My kids are playing a board game(*) whose rules have the following as their core. Paul throws 3 or 4 dice, each with 6 faces, 4 of which are red and 2 of which are purple. Alice does the same. Paul wins if the number of red faces in his throw is larger than the number of purple faces in Alice's throw. I am trying to compute the probability that Paul wins. I guessed that the probability to get $n$ red faces in a throw of $d$ dice is
$$\frac{\begin{pmatrix}d\\n\end{pmatrix}4^n 2^{d-n}}{6^d},$$
from which it is easy to compute the probabilities I want. But my combinatorials are weak! Is that correct? If so, what would be the simplest demonstration?
(*) based on the Miraculous TV series for the record!
As I mentioned in the comment this is true, and you can achieve the same result by:
Choose $n$ out of $d$ dice, on each of the $n$ dice you need a red face which has a probability of $2/3$ and on any other die you need a purple face which has a probability of $1/3$. Overall the probability is $$\binom{d}{n}\cdot \left(\frac{2}{3}\right)^n \left(\frac{1}{3}\right)^{d-n}$$