Dice roll game probability Suppose we play a 1 vs 1 game. I roll a 6 sided fair die, if I get a 1 I win, if not you roll. If you get a 6 you win, otherwise I go again.... What are my chances of winning and what are yours? 
 A: $\underline{A\; way\; without\; using\; an\; infinite\; geometric\; series}$
Either you win on the first roll, or you don't, and the turn passes to your opponent,who is now exactly in the same position you were at the start of the game.
Thus if P(you eventually win) $=p,\quad p + \frac56p = 1 \to p = \frac6{11} $ 
A: The probability of you winning immediately is $\frac{1}{6}$. The probability of you winning on the third roll (after your opponent rolls for the first time) is $\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}$. This represents the probability that you do not win on the first roll, your opponent does not win on the second roll (given you did not win on the second roll), and you win on the third roll (given no one won on the first two rolls).
We can repeat this process to calculate the probability of you winning on the fifth roll as: $\frac{5}{6}\times\frac{5}{5}\times\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}$.
We note that this is a geometric sequence with initial value $\frac{1}{6}$ and ratio $(\frac{5}{6})^2$. To find the probability that you will win the game, we want to sum:
$$\frac{1}{6} + (\frac{5}{6})^2\times\frac{1}{6}+(\frac{5}{6})^4\times\frac{1}{6}+...$$
$$=\frac{1}{6} \times(1+(\frac{5}{6})^2+(\frac{5}{6})^4+...)$$
$$=\frac{1}{6}\times\frac{1}{1-(\frac{5}{6})^2}$$
$$=\frac{1}{6}\times\frac{1}{1-\frac{25}{36}}$$
$$=\frac{1}{6}\times\frac{36}{11}$$
$$=\frac{6}{11}$$
The probability that you win is $\frac{6}{11}$. We notice that someone must eventually win the game, so the probability that your opponent wins will be $1-\frac{6}{11}=\frac{5}{11}$.
