I am struggling with a bit of intuition when it comes to poker dice. When finding the probability of two pairs in poker dice I understand there are (5 choose 2) ways to choose which of the 5 dice make up the first pair and (3 choose 2) ways to select which of the remaining 3 make up the second pair. (6 choose 2) would then be selecting the 2 numbers that make up the pairs and (4 choose 1) the options left for the single remaining dice. All divided by the total number of possible rolls.
Why then, when finding the probability of a single pair, is it not (5 choose 2) to select the dice that make up the pair, (6 choose 1) to select its number and then (5 choose 3) to select the three remaining dice as opposed to the (5 choose 1)(4 choose 1)(3 choose 1) the answer requires ?
Or alternatively, why is the (6 choose 2) in the two pairs not (6 choose 1)(5 choose 1)?
It feels like a similar situation in the two hands treated differently.
Thank you for any insight.
Once you have selected the dice which are paired, you wish to select numbers for them that avoid over counting common situations.
The set of permutations of $\rm XXYYZ$, is the same set as the permutations of $\rm YYXXZ$. So you only want to count the permutations for all $X<Y$.
Thus we just count ways to select an ordered pair of values for the pairs, an other value for the singleton, and the permutations for two pair and singleton. $$\binom 6 2\binom 41\cdot\binom 52\binom 32\binom 11$$
A single pair (and three singletons) is similarly counted: the ways to select a number for the pair, three numbers for the singletons, and arrange the dice.
$$\binom 6 1 \binom 5 3\cdot \binom 5 2\binom 3 1\binom 21\binom 11$$
Once you have selected which two from the five dice are the pair, there are $3!$ to select which dice are the smallest, middle, and largest singleton value. Which is $\tbinom 3 1\tbinom 21\tbinom 11$.