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
 A: To show that $$\sum_{k=3}^\infty\left[\left(\frac56\right)^{k-1}\cdot\frac16\binom{k-1}2\left(\frac12\right)^{k-3}\left(\frac12\right)^3\right]= \frac{25}{343},\tag1$$ it suffices to show that $$\sum_{k=3}^\infty\left(\frac5{12}\right)^k\binom{k-1}2= \frac{125}{343}\tag2$$ after combining constants. The binomial term can be rewritten as $(k-1)(k-2)/2$ so that $(2)$ is equivalent to $$\sum_{k=3}^\infty\left(\frac5{12}\right)^k(k-1)(k-2)= \frac{250}{343}\tag3.$$ Now the LHS can be split to give \begin{align}\small\sum_{k=3}^\infty\left(\frac5{12}\right)^k(k-1)(k-2)&=\small\sum_{k=3}^\infty k(k-1)\cdot\left(\frac5{12}\right)^k-2\sum_{k=3}^\infty k\cdot\left(\frac5{12}\right)^k+2\sum_{k=3}^\infty\left(\frac5{12}\right)^k\\&=\small\left(\frac5{12}\right)^2\sum_{k=3}^\infty k(k-1)\cdot\left(\frac5{12}\right)^{k-2}-2\cdot\frac5{12}\sum_{k=3}^\infty k\cdot\left(\frac5{12}\right)^{k-1}+\frac{125}{504}\tag4\end{align} using standard geometric series. Notice that for $|a|<1$, we have \begin{align}\sum_{k=3}^\infty k(k-1)\cdot a^{k-2}&=\sum_{k=3}^\infty\frac{d^2}{da^2} a^{k}=\frac{d^2}{da^2}\sum_{k=3}^\infty a^{k}=\frac{d^2}{da^2}\left[-\frac{a^3}{a-1}\right]=-2-\frac2{(a-1)^3}\end{align} and \begin{align}\sum_{k=3}^\infty k\cdot a^{k-1}&=\sum_{k=3}^\infty\frac{d}{da} a^{k}=\frac{d}{da}\sum_{k=3}^\infty a^{k}=\frac{d}{da}\left[-\frac{a^3}{a-1}\right]=-\frac{a^2(2a-3)}{(a-1)^2}.\end{align} Letting $a=5/12$, equation $(4)$ becomes $$\small\sum_{k=3}^\infty\left(\frac5{12}\right)^k(k-1)(k-2)=\frac{25}{144}\cdot\left(-2-\frac2{\left(\frac5{12}-1\right)^3}\right)-\frac56\cdot\left(-\frac{\left(\frac5{12}\right)^2\left(2\cdot\frac5{12}-3\right)}{\left(\frac5{12}-1\right)^2}\right)+\frac{125}{504}=\frac{250}{343}$$ which proves $(1)$.
A: 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 $x=\frac{5}{12}$.
