Probability of rolling a 2 before a 6 and a 1 in that order 
I am rolling a fair die. What is the probability that the first time I see a $2$ is before the first time I see a $6$ , and the first time I see a $6$ is before the first time I see a $1$ ?

So I thought about doing it with an infinite sum: 
\begin{align}
\frac{1}{6}\frac{1}{6} + \frac{1}{6}\frac{4}{6}\frac{1}{6} + \frac{1}{6}\big(\frac{4}{6}\big)^2\frac{1}{6} + \dots
\end{align}
where the first term is rolling a $2$ and then rolling a $6$ , the second term is rolling a $2$ , rolling anything but a $1$ or $6$ and then rolling a $6$ , etc. but I don't believe this is correct. 

Your roll will look like this:
$\qquad (\text{1})$: $k$ rolls, each roll being one of $3,4,5$
$\qquad (\text{2})$: a $2$
$\qquad (\text{3})$: $m$ rolls, each roll being one of $2,3,4,5$
$\qquad (\text{4})$: a $6$
$\qquad (\text{5})$: $n$ rolls, each roll being one of $2,3,4,5,6$
$\qquad (\text{6})$: a $1$
For each $k$, $(1)$ has probability $(1/2)^k$.
For each $m$, $(3)$ has probability $(2/3)^m$.
For each $n$, $(5)$ has probability $(5/6)^n$.
Steps $(2)$, $(4)$ and $(6)$ each have probability $1/6$.
All in all, that gives you
\begin{align}
\left[\sum_{k\geq 0} {\left(\frac12\right)}^k\right]
\cdot \frac16\cdot
\left[\sum_{m\geq 0} {\left(\frac23\right)}^m\right]
\cdot \frac16\cdot
\left[\sum_{n\geq 0} {\left(\frac56\right)}^n\right]
\cdot \frac16
&=
\frac{1}{\left(1-\frac12\right)\,\left(1-\frac23\right)\,\left(1-\frac56\right)\,6^3}
\\&=
\frac{1}{\frac12\cdot\frac13\cdot\frac16\cdot 6^3}
\\&=
\frac16
\end{align}
