I would like to calculate the probability distribution of the sum of all the faces of $N$ dice rolls. The face probabilities ${p_i}$ are know, but are not $1 \over 6$.
I have found answers for the case of a fair dice (i.e. $p_i={1 \over 6}$) here and here
For large $N$ I could apply the central limit theorem and use a normal distribution, but I don't know how to proceed for small $N$. (In particular, $N=2,4, 20$)
Best Answer
You can use generating functions.
Let $P=p_1x+p_2x^2+p_3x^3+p_4 x^4+p_5 x^5 +p_6 x^6$ where $p_i$ is the probability of $i$ occurring when rolling the die once.
Then the coefficient of $x^k$ in $P^N$ gives the probability of rolling a sum of $k$ when rolling the die $N$ times and summing.
For example, suppose $P=\frac{1}{7}x+\frac{1}{7}x^2+\frac{1}{7}x^3+\frac{1}{7}x^4+\frac{1}{7}x^5 + \frac{2}{7}x^6$.
Then, using a computer algebra system (I like PARI/GP), we find $P^3 = \frac{8}{343} x^{18} + \frac{12}{343} x^{17} + \frac{18}{343} x^{16} + \frac{25}{343} x^{15} + \frac{33}{343} x^{14} + \frac{6}{49} x^{13} + \frac{40}{343} x^{12} + \frac{39}{343} x^{11} + \frac{36}{343} x^{10} + \frac{31}{343} x^9 + \frac{24}{343} x^8 + \frac{15}{343} x^7 + \frac{10}{343} x^6 + \frac{6}{343} x^5 + \frac{3}{343} x^4 + \frac{1}{343} x^3$.
From this, we can conclude, for instance, that the probability of a sum of $10$ when rolling $3$ times (or rolling once with three identical copies of this die) is $\frac{36}{343}.$
(Using Bruce's example, we get $P^3=\frac{1}{64} x^{18} + \frac{3}{64} x^{17} + \frac{3}{32} x^{16} + \frac{1}{8} x^{15} + \frac{9}{64} x^{14} + \frac{9}{64} x^{13} + \frac{25}{192} x^{12} + \frac{7}{64} x^{11} + \frac{5}{64} x^{10} + \frac{91}{1728} x^9 + \frac{19}{576} x^8 + \frac{11}{576} x^7 + \frac{1}{108} x^6 + \frac{1}{288} x^5 + \frac{1}{576} x^4 + \frac{1}{1728} x^3$, and so the probability of $10$ is $\frac{5}{64}=0.078125$ (exactly).)