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?
 A: I'm going to assume P goes first.  The question should have said so, unless something else was intended, in which case it's unclear.
Let (lower-case) $p$ be the probability that P ultimately wins.  Then
$$
\begin{align}
p & = \Pr(\text{P wins on 1st trial}) + \Pr(\text{P loses on first trial and ultimately wins)} \\[8pt]
& = \frac 1 6 + \frac 5 6 (1-p).
\end{align}
$$
So solve the following:
$$
p =\frac 1 6 + \frac 5 6 (1-p).
$$
A: Let $p$ be the probability that the first player $P$ (ultimately) wins, and let $q$ be the probability that $Q$ ultimately wins. It is clear that with probability $1$, the game terminates, so $p+q=1$.
We condition on $P$'s first throw. With probability $\frac{1}{6}$, she gets a sum of $7$, and wins immediately. 
Another way she can ultimately win is if she tosses something other than $7$, but $Q$ does not ultimately win. The probability $P$'s first toss is not a $7$ is $\frac{5}{6}$. Given that this has happened, the probability that $Q$ does not win is $1-p$. Thus
$$p=\frac{1}{6}+\frac{5}{6}(1-p).$$
We have a linear equation for $p$. Solve. 
A: Note that the probability of a seven is $1/6$.  So the answer is
$$(1/6) + (5/6)\times (5/6) \times (1/6) + \dots=
\frac{1}{6}\sum_{j=0}^\infty \Bigl(\frac{25}{36}\Bigr)^j = \frac{6}{11}.$$
A: Suppose that player $P$ begins the game. Probability that player $P$ wins after the first try is $1/6$, probability that player $P$ does not win after the first try is $5/6$. If player $P$ does not win after the first try, the probability that player $P$ can try to win again is $5/6$. So probability that player $P$ will win after the second try is $\frac16(\frac56\frac 56)$. The probability that player $P$ will win after the third try is $\frac16(\frac56\frac 56)^2$. The probability that player $P$ will win is then equal to
$$
\sum_{n=0}^\infty\frac16\biggl(\frac56\cdot\frac56\biggr)^n=\frac16\sum_{n=0}^\infty\biggl(\frac{25}{36}\biggr)^n=\frac6{11}.
$$
A: Think about this problem as if it were a single player rolling dice, and we stop the game when she rolls a $7$. This is a geometric distribution, with the probability of "success" being $p=1/6$, so we can write the pmf describing the number of rolls $k$ until the first $7$ appears:
$$
P(X=k)=(1-p)^{k-1}p
$$
Now, first player wins the game if the game stops on the first, third, fifth roll and so on, in other words $k$ must be odd. Let $k=2n-1, n=1,2,\dots$ Now we have
$$
P(P \text{ wins})=\sum_{n\ge 1}(1-p)^{2n-2}p=\dfrac{1}{2-p}
$$
Substitute $p=1/6$ and get that your answer is $\dfrac{6}{11}$ (assuming P rolls first; otherwise it's just the complement). But the nice intuition here is that rolling first is advantageous!
A: 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
$$ \large\dfrac{\dfrac{1}{6}}{\left( 1 - \dfrac{5}{6}^2 \right)} = \boxed{\dfrac{6}{11}} $$
