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Printjmc
counting and probability senior
Problem
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)?
Solution
There are a total of 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 fifth die showing something different.
In the first case, there are ways to pick which number makes a pair and ways to pick which of the dice show that number. Out of the other three dice, there are ways to pick a value for the first die so that that die doesn't match the pair, ways to pick a value for the second one so it doesn't match that die or the pair, and ways to pick a value for the last die so that it doesn't match any of the others. So there are ways to roll this case.
In the second case, to form two pairs and one die not part of those pairs, there are ways to pick which two numbers make the pairs, then ways to pick a value for the last die so that it doesn't match either of those pairs. There are ways order the five dice (equal to the number of ways to order XXYYZ), so that makes a total of ways to roll this case.
This makes a total of ways to roll a pair without rolling a three-of-a-kind. So, the probability is
In the first case, there are ways to pick which number makes a pair and ways to pick which of the dice show that number. Out of the other three dice, there are ways to pick a value for the first die so that that die doesn't match the pair, ways to pick a value for the second one so it doesn't match that die or the pair, and ways to pick a value for the last die so that it doesn't match any of the others. So there are ways to roll this case.
In the second case, to form two pairs and one die not part of those pairs, there are ways to pick which two numbers make the pairs, then ways to pick a value for the last die so that it doesn't match either of those pairs. There are ways order the five dice (equal to the number of ways to order XXYYZ), so that makes a total of ways to roll this case.
This makes a total of ways to roll a pair without rolling a three-of-a-kind. So, the probability is
Final answer
\frac{25}{36}