# 2014 AMC 10A Problems/Problem 17

## Problem

Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die? $\textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29$

## Solution 1

First, we note that there are $1, 2, 3, 4,$ and $5$ ways to get sums of $2, 3, 4, 5, 6$ respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is $$\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.$$ Since there are $\dbinom31$ ways to choose which die will be the one with the sum of the other two, our answer is $3 \cdot \dfrac{5}{72} = \boxed{\textbf{(D)} \: \dfrac{5}{24}}$.

## Solution 2

Since there are $6$ possible values for the number on each dice, there are $6^3=216$ total possible rolls.

The possible results of the 3 dice such that the sum of the values of two of the die is equal to the value of the third die are, without considering the order of the die, $$(1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6), (3, 3, 6)$$

There are $\frac{3!}{2}=3$ ways to order the first, sixth, and ninth results, and there are $3!=6$ ways to order the other results.

Therefore, there are a total of $3\times3+6\times6=45$ ways to roll the dice such that 2 of the dice sum to the other die, so our answer is $$\frac{45}{216}=\boxed{\textbf{(D)} \ \frac{5}{24}}$$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 