You have two six-sided dice and you roll them until you have a sum of 12? What is the expected number of rolls you have to throw?
I think if I throw a twelve, and I keep throwing, I have to throw 36 times to get another 12. But is seems weird if I do it the first time. Is this true?
Is there also a general solutions for this problem: you throw $n$ dice and you want a sum of $x$, how often do you have to roll?
The expected number of (independent) trials to achieve success is simply the reciprocal of the probability of success (here $36$ as you say).
To see this, note that the probability of taking at least $r$ trials is $(1-p)^{r-1}$ - it is just the probability that your first $r-1$ attempts all fail.
Now the expectation of a random variable taking values in $\{1,2,\ldots\}$ is given by $$\sum_{r=1}^\infty r\Pr(X=r)=\Pr(X\geq 1)+\sum_{r=2}^\infty (r-1)\Pr(X=r)\\=\Pr(X\geq 1)+\Pr(X\geq 2)+\sum_{r=3}^\infty(r-2)\Pr(X=r)\\ =\ldots$$ so the expectation equals $\sum_{r=1}^\infty\Pr(X\geq r)$. In this case, that is just $\sum_{r=1}^\infty(1-p)^{r-1}$. That is a geometric series having first term $1$ and common ratio $(1-p)<1$, so has value $\frac{1}{1-(1-p)}=\frac 1p$.