2012 AMC 10A Problem 9

1. Each value appears on two of the six faces, so the even die shows $2$, $4$ or $6$, each one-third of the time. The odd die shows $1$, $3$ or $5$, each one-third of the time. In effect, each die has only three equally likely values.
2. So there are $3 \times 3 = 9$ equally likely pairs of values. Here they are, with the sum in each box:

1. Three of the nine boxes have sum $7$: $2 + 5$, $4 + 3$ and $6 + 1$.
2. The probability is $\frac39 = \frac13$.

The answer is $\boxed{\textbf{(D)}\ \frac{1}{3}}$.

Two faces with the same number act like one face that comes up twice as often. As long as every value is equally likely, you can work with values instead of faces, and the table shrinks from $36$ boxes to $9$. The table also shows a shortcut: each row has exactly one $7$, so whatever the even die shows, the chance of making $7$ is $\frac13$.

Count all $6 \times 6 = 36$ face pairs. The value pair $2$ and $5$ can happen in $2 \times 2 = 4$ ways, because there are two faces showing $2$ and two showing $5$. The same goes for $4$ and $3$, and for $6$ and $1$. That makes $3 \times 4 = 12$ good face pairs, and $\frac{12}{36} = \frac13$.

Because each value appears on two of the six faces, the even die shows $2$, $4$ or $6$ each with probability $\frac13$, and the odd die shows $1$, $3$ or $5$ each with probability $\frac13$. So the $3 \times 3 = 9$ value pairs are equally likely.

List the pairs whose sum is $7$:
$$(2, 5), \quad (4, 3), \quad (6, 1).$$
Every even value has exactly one odd partner, so there are $3$ favorable pairs out of $9$.

The probability is $\frac{3}{9} = \frac{1}{3}$.

The answer is $\boxed{\textbf{(D)}\ \frac{1}{3}}$.