TITLE: What's the expected number of times I have to roll two die until they both sum $7$?
QUESTION [14 upvotes]: Here is my guess: the probability of summing $7$ on two rolls is $\frac 16$. This means if I repeat the experiment many times I'll roll $7$ one sixth of them (approximately). Hence,
$$N \cdot \bigg(\cfrac 16\bigg) \cdot 7 = 7$$  
where $N$ is the total number of rolls. That gives me a total number of $6$ rolls on average to sum $7$. 
I'm not quite sure so I'm all open to suggestions! Thanks in advance.

REPLY [21 votes]: If $X$ is the number of rolls to get $7$ then the expected (or average) value of $X$ satisfies:
$$E(X)=1+\frac{5}{6}E(X)$$
That is, we always start with one roll, and $5/6$ of the time, we just start all over again. So $E(X)=6.$

Technically, as Heinrich comments below, this only proves that either $E(X)=6$ or $E(X)=+\infty.$ You might actually need some trick to prove that the expected value must be finite.