Probability of first actor winning a "first to roll seven with two dice" contest? Two players P and Q take turns, in which they each roll two fair and independent dice. P rolls the dice first.
The first player who gets a sum of seven wins the game. What is the probability that player
P wins the game?

The probability that $ P $ eventually wins the game can be represented as the sum of an infinite geometric series. 
For example, 
The probability of $  P$  winning in the first round itself is  $ \dfrac{1}{6} $. (Since out of the $ 36 $ possible outcomes, there are $ 6 $ outcomes in which the sum of the dice is equal to $ 7 $.)
The probability of $ P $ winning in the second round is $ \dfrac{5}{6} \times \dfrac{5}{6}\times \dfrac{1}{6} $. This means that for $ P $ to win th second round, he has to lose the first round, then $ Q $ has to lose his turn and finally $ P $ wins. This goes on.
The final answer is
$$ \large\dfrac{1}{6} + \left( \dfrac{5}{6}^{2} \times \dfrac{1}{6} \right) + \left( \dfrac{5}{6}^{4} \times \dfrac{1}{6} \right) + \dots = \displaystyle\sum\limits_{i=0}^{\infty} \dfrac{1}{6} \times \dfrac{5}{6}^{2n}$$
This is an infinite geometric series and its sum is equal to
$$ \large\dfrac{\dfrac{1}{6}}{\left( 1 - \dfrac{5}{6}^2 \right)} = \boxed{\dfrac{6}{11}} $$
