9. (20,9-11) Aся and Vasya are cutting rectangles out of graph paper. Vasya is lazy; he rolls a die once and cuts out a square with a side equal to the number of points that come up. Aся rolls the die twice and cuts out a rectangle with length and width equal to the numbers that come up. Whose expected area of the rectangle is greater?
# Solution.

Let $A$ and $B$ be independent random variables taking values 1, 2, ..., 6 with equal probabilities. Then the expected value of the area of A's rectangle is $\mathrm{M} A B=\mathrm{M} A \cdot \mathrm{M} B$ (since $A$ and $B$ are independent). The expected value of the area of Vasya's rectangle is $\mathrm{M} A^{2}$. Let's compare these two expectations.

$$
\begin{gathered}
\mathrm{M} A=\mathrm{M} B=\frac{1}{6}(1+2+\ldots+6)=3.5 ; \mathrm{M} A B=3.5 \cdot 3.5=12.25 \\
\mathrm{M} A^{2}=\frac{1}{6}(1+4+\ldots+36)=15 \frac{1}{6}>12.25
\end{gathered}
$$

Thus, on average, the area of Vasya's square is greater than the area of A's rectangle.

Answer: Vasya.