Probability question involving colored dice This question is really getting to me, a friend of mine has suggested a solution but I believe it is incorrect.
Anyway here's the question: 

You throw 9 coloured dice. Every dice has 2 white sides, 2 red sides and 2 blue sides. What is the chance to get every color three times?

I'd be over the moon if you can help me find the solution as it's been getting to me. Thanks.
 A: Since each colour is on two sides, the probability for each colour on each die is $1/3$. (Unfortunately we can't say that the dice are equivalent to $3$-sided dice because there's no such thing.)
The probability to get every colour thrice is the number of ways of colouring nine dice with three dice coloured in each colour, over the number of ways of colouring nine dice with three colours. The latter is simply $3^9$, and the former is the multinomial coefficient
$$
\left({9\atop3,3,3}\right)=\frac{9!}{3!3!3!}=1680\;,
$$
so the probability is
$$
\frac{1680}{3^9}=\frac{560}{3^8}=\frac{560}{6561}\approx0.085\;.
$$
A: The probability we get exactly $3$ white is, by the standard "binomial distribution" formula, equal to 
$$\binom{9}{3}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^6.$$
Given that we got exactly $3$ blue, the number of red is equal to $3$ with probability
$$\binom{6}{3}\left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^3.$$
Multiply. There is some simplification, since the $\left(\frac{1}{2}\right)^6$ in the second expression cancels the $2$'s in the first. And the product 
$\binom{9}{6}\binom{6}{3}$ "simplifies" to $\dfrac{9!}{(3!)^3}$.  
