I wan't to obtain the probability of obtaining 2 pairs when throwing 6 dice. For example if the throw was (1, 1, 3, 3, 4, 6) there would be two pairs (the one of 1s and the one of 3s).
The number of posible outcomes (ordered) is $6^6$.
To calculate the number of outcomes where there are two pairs I would do:
$$ \binom{6}{2} \binom{6}{1} \binom{4}{2} \binom{5}{1} \binom{4}{1} \binom{3}{1}$$
First I choose 2 from 6 for the first pair, and one dice result from the possible 6. Then I choose from the remaining 4 another 2 for the other pair, and choose a dice result for the pair from the remaining 5. Finally I choose a result for the remaining throws from 4 and 3 remaining outcomes.
When I divide the above expression by $6^6$ I don't get the correct answer.
There is something about the ordering. In $6^6$ the expression gives all the ordered possibilities, and I think I'm computing the two-pair unordered possibilities. But my result is $\frac{25}{36}$ and the correct result is $\frac{25}{72}$, so if I consider ordered posibilites my answer would be further away of the correct one.
Best Answer
You write favorable outcomes as $ \displaystyle \color {blue}{\binom{6}{2} \binom{6}{1} \binom{4}{2} \binom{5}{1}} \binom{4}{1} \binom{3}{1}$
The part highlighted in blue is where you are double-counting. I will name dice as $A - F$ and each dice shows a number between $1$ and $6$. Look at it this way - I first choose $2$ numbers out of $6$ possible numbers for two pairs of dice and that is $\displaystyle {6 \choose 2}$. Now when I choose $2$ pairs of dice $ \displaystyle {6 \choose 2} {4 \choose 2}$, they are already ordered.
To explain further, say, the two chosen numbers are $1$ and $6$, and we assign $1$ to first two chosen dice and $6$ to next two chosen dice. Now say, I choose dice $A$ and $B$ first (so assign $1$ to them) and then, say, we choose dice $C$ and $D$ (so assign $6$ to them) but ${6 \choose 2} {4 \choose 2}$ will also count cases where I choose dice $C$ and $D$ first and then dice $A$ and $B$.
So if we are choosing two pairs of dice as $\displaystyle {6 \choose 2} {4 \choose 2}$, we should multiply it by $\displaystyle {6 \choose 2}$ instead of by $\displaystyle {6 \choose 1} {5 \choose 1}$.