2018 AMC 10B Problem 9

On a standard die, opposite faces add to $7$: $1$ and $6$, $2$ and $5$, $3$ and $4$. Use that to pair up rolls.

1. Flip each die. Take any roll of the $7$ dice and replace each face $k$ by the opposite face $7 - k$.
2. What happens to the sum? Each die's value changes from $k$ to $7 - k$. Over $7$ dice the new sum is
$$7 \times 7 - (\text{old sum}) = 49 - (\text{old sum}).$$
For example, the roll $1,1,1,1,1,1,4$ has sum $10$. Flipped, it becomes $6,6,6,6,6,6,3$, with sum $39$.
3. Why this gives equal chances. Every roll with sum $10$ flips to a different roll with sum $39$, and flipping again brings it back. So the two sums have exactly the same number of rolls, and they are equally likely.

The other sum is $49 - 10 = 39$.

The answer is $\boxed{\textbf{(D)}\ 39}$.

The smallest sum is $7$ and the largest is $42$, and the chances are symmetric around the middle. The sum $10$ is $3$ above the smallest, so its partner is $3$ below the largest: $42 - 3 = 39$. When a question asks for "another value with the same probability," look for a way to flip outcomes into each other.

A sum of $10$ is $3$ more than the lowest sum $7$ (all ones), so it comes from sharing $3$ extra points among the $7$ dice. A sum of $39$ is $3$ less than the highest sum $42$ (all sixes), so it comes from sharing $3$ missing points among the $7$ dice. Those are the same counting problem, so the chances match.

Pair every outcome with its "flipped" outcome, in which each die showing $k$ is replaced by a die showing $7 - k$. This is a one-to-one correspondence on the $6^7$ equally likely outcomes, and it turns a total of $S$ into a total of
$$7 \cdot 7 - S = 49 - S.$$
So the number of outcomes with sum $S$ equals the number with sum $49 - S$, and the two sums are equally likely.

For $S = 10$ the partner sum is $49 - 10 = 39$.

The answer is $\boxed{\textbf{(D)}\ 39}$.