2014 AMC 10A Problem 17

There are $6^3 = 216$ equally likely ordered rolls. Since every face is positive, if two dice add to the third, the third must be the largest value, so we count ordered triples in which one die equals the sum of the other two.

Step 1: choose which of the three dice is the "total." There are $3$ choices.

Step 2: the remaining two dice, in order, must add to that total $s$, and $s \le 6$. The number of ordered pairs of faces adding to $s$ is $s - 1$:

That is $1 + 2 + 3 + 4 + 5 = 15$ ordered pairs.

Step 3: no roll is counted twice. If die $1$ equals die $2$ plus die $3$ and also die $2$ equals die $1$ plus die $3$, then die $3$ would be $0$, impossible.

Favorable rolls: $3 \cdot 15 = 45$. The probability is
$$\frac{45}{216} = \frac{5}{24}.$$

The answer is $\boxed{\textbf{(D)}\ \dfrac5{24}}$.

A condition like "two of them add to the third" is really a condition about the maximum, which removes the ambiguity of which die plays which role. Counting ordered pairs with a fixed sum ($s-1$ pairs for $s \le 7$) is a standard dice fact worth memorizing. Keeping everything ordered, both in the total $216$ and in the favorable count, avoids the overcount/undercount mess of unordered outcomes.

Count unordered multisets and weight by arrangements. Pairs $\{x, y\}$ with $x + y = s \le 6$: with $x = y$ there are $3$ (sums $2, 4, 6$), each giving a triple like $(1,1,2)$ with $3$ arrangements; with $x \ne y$ there are $6$, each giving a triple of distinct values with $6$ arrangements. Total $3 \cdot 3 + 6 \cdot 6 = 45$.

• Forgetting the factor of 3 for which die is the sum, or treating the dice as unordered while still dividing by 216.
• Including sums $s = 7$ or more, which no die can show; the count of pairs must stop at $s = 6$.
• Worrying about "double counting" triples such as $(1,1,2)$; the position of the total die is fixed, so each roll is counted exactly once.