An alternate approach would be to find the probability of the complementary event:
$\textbf{1)}$ The probability of getting no 6's is given by $\big(\frac{5}{6}\big)^3\cdot\big(\frac{5}{6}\big)^3=\big(\frac{5}{6}\big)^6$
$\textbf{2)}$ The probability of getting exactly one 6 is given by
$\hspace{.2 in}\big(\frac{5}{6}\big)^3\cdot3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2+3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2\cdot\big(\frac{5}{6}\big)^2=\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5+\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4$
Therefore the probability of getting at least two 6's is given by
$\hspace{.2 in} 1-\big(\frac{5}{6}\big)^6-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4=\frac{5203}{23328}\approx.223$
The generating function approach is to ask for the coefficients of $x^{n}$ in the expansion of:
$$(x+x^2+x^3+x^4+x^5+x^6)^{3}=x^3\left(\frac{1-x^6}{1-x}\right)^3$$
Now, $$\frac{1}{(1-x)^3}=\sum_{k=0}^{\infty}\binom{k+2}{2}x^k$$
And $$(1-x^6)^3=1-3x^6+3x^{12}-x^{18}$$
So the product of these and $x^3$ has, for the coefficient $x^n$:
$$\binom{n-1}{2}-3\binom{n-7}{2}+3\binom{n-13}{2}-\binom{n-19}{2}$$
Trick is to treat $\binom{j}{2}=0$ when $j<2.$
So, for example, when $n=11$, you get $\binom{10}{2}-3\binom{4}{2}=45-18=27$.
This also gives you a hint of the value when dealing with $k$ dice. Then:
$$c_n = \sum_{i=0}^{k}(-1)^i\binom{k}{i}\binom{n-(1+6i)}{k-1}$$
If you don't like generating functions, this can be proven via inclusion-exclusion.
There's another approach that is a little faster for computing all values.
If $(1+x+x^2+\cdots+x^5)^3=a_0+a_1x+\cdots+a_nx^n+\cdots$ then you get that:
$$(a_0+a_1x+\cdots+a_nx^n+\cdots)(1-3x+3x^2-x^3)=(1-x^6)^3=1-3x^6+3x^{12}-x^{18}$$
What this means is that (setting $a_n=0$ when $n<0$:
$$a_{n}-3a_{n-1}+3a_{n-2}-a_{n-3}=\begin{cases}
1&n=0\\
-3&n=6\\
3&n=12\\
-1&n=18\\
0&\text{otherwise}
\end{cases}$$
or:
$$a_{n}=3\left(a_{n-1}-a_{n-2}\right)+a_{n-3}+\begin{cases}
(-1)^k\binom{3}{k}&n=6k\\
0&\text{otherwise}
\end{cases}$$
Then the final coefficient of $x^n$, after we multiply by $x^3$ again, is $a_{n-3}$.
So you get:
$$\begin{align}a_0&=0+(-1)^{0}\binom{3}{0}=1\\
a_1&=3\left( a_0-a_{-1}\right)=3\\
a_2&=3\left( a_1 - a_0\right)=6\\
a_3&=3\left(a_2 - a_1\right)+a_0=10\\
a_4&=3\left(a_3-a_2\right)+a_1=15\\
a_5&=3\left(a_4-a_3\right)+a_2=21\\
a_6&=3\left(a_5-a_4\right)+a_3+(-1)^{1}\binom{3}{1}=25\\
&\cdots
\end{align}$$
There's a special trick that can be applied here: $a_{n}=3a_{n-1}-3a_{n-2}+a_{n-3}$ is known to be a quadratic polynomial in $n$. So, when $n$ is not a multiple of $6$, you get that:
$$a_{n-2}-a_{n-3},a_{n-1}-a_{n-2},a_{n}-a_{n-1}$$ must be an arithmetic progression for any $n$ not a multiple of $6$.
So we have that $a_5-a_4 = 6, a_6-a_5=4$ and thus $a_7-a_6=2$, or $a_7=a_6+2=27.$ $a_8=27+0=27, a_9=27-2=25,a_{10}=25-4=21,a_{11}=21-6=15,a_{12}=15-8+(-1)^2\binom{3}{2}=10$.
Best Answer
If we have a blue die and a green die, each outcome can be written as an ordered pair $(b, g)$, where $b$ is the number that appears on the blue die and $g$ is the number that appears on the green die. Since there are six possible outcomes for each die, there are $6 \cdot 6 = 36$ such ordered pairs.
Notice that $(2, 1) \neq (1, 2)$ since $(a, b) = (c, d) \iff a = c$ and $b = d$. On the other hand, $(2, 2) = (2, 2)$ since $a = c$ and $b = d$.
Clearly, the events $(2, 1)$ and $(1, 2)$ are distinguishable (to people who can distinguish blue from green) since $(2, 1)$ means we have a $2$ on the blue die and a $1$ on the green die, while $(1, 2)$ means we have a $1$ on the blue die and a $2$ on the green die. On the other hand, the only way $(2, 2)$ can occur is if we have a $2$ on both dice. Thus, there are two ways to get a $1$ and a $2$ when we roll two distinguishable dice, but only one way to get two $2$'s.