Occurrences of 1,2, and 3 out of 4d6

Question

I have 4d6 and I'm concerned with the odds of the numbers of ones, twos, and threes that show up in a roll of 4d6. Would I be correct in calculating the odds of one of these combinations, Say 1 one, 0 twos and 0 threes, by $\left({\frac{1}{6}} \right)\left({\frac{3}{6}} \right)\left({\frac{3}{6}} \right)\left({\frac{3}{6}} \right) = \left({\frac{1}{48}} \right)$ or am I missing something super important that complicates things?

Answer 1

You are talking about a multinomial distribution distribution. Suppose you have $n$ independent trials where each trial results in one of $k$ types with respective probabilities $p_1,\dots, p_k$. Then the chance that type 1 shows up $x_1$ times, type 2 shows up $x_2$ times, etc. is
$$\mathbb{P}(X_1=x_1,\dots, X_k=x_k)={n!\over x_1!\cdots x_k!}\, p_1^{x_1}\cdots p_k^{x_k}$$

Now in your problem, you have four types: "ones", "twos", "threes", and "others" with respective probabilities $p_1=1/6$, $p_2=1/6$, $p_3=1/6$, $p_4=3/6$. The probability of rolling four dice and getting one one, and no twos or threes is $${4!\over 1!\, 0!\, 0!\, 3!}\left({1\over6}\right)^1 \left({1\over6}\right)^0\left({1\over6}\right)^0\left({3\over6}\right)^3={1\over 12}=.083333$$