Probability throwing 16 or less with 12 dice

Question

We've got 12 dice with the numbers 0,1,1,2,2,3 on it. What's the chance you throw 16 or less?

I've trouble calculating it because of the different probabilities of throwing 0 ($\frac{1}{6}$) or throwing 1 ($\frac{1}{3}$). It's too long to solve with a probability tree, another method could be using $(\frac{correct-outcomes}{total-outcomes})$ but I don't know how to get all the correct answers without using computer scripts.

I would like to understand the method behind the calculation so I can adjust it when we're doing different calculations with different numbers, f.e.: when throwing 10 dices, what's the chance of throwing 18 or less.

Answer 1

The probability generating function (PGF) for a single die is $$G_X(t)=\frac16+\frac13t+\frac13t^2+\frac16t^3$$ So the PGF for the sum of $12$ dice is given by $$G_Y(t)=\Big(\frac16+\frac13t+\frac13t^2+\frac16t^3\Big)^{12}$$ You want the probability that $Y\le16$ which is given by the sum of the first $17$ coefficients of the above expansion. One can either expand the above function to read each coefficient or use the following equation to find each probability seperately. $$P(Y=k)=\frac{G_Y^{(k)}(0)}{k!}$$ Either way we have $$P(Y\le16)=\frac{377773}{3779136} + \frac{21826079}{272097792} + \frac{666827}{11337408} + \frac{1783145}{45349632} + \frac{52010479}{2176782336} + \frac{397423}{30233088} + \frac{147719}{22674816} + \frac{524161}{181398528} + \frac{25795}{22674816} + \frac{11891}{30233088} + \frac{128161}{1088391168} + \frac{2717}{90699264} + \frac{143}{22674816} + \frac{575}{544195584} + \frac{1}{7558272} + \frac{1}{90699264} + \frac{1}{2176782336}$$ $$\boxed{P(Y\le16)=\frac{355296929}{1088391168}\approx0.3264423118}$$

Answer 2

Getting an exact answer requires the use of generating functions or the equivalent, as demonstrated in Peter Foreman's answer. However, one can get a good estimate by employing a normal approximation. We observe that a single roll of this die has first and second moments of $3/2$ and $19/6$, respectively, so the variance is $19/6 - 9/4 = 11/12$.

Normally, I'd say $12$ rolls isn't enough for a good normal approximation, but the die's native distribution already has a normal-ish kind of shape, so $12$ is probably good enough. We get a mean value of $12(3/2) = 18$, and a variance of $12(11/12) = 11$, and therefore a standard deviation of $\sqrt{11} \approx 3.316$. Using a discrete approximation bins this normal approximation, so that a roll of $16$ corresponds to a value between $15.5$ and $16.5$ in the continuous approximation. Therefore, we want the probability that a normal random variable with a mean of $18$ and a standard deviation of $\sqrt{11}$ falls below $16.5$. This is

$$ P\left(z < \frac{16.5-18}{\sqrt{11}}\right) \approx 0.3255 $$

which is about $0.3$ percent low.