Expected duration of a die game In a 2 person game, the player who first obtains a $6$ wins. I'm trying to determine the expected number of die rolls needed before a winner is determined. (One turn
consists of two die rolls, assuming that neither player won in that turn). 
$\textbf{My attempt:}$
This problem can be modeled by a geometric distribution with probability $p$, and by virtue of this the expectation, $E[X] = \frac{1}{p}$. Now since, the probability of getting a $6$ is $\frac{1}{6}$, then $E[X] = 6$. I'm wondering if this is correct, or am I missing something.
 A: We have to find the expected number of turns for the game to end. That is given by:
$$ \sum_{r=0}^\infty rP_r\;\;\;\;\;\;\;\text{(by definition)} $$
where $ P_r $ is the probability for the game to end in the $r^\text{th}$ turn. 
Now, let's see the probability for the game to end in the zeroth turn (that is, the players have just begun playing).
The game will end in the zeroth turn iff either the first player wins or the second player wins. Thus, the probability is $ P_0 = \dfrac 16 + \left(\dfrac 56\right)\dfrac 16 $
Similarly, $ P_1 = \left(\dfrac 56\right)^2\dfrac 16 + \left(\dfrac 56\right)^3\dfrac 16 $
It's easy to see that $ P_r = \left(\dfrac 56\right)^{2r}\dfrac 16 + \left(\dfrac 56\right)^{2r+1}\dfrac 16 = \dfrac{275}{1296}\left(\dfrac{25}{36}\right)^{r-1} $
Thus, the expected number of turns of the game is:
$$\begin{align}
&\sum_{r=0}^\infty r\dfrac{275}{1296}\left(\dfrac{25}{36}\right)^{r-1} \\\\
&=\dfrac{275}{1296} \sum_{r=0}^\infty r\left(\dfrac{25}{36}\right)^{r-1}\\\\
&=\dfrac{275}{1296}\dfrac{1}{\left(1-\dfrac{25}{36}\right)^2} = \boxed{\boxed{\dfrac{25}{11}}}
\end{align}$$
where in the last step, I've used the fact that:
$$ \sum_{r=0}^\infty rx^{r-1} = \dfrac{1}{(1-x)^2} \forall |x| < 1 $$
So, the expected number of turns turns out to be (sic) is $\dfrac{25}{11}$.
EDIT:
As suggested by Ross Millikan, it's better to start the counting from $1$. In that case, there's a slight change. Our $P_r$ becomes $ \dfrac{11}{36} \left(\dfrac{25}{36}\right)^{r-1}$.
Hence, the 'new' expected number of turns is $\dfrac{36}{11}$.
