You roll a standard six-sided die, then roll n more six-sided die, where n was the first roll. What is the expected value of the sum of all the die you rolled?
I did some calculations, but I'm not sure. Let expected value of the sum of all the die rolls be E[X], then $$E[X] = \frac{3.5}{6}+2\frac{3.5}{6}+3\frac{3.5}{6}+4\frac{3.5}{6}+5\frac{3.5}{6}+6\frac{3.5}{6}$$ because the expected value of each single die is 3.5. Is it correct?
Let $N$ be the outcome of the first six-sided die. Then you want to find $$ \mathbb{E}[N + \sum_{i = 1}^N X_i] $$ where $X_i$ is the outcome of the $i$-th six-sided die. The trick is to condition on the value of $N$. Assuming that $N$ can be any value between 1,..,6 we have: $$ \mathbb{E}[N + \sum_{i = 1}^N X_i] = \sum_{j = 1}^6 \mathbb{E}[N + \sum_{i = 1}^N X_i | N = j]\mathbb{P}(N = j) = \sum_{j = 1}^6 (j + \mathbb{E}[\sum_{i = 1}^j X_i])\mathbb{P}(N = j), $$ Then you can fill in the known $\mathbb{P}(X = j)$ and use that $$ \mathbb{E}[\sum_{i = 1}^j X_i] = \sum_{i = 1}^j \mathbb{E}[X_i] $$ when $i,j$ are known.