Probability breakdown of a certain roll There are six regular (6-sided) dice being rolled. However, each dice has one side colored gold. The 1st has a gold 1, the 2nd has a gold 2... and the 6th has a gold 6.
I calculate the probability of rolling two sets of three with 6 dice:
$$\frac{{6 \choose 2} {6 \choose 3} {3 \choose 3}}{6^6} = \frac{300}{6^6}$$
As there are ${6 \choose 2}$ ways to assign the numbers, ${6 \choose 3}$ ways to arrange the 1st set of 3 dice into the available spaces and ${3 \choose 3}$ for the 2nd set.
An example of this kind of roll would be 1 1 1 2 2 2 or 3 5 3 5 5 3 (i.e. two groups of 3 of a kind).
How can I determine how many of these 300 rolls have 0 gold sides, how many have 1 gold side, and how many have 2 gold sides.
For instance, the roll 2 1 2 1 2 1 would have 0 gold sides, the roll 1 1 1 2 2 2 would have 1 gold side (the 1st dice), and the roll 1 2 1 2 1 2 would have 2 gold sides (the 1st and 2nd dice).
 A: It seems to be $\displaystyle {4 \choose 2}{6 \choose 2}*2=180$.
You choose which 2 of the 6 numbers to be the gold faces. Then you choose which 2 of the 4 remaining numbers will be the first number (none of them possibly being a gold face), and the other two are left for the second number.
E.g.
1 x x 4 x x
1 1 x 4 x 1
1 1 4 4 4 1
Lastly, I think you may have made a mistake in your calculation. There are ${6 \choose 3}{6 \choose 2}*2=600$ ways to get three each of two numbers. You need to multiply by 2 because after you’ve selected the positions and the numbers, you can put the numbers in one way, or the opposite.
_ _ * _ * *
3,5
3 3 5 3 5 5
5 5 3 5 3 3
A: Note: "WLOG" means "Without Loss of Generality"
WLOG, the roll is 3-"1"'s and 3-"2"'s, sequenced as "111222".
WLOG, D-1 is the die with a gold "1", D-2 is the die with a gold "2".
Chance of D-1 being in the last 3 numbers times chance of D-2 then being in the first 3 numbers = chance of Gold = 0 ($G_0$)
$$G_0 = \frac{3}{6} \times \frac{3}{5}.$$
Chance of D-1 being in the first 3 numbers times chance of D-2 then being in the second 3 numbers = chance of Gold = 2 ($G_2$)
$$G_2 = \frac{3}{6} \times \frac{3}{5}.$$
Obviously,
$$G_1 = 1 - (G_0 + G_2).$$
$(G_x \times n) $ where $n$ is the number of throws gives the number of throws that produce exactly $x$ gold numbers.
