Probability of 3 of a kind - Probability of Pair

Question

Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)?

Here is my thinking. There are $6$ possible values for the pair. (e.g $1$,$1$,$2$,$2$...$6$,$6$). The other four dice can be any number other than the number of the pair. Numerator = $6\cdot5\cdot5\cdot4\cdot4$. Denominator = $5^6$. So I thought that the answer was $\frac{2400}{46656}$. Can this be verified? Help is greatly appreciated.

Answer 1

Not quite.

Two mistakes:

Yes, there are $6$ options for the first pair, after which you have $5$ options for the third die, and another $5$ for the fourth ... and then there is only one more die left .. so you certainly don't get two more $4$'s in the numerator. I think you confused yourself by having that single $6$ ... which seems to suggest it trepresents one die, but it represents two. In fact, if you want it to do die for die, you can think: OK, the first die can be anything, but then the second needs to be the same as the first. So, you get $6 \cdot 1 \cdot 5 ...$ ... and now tyou wouldn;t have made the mistake of putting an extra $4$ in there.

But .. you don't get a single $4$ either! ... for that would only be true if the third and fourth dice are the same. So, consider the case where the third and fourth are the same, and the case where they are not, and add those up.

Answer 2

There are a total of $6^5=7776$ possible sets of dice rolls. To get a pair without a three-of-a-kind, we can either have one pair and the other three dice all showing different numbers, or we have two pairs and the die showing something different.

In the first case, there are $6$ ways to pick which number makes a pair and $\binom{5}{2}=10$ ways to pick which $2$ of the $5$ dice show that number. Out of the other three dice, there are $5$ ways to pick a value for the first die so that that die doesn't match the pair, $4$ ways to pick a value for the second one so it doesn't match that die or the pair, and $3$ ways to pick a value for the last die so that it doesn't match any of the others. So there are $$6\cdot 10\cdot 5 \cdot 4 \cdot 3 = 6^2 \cdot 100$$ ways to roll this case.

In the second case, to form two pairs and one die not part of those pairs, there are $\binom{6}{2}=15$ ways to pick which two numbers make the pairs, then $4$ ways to pick a value for the last die so that it doesn't match either of those pairs. There are $$\frac{5!}{2!\cdot 2!\cdot 1!}=30$$ ways order the five dice (equal to the number of ways to order XXYYZ), so that makes a total of $$15\cdot 4 \cdot 30 = 6^2\cdot 50$$ ways to roll this case.

This makes a total of $$6^2 \cdot 100 + 6^2 \cdot 50 = 6^2 \cdot 150 = 6^3 \cdot 25$$ ways to roll a pair without rolling a three-of-a-kind. So, the probability is $$\frac{\text{successful outcomes}}{\text{total outcomes}}=\frac{6^3 \cdot 25}{6^5}=\frac{25}{6^2}=\boxed{\frac{25}{36}}.$$