199. $n$ dice are thrown. Find the expected value of the sum of the number of points that will fall on all faces.
Solution. Let $X$ be the sum of the number of points that will fall on all faces, and $X_{i}(i=1,2, \ldots, n)$ be the number of points that fall on the $i$-th die face. Then, obviously,

$$
X=X_{1}+X_{2}+\ldots+X_{n}
$$

Therefore,

$$
\begin{aligned}
& M(X)=M\left(X_{1}+X_{2}+\ldots+X_{n}\right)= \\
& =M\left(X_{1}\right)+M\left(X_{2}\right)+\ldots+M\left(X_{n}\right) .
\end{aligned}
$$

Obviously, all quantities $X_{i}$ have the same distribution, and therefore the same numerical characteristics, and in particular, the same mathematical expectations, i.e., $M\left(X_{1}\right)=M\left(X_{2}\right)=$ $=\ldots=M\left(X_{n}\right)$.

By (*) we get

$$
M(X)=n M\left(X_{1}\right)
$$

Thus, it is sufficient to calculate the mathematical expectation of the quantity $X_{1}$, i.e., the mathematical expectation of the number of points that can fall on the first die. For this, we will write the distribution law of $X_{1}$:

| $X_{i}$ | 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: |
| $p$ | $1 / 6$ | $1 / 6$ | $1 / 6$ | $1 / 6$ | $1 / 6$ | $1 / 6$ |

Let's find $M\left(X_{1}\right)$:

$$
M\left(X_{1}\right)=1 \cdot 1 / 6+2 \cdot 1 / 6+3 \cdot 1 / 6+4 \cdot 1 / 6+5 \cdot 1 / 6+6 \cdot 1 / 6=7 / 2 . \quad(* * *)
$$

Substituting (***) into (**), we finally get

$$
M(X)=(7 / 2) n
$$