PMF for total on a random number of dice? I am not sure how to do this, I know the PMF of rolling a dice is:
$$
P(x)=\begin{cases}
1/6 &x\in\{1,2,3,4,5,6\}\\ 
0& \text{Otherwise.}
\end{cases}
$$
 A: Hint:
The mean can be found on base of:$$\mathbb E(Y)=\sum_{i=1}^6\mathbb E(Y\mid N=i)P(N=i)$$
A: You can use generating functions to see the pmf of the sum. The pmf, when you increase the number of addends, tends to be a normal shape, what is the content of the central limit theorem. 
If you represent the sum $Y$ as a random variable composed of $N_i$ identical random variable and you apply the theorem about linearity of expectation
$$\Bbb E[Y]=\Bbb E(N_1 +N_2 + N_3+...)=\Bbb E[N_1]+\Bbb E[N_2]+\Bbb E[N_3]+...$$
A: Comment:
This seems to 'a random sum of random numbers': $Y = \sum_{i=1}^N D_i,$
where $N$ is found by rolling a fair die and, independently the $D_i$ are also found by rolling fair dice. Then standard formulas are
$E(Y) = E(N)E(D)$ and $V(Y) = E(N)V(D) + V(N)[E(D)]^2,$ which can
be proved by conditioning methods as in another hint. 
A simple simulation in R of 100,000 such sums gives the following
numerical and graphical results, which you can compare with
your analytic answer:
 m = 10^5;  y = numeric(m)
 for (i in 1:m) {
    n = sample(1:6, 1)
    y[i] = sum(sample(1:6, n, repl=T)) }
 mean(y);  var(y)
 ## 12.23903  # approx E(Y) = 3.5(3.5) = 12.25
 ## 46.13508  # approx V(Y)

The simulated PDF is 'lumpier' than I guessed, but several
runs of the simulation gave similar results.

