# 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.

• Take a look at the 'geometric distribution'. The wikipedia page has more than enough information on it for your needs. – Marc Jun 7 '15 at 22:09
• The moral is that since the probability of it happening on a single roll is $1/36$, you're expected to use on average $36$ throws to get to one. – Arthur Jun 7 '15 at 22:09

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 $x=\frac{35}{36}$.