2014 AMC 10B Problems/Problem 16
Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?
We split this problem into cases.
First, we calculate the probability that all four are the same. After the first dice, all the numbers must be equal to that roll, giving a probability of .
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 orders to roll the different dice, giving .
Adding these up, we get , or .
Note that there are two cases for this problem
: Exactly three of the dices show the same value.
There are values that the remaining die can take on, and there are ways to choose the die. There are ways that this can happen. Hence, ways.
: Exactly four of the dices show the same value.
This can happen in ways.
Hence, the probability is
We solve using PIE.
We first calculate the number of ways that we can have dice be the same and the other dice be anything. We therefore have ways to have at least dice be the same.
But wait! We have overcounted the case where all dice are the same! Since the previous case occurs in each of these cases times, we must subtract the -dice total three times in order to have them counted once. There are ways to have four dice be the same, so we our total count is .
Therefore, our probability is , which is answer choice .
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