2014 AMC 10A Problem 17

Treat the dice as three different colours, say red, white and blue. Then there are $6 \cdot 6 \cdot 6 = 216$ rolls, all equally likely. Count the rolls where one die equals the sum of the other two.

1. Which die is the total? It can be the red, the white or the blue one: $3$ choices.
2. The other two dice. They must add up to the total die, and a die shows at most $6$, so their sum is at most $6$. Count the ordered ways for each sum:

That is $1 + 2 + 3 + 4 + 5 = 15$ ways. Once these two dice are chosen, the total die is fixed.
3. No roll counted twice. Could one roll have the red die as the total and also the white die as the total? Then red $=$ white $+$ blue and white $=$ red $+$ blue, which forces blue to be $0$. A die never shows $0$, so this cannot happen.
4. Favourable rolls: $3 \cdot 15 = 45$.
5. Probability: $\frac{45}{216}$. Divide top and bottom by $9$: $\frac{5}{24}$.

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

"Two of them add to the third" really says the largest die is the sum of the other two. Choosing that die first, then counting pairs with a given sum, keeps the counting clean. Keep the dice ordered (different colours) both in the total $216$ and in the good cases, so both counts are measured the same way.

Count sets of values first, then count how many ways each set can be arranged on the three dice.

1. Two equal small dice: $(1,1,2)$, $(2,2,4)$, $(3,3,6)$. Each has $3$ arrangements (choose where the odd one out goes). That gives $3 \cdot 3 = 9$.
2. Three different values: $(1,2,3)$, $(1,3,4)$, $(1,4,5)$, $(1,5,6)$, $(2,3,5)$, $(2,4,6)$. Each has $6$ arrangements. That gives $6 \cdot 6 = 36$.
3. Total $9 + 36 = 45$, and $\frac{45}{216} = \frac{5}{24}$.

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