the usual formula for the sum of an infinite geometric series, this is $\dfrac{a}{1-r}$, which simplifies to $\dfrac{2}{5}$.
