2004 AMC 10B Problem 11

There are $8 \times 8 = 64$ equally likely rolls. Most of them should work, so count the ones that fail.

1. A die shows $1$. Say the rolls are $1$ and $5$: product $5$, sum $6$. The product is always smaller, since $1 \times n = n$ but the sum is $n + 1$. So every roll with a $1$ fails.
- First die is $1$: $8$ rolls. Second die is $1$: $8$ rolls. The roll $(1,1)$ is on both lists, so there are $8 + 8 - 1 = 15$ such rolls.
2. Both dice are $2$ or more. Try the smallest cases:
- $2$ and $2$: product $4$, sum $4$. A tie, so it fails (the product must be greater).
- $2$ and $3$: product $6$, sum $5$. Works.
- $3$ and $3$: product $9$, sum $6$. Works.

Once both numbers are at least $2$ and not both equal to $2$, the product wins. For example, with $2$ and $n$: the product is $n + n$ and the sum is $2 + n$, and $n + n$ is bigger as soon as $n > 2$. Larger first numbers only make the product grow faster.
3. So the failing rolls are the $15$ with a $1$, plus $(2,2)$: $16$ rolls.
4. The rolls that work: $64 - 16 = 48$. The probability is
$$\frac{48}{64} = \frac{3}{4}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{3}{4}}$.

When almost everything works, count the few cases that fail and subtract. Trying the smallest numbers first shows where the trouble is: a $1$ makes the product small, and $2$ and $2$ is the one exact tie. Watch for rolls like $(1,1)$ that land on two lists at once.

Use algebra. With rolls $m$ and $n$, the condition $mn > m + n$ becomes $mn - m - n + 1 > 1$, and the left side factors as $(m-1)(n-1) > 1$. Now $m - 1$ and $n - 1$ are whole numbers from $0$ to $7$. Their product is $0$ when a die shows $1$ ($15$ rolls) and is exactly $1$ only when both dice show $2$ ($1$ roll). Every other roll works: $64 - 16 = 48$, giving $\frac{3}{4}$.