Stuck on a probability law problem I'm currently trying to solve a problem, I completed the first question but I am stuck at the second one, here is the problem:
One person roll a die until the result is a $1$, a second person toss a coin until he gets $3$ tails.


*

*How many tries are they going to make on average.


$X$ = number of die rolls
$Y$ = number of time we toss a coin
$X$ follows a geometric law with parameter $1/6$, $E(X) = 6$ and $V(X) = 30$.  $Y$ follows a negative binomial law with $n = 3$ and $p = 1/2,$ $E(Y) = 3/(1/2) = 6$ and $V(Y) = n(1-p)/p^2 = 6.$


*What is the probability that they both stop at the same time $(p(X = Y)).$
I found $p(X=Y) =  \sum_{k=3}^\infty p(X=k, Y=k) $  so the same sum with $p(X=k)p(Y=k)$ since they are independent. Then $p(X=k)p(Y=k)  = (5/6)^{k-1} \cdot\frac16 \begin{pmatrix} k-1 \\ 2 \\ \end{pmatrix} 
 \left(\frac12\right)^{k-3}\left(\frac12\right)^3$
But now i don't know how to compute it so i can get the value of that probability? Any help would be very appreciated.
Thanks

We have $$\begin{align}
\Pr(X=Y)&=\sum_{k=3}^\infty\frac16\left(\frac56\right)^{k-1}\left(\frac12\right)^k\binom{k-1}{2}\\
&=\frac15\sum_{k=3}^\infty\left(\frac{5}{12}\right)^k\binom{k-1}{2}\\
&=\frac{1}{10}\sum_{k=3}^\infty\left(\frac{5}{12}\right)^k(k-1)(k-2)\\
&=\frac{1}{24}\sum_{k=3}^\infty\left(\frac{5}{12}\right)^{k-1}(k-1)(k-2)
\end{align}$$
Now write $$f(x)=\frac{1}{24}\sum_{k=3}^\infty x^{k-1}$$
Differentiating twice,
$$f''(x)=\frac{1}{24}\sum_{k=3}^\infty (k-1)(k-2)x^{k-3}$$ and
$$x^2f''(x)=\frac{1}{24}\sum_{k=3}^\infty (k-1)(k-2)x^{k-1}$$ 
So, sum the geometric series for $f$, differentiate twice, multiply by $x^2$, and then set $x=\frac{5}{12}$.
