According to $f(k)=\binom{n}{k}p^k(1-p)^{n-k}$ and because $p = \frac16$ for each die we calculate the probability of two sixes with three dice thrown $f(2)=3\frac16^2(1-\frac16)=\frac1{12}\frac5{6}=\frac5{72}$.
Right now I'm facing an unfair dice problem, e. g. the probabilities of the dice for a six would be $p_1=\frac16,\quad p_2=\frac2{11},\quad p_3=\frac3{20}$.
How do we now calculate the probability of two sixes, suspecting that $p = p_1p_2p_3$ would not be the right solution?
Probably we have to calculate $p$ differently or use a nested equation like $f(k)=f_1(k)f_2(k)f_3(k)$? So far my ideas.