Asya 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. Asya 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

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

Since $0(\mathrm{E} A)^{2}$.

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

## Answer

Vasya.

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