# 2014 AMC 10B Problems/Problem 16

## Problem

Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value? $\textbf {(A) } \frac{1}{36} \qquad \textbf {(B) } \frac{7}{72} \qquad \textbf {(C) } \frac{1}{9} \qquad \textbf {(D) } \frac{5}{36} \qquad \textbf {(E) } \frac{1}{6}$

## Solution

We split this problem into $2$ cases.

First, we calculate the probability that all four are the same. After the first dice, all the number must be equal to that roll, giving a probability of $1 \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{216}$.

Second, we calculate the probability that three are the same and one is different. After the first dice, the next two must be equal and the third different. There are $4$ orders to roll the different dice, giving $4 \cdot 1 \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{5}{6} = \dfrac{5}{54}$.

Adding these up, we get $\dfrac{7}{72}$, or $\boxed{\textbf{(B)}}$.

## Solution 2

Note that there are two cases for this problem $\textbf{Case 1}$: Exactly three of the dices show the same value.

There are $5$ values that the remaining die can take on, and there are $\binom{4}{3}=4$ ways to choose the die. There are $6$ ways that this can happen. Hence, $6\cdot 4\cdot5=120$ ways. $\textbf{Case 2}$: Exactly four of the dices show the same value.

This can happen in $6$ ways.

Hence, the probability is $\frac{120+6}{6^{4}}=\frac{21}{216}\implies \frac{7}{72}\implies \boxed{\textbf{(B)}}$

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