Probability of a certain roll with six dice, each with a single marked side 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 three sets of two with 6 dice:
$$\frac{{6 \choose 3} {6 \choose 2} {4 \choose 2} {2 \choose 2}}{6^6} = \frac{1800}{6^6}$$
As there are ${6 \choose 3}$ ways to assign the numbers, ${6 \choose 2}$ ways to arrange the 1st pair of dice into the available positions, then ${4 \choose 2}$ for the 2nd pair and then ${2 \choose 2}$ for the 3rd pair.
An example of this kind of roll would be "112233" or "242554".
How can I determine how many of these rolls have 0 gold sides and how many have 1 gold side?
Finding how many have 2 gold sides is:
$${3 \choose 2} \times \frac{2}{6} \times \frac {2}{5} \times \frac {2}{4} \times 1800 = 360$$
As there are ${3 \choose 2}$ ways to assign the gold sides, $\frac{2}{6}$ chance the dice with the gold A is an A, then $\frac{2}{5}$ chance the dice with the gold B is a B, and then $\frac{2}{4}$ chance the dice with the gold C is not a C (in the context "AABBCC", where A, B, and C are different numbers).
And similarly 3 gold sides:
$${3 \choose 3} \times \frac{2}{6} \times \frac {2}{5} \times \frac {2}{4} \times 1800 = 120$$
But for 0-1 gold sides it's more complicated as the chance for the dice with the gold B depends on the position of the dice with the gold A, and so on for the gold C.
 A: Actually I didn't get how you calculated the number of possibilities for $2,3$ gold sides (though I have the same answers) but anyways this is my method (we'll solve for $2,3$ gold sides and then go for $1,0$ gold sides):
First of all, let's partition our $6$ dices into $3$ pairs of dices. Clearly there are $\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}=15$ ways to do that. Throughout this solution, the number of the two dices in each pair shall be equal and the number of each dice means the number of its golden side. Also we'll name pairs with a dice with gold side as a golden pair. (Trivially no golden pair has more than one dice with gold side)
Now for the case with $3$ gold sides, exactly one dice in each pair shall be golden. Hence, there are $2^3=8$ ways to choose the number of the dices of each pair. So there are $8*15=120$ ways for that.
For the case with $2$ gold sides, we'll first choose the pairs that shall have a golden side. There are $\binom{3}{2}=3$ choices for this. Then there are two choices for each of these pairs to choose which one of the dices of each of these pairs shall have golden side. So there are $3*2^2=12$ choices for this and finally for the third pair, there are $2$ possibilities for the number of the dices of this pair. (Excluding two numbers because it shouldn't be a golden pair and excluding another couple of numbers because the dices of this pair shall have different numbers than the number of first and second pairs) So finally there are $15\cdot12\cdot2=360$ choices in this case.
For the case for only one golden side, there are $3$ ways to choose the golden pair and $2$ choices to choose the dice with golden side. Now for the two other pairs, if the number of the second pair is the number of one of the dices in the third pair, then there are two choices for the number of the second pair and $3$ choices for the number of the third pair. Otherwise there is only $1$ choice for the number of the second pair, and $2$ choices for the number of the third pair. Therefore there are $8$ choices for the numbers of the two other pairs. So at last there are $15\cdot3\cdot2\cdot8=720$ possibilities for this case.
Finally for no golden sides, there are $4$ choices to choose the number on the dices of the first pair and WLOG assumer that this number is the number of one of the dices in the third pair. Then like the previous case, if the number of the second pair is in the third pair, then there is only $1$ possibility for the number of the second pair and $4$ choices for the number of the third pair. Otherwise there are $2$ possibilities for the number of the second pair and $3$ choices for the number of the third pair. Therefore there are $10$ different possibilities for choosing the number of the two other pairs. Hence there are $15\cdot4\cdot10=600$ different possibilities for this case.
P.S. You can verify the answer by noting that $120+360+720+600=1800$.
