What's the expected number of times I have to roll two die until they both sum $7$? 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. 
 A: You are correct that it will take on average $6$ rolls, IF we are considering the tossing of both dice together as one roll. If we let $X$ be the number of rolls until we reach a sum of $7$, then we can model this using a Geometric Distribution.
You have already calculated that the probability of rolling a sum of $7$ to be $1/6$. Therefore the probability that you don't roll a sum of $7$ is $5/6$.
The distribution for the probability that it will take $k$ rolls to reach a sum of $7$ will be $$P(X=k) = (5/6)^{k-1} * (1/6)$$
You can then find the mean of this distribution, which turns out to be $$\frac{1}{1/6} =6$$  I'll leave the derivation of this for you to look up :)
A: 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.
A: The probability of doing it after one roll is $1/6$, in two is $5/6 \times 1/6$, in three $(5/6)^2 1/6$ and so on ... we get 
\begin{eqnarray*}
E(7)=1 \times \frac{1}{6} + 2 \times \frac{5}{6} \times\frac{1}{6} + 3 \times \left(\frac{5}{6}\right)^2 \times\frac{1}{6}+\cdots = \frac{1}{6} \sum_{i=1}^{\infty} i \left( \frac{5}{6} \right)^i \\
\end{eqnarray*}
Now recall that 
\begin{eqnarray*}
\sum_{i=1}^{\infty} i x^{i-1} =\frac{1}{(1-x)^2}.
\end{eqnarray*}
So
\begin{eqnarray*}
E(7)= \frac{1}{6} \sum_{i=1}^{\infty} i \left( \frac{5}{6} \right)^{i-1} =\frac{1}{6} \frac{1}{(1-\frac{5}{6})^2}=6
\end{eqnarray*}
So the expected value is $\color{red}{6}$ as expected.
