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?

Solution 1:

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: $$ p =\frac 1 6 + \frac 5 6 (1-p). $$

Solution 2:

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.