# 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? $\textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29$

~ pi_is_3.14

## 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 (Bashy) $(1, 2, 3); (1, 3, 4); (1, 4, 5); (1, 5, 6); (2, 3, 5); (2, 4, 6)$ have $6$ ways to rearrange them for a total of $36$ ways. $(1, 1, 2); (2, 2, 4); (3, 3, 6)$ have $3$ ways to rearrange them for a total of $9$ ways. Adding them up, we get $45$ ways. We have to divide this values by $6^3$ because there are 3 dice. $\dfrac{45}{216}=\boxed{\dfrac{5}{24}}$.

~MathFun1000

## Solution 3 (Summary of Solution 2)

Start by listing all possible sums. Then find the ways to arrange them and sum them up and divide by $6^3$ for 3 dice. $\dfrac{45}{216}$ is $\boxed{\textbf{(D)} \: \dfrac{5}{24}}$.

-aopspandy

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