<s>[INST] 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. 
 [/INST] 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.
</s>