In a rudimentary manner, favorable ways $=\dbinom{20}{x}\cdot 1^x\cdot 11^{20-x}$, against $12^{20}$ total ways,
or you could use the familiar binomial distribution,
$P(X = x) = \dbinom{20}{x}\cdot\left(\dfrac{1}{12}\right)^x\cdot\left(\dfrac{11}{12}\right)^{20-x}$
There are ${8 \choose 2}$ ways to place the threes, then ${6 \choose 3}$ ways to place the ones, and then ${3 \choose 3}$ to place the remaining sixes.
All together we can arrange these groups in
$${8\choose 2}\cdot{6\choose3}=\frac{8!}{2!\cdot6!}\cdot\frac{6!}{3!\cdot3!}=\frac{8!}{2!\cdot3!\cdot3!}$$
different ways.
Notice that this is a multinomial distribution which takes the form
$$P(X_1=x_1,...,X_k=x_k)=\frac{n!}{x_1!\cdots x_k!}p_1^{x_1}\cdots p_k^{x_k}$$
Let $X_3,X_1,X_6$ denote the numbers of threes, ones, and sixes observed, respectively. Then
$$P(X_3=2, X_1=3,X_6=3)=\frac{8!}{2!\cdot3!\cdot3!}\left(\frac{1}{6}\right)^8\approx3.334\cdot10^{-4}$$
where we do not have to take into account $X_2=X_4=X_5=0$ since $0!=1$ and $\frac{1}{6}^0=1$
R Simulation:
dice=c(1,2,3,4,5,6)
u = replicate(10^7,sample(dice,8,repl=T))
one=colSums(u==1)
two=colSums(u==2)
three=colSums(u==3)
four=colSums(u==4)
five=colSums(u==5)
six=colSums(u==6)
mean(three==2 & one==3 & six==3)
[1] 0.0003365
which agrees with our result fairly accurately.
Best Answer
The probability of throwing a 1 at each roll is 1/3. So, even if the first two rolls were 1, the probability of the next role being 1 is still 1/3 - the fact that you already threw that twice makes no difference. (See the gambler's fallacy for more on that).
That being said, the probability of all of the rolls being 1/3 does change because there are a lot more possible outcomes to choose from. In fact, as previously indicated, $(\frac{1}{3})^3$ is correct here. Think about it this way: for the first roll, there are 3 possible results (1, 2, or 3). For two rolls, there are nine possibilities because for each possible first roll, there are 3 possible first rolls and three possible second rolls, so there are 9 possible results: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3). For three dice, there are 27 possible solutions: e.g. (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 1), etc. The probability of rolling 3 consecutive 3s is therefore $\frac{1}{27} = \frac{1}{3}^3$.