2009 AMC 10A Problem 22

The labels are attached to the twelve faces uniformly at random, and the roll then picks one face of each die. By symmetry, the label on the first die's top face is equally likely to be any of the $12$ labels in the bag, and the label on the second die's top face is equally likely to be any of the remaining $11$. So the pair of top labels is a uniformly random ordered pair of two distinct labels from the multiset $\{1,1,2,2,3,3,4,4,5,5,6,6\}$.

Condition on the first label, $x$. The sum is $7$ exactly when the second label is $7 - x$. Because $7$ is odd, $7 - x \ne x$, so removing $x$ leaves both copies of $7 - x$ in play: $2$ favorable labels out of $11$.

This is the same for every $x$, so the probability is $\dfrac{2}{11}$.

The answer is $\boxed{\textbf{(D)}\ \frac{2}{11}}$.

Randomly assigning labels and then randomly choosing a face is the same as randomly choosing labels directly; the cubes' geometry is irrelevant. Once the problem is "draw two labels without replacement from a bag of twelve," it is a one-line conditional count. Look for this kind of symmetry whenever a random relabeling is followed by a random selection.

Count unordered pairs of labels: $\binom{12}{2} = 66$ equally likely pairs of faces (one on each die is forced by the roll, but by symmetry any two labels are equally likely to be the pair). Pairs summing to $7$: the value pairs $\{1,6\}, \{2,5\}, \{3,4\}$, each realized by $2 \times 2 = 4$ label pairs, for $12$ total. Probability $\frac{12}{66} = \frac{2}{11}$.

• Assuming each die still shows 1 through 6 and answering the usual 1/6, or forgetting the first label is gone and answering 2/12.
• Trying to enumerate face assignments ($12!$ of them); the symmetry argument makes that entirely unnecessary.