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.
Best Answer
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 $$P(Y\le16)=\frac{377773}{3779136} + \frac{21826079}{272097792} + \frac{666827}{11337408} + \frac{1783145}{45349632} + \frac{52010479}{2176782336} + \frac{397423}{30233088} + \frac{147719}{22674816} + \frac{524161}{181398528} + \frac{25795}{22674816} + \frac{11891}{30233088} + \frac{128161}{1088391168} + \frac{2717}{90699264} + \frac{143}{22674816} + \frac{575}{544195584} + \frac{1}{7558272} + \frac{1}{90699264} + \frac{1}{2176782336}$$ $$\boxed{P(Y\le16)=\frac{355296929}{1088391168}\approx0.3264423118}$$