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.

The key point to realise is that if a 6 is rolled, then the probability of each player winning after that roll is the same as it was before.  So if $p_A$ and $p_B$ denote the probability of each player winning, then we have $$
p_A = \frac 26 + \frac 16 p_A\\
p_B = \frac 36 + \frac 16 p_B
$$
which is easy to solve.
