Dice probability of sum on different sided dice

Question

I need some help here: How do I calculate the probabilities of getting, say, the sum of 38 when rolling 2d4 3d3 8d12 (for those who does not know, that means: 2 dice of 4 sides, 3 of 3 sides and 8 of 12 sides). Can somebody help me with that?

Answer 1

It's the coefficient of $x^{38}$ in

$$p(x) = (x+x^2+x^3+x^4)^2 (x+x^2+x^3)^3(x+x^2+ x^3 + x^4+ \cdots +x^{12})^8$$

Can you see why? Every throw with that sum corresponds to a choice of a term in these polynomials and all such combinations give rise to that coefficient. Expand $(x+x^2+x^3+ x^4+x^5+x^6)^2$ for a pair of D6, to see this happening.

Wolfram Alpha can expand these sums and form there I got the answer 314152685 (link to computation).

But by hand: get out the $x$ from each term, and use $$(1+x+\cdots +x^n) = \frac{1-x^{n+1}}{1-x} = (1-x^{n+1})(1-x)^{-1}$$

and $$(1-x)^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} 1^{n-k}(-x)^{k} =\sum_{k=0}^\infty (-1)^k \binom{n+k+1}{k}(-x)^k = \sum_{k=0}^\infty\binom{n+k-1}{k} x^k$$

by the negative binomial series

Using this:

$$p(x) = x^{13}(1-x^5)^2\left(\sum_{k\ge 0} \binom{k+1}{k}x^k\right)(1-x^4)^3 \left(\sum_{k \ge 0} \binom{k+2}{k}x^k\right) (1-x^{13})^{8}\left( \sum_{k \ge 0} \binom{k+7}{k} x^k\right)$$

and expand the positive powers with the binomial formula, leaving relatively few terms to consider for $x^{38}$, try it.