Emily and Greg play a dice-throwing game. They take turns to throw a fair die starting with Emily. All throws are independent. Both players start with zero points. When Emily throws a six she scores two points. When Greg throws a five or a six he scores one point.
The winner is the first player to score two points.
(a) For each integer $k\geq 2$, what is the probability that Greg wins on
his $k$-th throw?
(b) What is the overall probability that Greg wins?
I've already attempted part (a) and using the fact that Greg must have 1 success in k-1 throws and a success on k-th throw and also that Emily must fail to throw a 6 k times. With the formula: $P(X=k) = \binom{k-1}{r-1} * (1-p)^{k-r} * p^r$ and got
$P(X=k) = ((2/3)^{k-2} * (5/6)^k * (k-1))/9$ and this simplifies to:
$P(X=k) = 1/4 * (5/9)^k * (k-1)$
Is my answer to part (a) correct?
If yes how do I tackle part (b) and if not where did I go wrong?
Many thanks.
 A: To determine the overall probability that Greg wins:
Each round begins with some state, depending on whether Greg has scored a point or not.  That is, we have two states:  $S_0,S_1$.  Accordingly, we let $p_0,p_1$ denote the probability that Greg wins assuming we are starting from the associated state.  As we start in $S_0$ the answer we want is $p_0$.
Starting in $S_1$ we see that the possible outcomes for the round are:  Emily wins (probability $\frac 16$), Greg wins (prob $\frac 56 \times \frac 13=\frac 5{18}$), we end up back in $S_1$ (prob $1-\frac 16 - \frac 5{18}=\frac 59$).  Thus $$p_1=\frac 16\times 0 +\frac 5{18} \times 1 +\frac 59 \times p_1\implies p_1=\frac 58$$ 
Starting in $S_0$ we see that the possible outcomes for the round are:  Emily wins (probability $\frac 16$), Greg scores and we move to $S_1$ (prob $\frac 56 \times \frac 13=\frac 5{18}$), we end up back in $S_0$ (prob $\frac 59$ as before). Thus $$p_0=\frac 16\times 0 +\frac 5{18}\times p_1 + \frac 59 \times p_0\implies p_0=\frac {25}{64}$$
Note:  Using the formula $p(X=k)=\left( \frac 23 \right)^{k-2}\times \left( \frac 56 \right)^k \times \frac {k-1}9$ and summing over $k$ we confirm this probability.
A: As OP asks (since Markov method is unfamiliar) the method for summing the series in this case applies when it is of the form $A=\sum_{n=0}^\infty n x^{n-1}.$ ) Of course here one doesn't need the first term which is zero, and the series we have converges provided $|x|<1.$
One starts with the known geometric series $G=\sum_{n=0}^\infty x^n=1/(1-x)$ and differentiates each side. On the left side this derivative is our series $A,$ while on the right it is $\frac{1}{(1-x)^2}.$
To finish using this in your example requires [using the series as in lulu's answer] manipulating the latter so one can apply the above to it. One factors out $(1/9)(5/6)^2$ so the rest is $(k-1)[(2/3)(5/6)]^{k-2}$ and then since this $k$ goes from $1$ to $\infty$ a shift of index shows our above formula should give $(1/9)(5/6)^2[(9/4)]^2=25/64$ as expected. Note the inside $x=(2/3)(5/6)=5/9,$ so $1-x=4/9,$ which gets "reciprocalled" and squared in the formula.
