5 fair dice are rolled. A pair is defined to be any number that shows up twice, while the rest of the dice show different numbers (to the number on the pair and to each other). I am looking for the probability of a pair occurring.
My thoughts are that the number of choices for the repeated number (the pair) is $6$, the number of choices for the next number is $5$ then the next number $4$, then $3$. So the total number of choices I have for the possible combination of 6 unique numbers for 5 dice (with $1$ repeated) is $6\cdot5\cdot4\cdot3$. Therefore for combination above, the number of ways to rearrange the the two identical numbers that form the pair is $\binom{5}{2}$ so the total number of pairs that exist is: $$\binom{5}{2}6\cdot5\cdot4\cdot3$$ I am wondering if this is the right way to count the number of combinations that form a pair?
There are ${6\choose4}$ ways to choose the four numbers which appear. Once this choice is made, there are $4$ ways to choose the number appearing twice. Once these choices are made, there are ${5\choose2}$ ways to choose the places where the number appearing twice appears. Once all these choices are made, there are $3!$ ways to choose the places where the other three numbers appear.
Since there is a total of $6^5$ results, the desired probablility is $$\frac1{6^5}\cdot{6\choose4}\cdot4\cdot{5\choose2}\cdot3!=\frac{25}{54}\approx46.3\%.$$ This is the number you suggest in your question, only, motivated differently.