2004 AMC 10B Problem 11

Call the rolls $m$ and $n$; there are $8 \cdot 8 = 64$ equally likely ordered pairs. The condition $mn > m + n$ rearranges as $mn - m - n > 0$; adding $1$ to both sides makes the left side factor:
$$(m - 1)(n - 1) > 1.$$

Both factors are integers from $0$ to $7$, so the inequality fails in exactly two situations:

• a factor is $0$, i.e. $m = 1$ or $n = 1$: $8 + 8 - 1 = 15$ pairs (the pair $(1,1)$ is counted once);
• the product is exactly $1$, i.e. $m = n = 2$: $1$ pair.

That is $16$ failing pairs, so $64 - 16 = 48$ pairs satisfy the condition, and the probability is
$$\frac{48}{64} = \frac{3}{4}.$$

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

Adding a constant to turn $mn - m - n$ into $(m-1)(n-1)$ is Simon's Favorite Factoring Trick; it converts a messy comparison into a product of nonnegative integers, whose small values are easy to list. Counting the failures is much faster than counting the successes, because the failures are concentrated on the edge of the grid.

Reason directly: if either die shows $1$, the product equals the other number, which is less than the sum. If both dice show at least $2$, then $mn \ge 2 \cdot \max(m,n) \ge m + n$, with equality only when $m = n = 2$. So the bad pairs are the $15$ containing a $1$ plus $(2,2)$, as before.

• Counting the pairs containing a 1 as 16 (double-counting (1,1)) or overlooking that (2,2) gives equality, landing on 47/64 or 55/64.
• Treating $(m, n)$ and $(n, m)$ as the same outcome for the total but not for the favorable count, which distorts the ratio.