Expected number of dice rolls for a sequence of dice rolls ending at snake eyes If I roll a pair of dice repeatedly and stop only when I get snake eyes (both dice show 1), what is the expected number of dice rolls that will occur? I know the answer is 36, but I'm having trouble understanding why that is the answer.

That happens because the mean of a geometric distribution with $p=\frac{1}{36}$ is exactly $\frac{1}{p}=36$.
The probability that a double one occurs at the $k$-th throw is given by:
$$ \mathbb{P}[X=k] = \frac{1}{36}\left(1-\frac{1}{36}\right)^{k-1},\tag{1}$$
hence:
$$ \mathbb{E}[X]=\sum_{k\geq 1}k\cdot\mathbb{P}[X=k]=\frac{1}{36}\sum_{k=1}^{+\infty}k\left(\frac{35}{36}\right)^{k-1},\tag{2}$$
but since for any $|x|<1$ we have:
$$ \sum_{k\geq 0}x^k = \frac{1}{1-x},\tag{3}$$
by differentiating both sides of $(3)$ with respect to $x$ we have:
$$ \sum_{k\geq 1}k x^{k-1} = \frac{1}{(1-x)^2}\tag{4}$$
so the claim follow by evaluating $(4)$ at $x=\frac{35}{36}$.
