Probability of Sum of Different Sized Dice I am working on a project that needs to be able to calculate the probability of rolling a given value $k$ given a set of dice, not necessarily all the same size.  So for instance, what is the distribution of rolls of a D2 and a D6?
An equivalent question, if this is any easier, is how can you take the mass function of one dice and combine it with the mass function of a second dice to calculate the mass function for the sum of their rolls?
Up to this point I have been using the combinatorics function at the bottom of the probability section of Wikipedia's article on dice, however I cannot see how to generalize this to different sized dice.
 A: You can use generating functions.
I presume D2 means dice with numbers 1 and 2.
In which case the probability generating function is
$$(x/2 + x^2/2)(x/6 + x^2/6 + x^3/6 + x^4/6 + x^5/6 + x^6/6) = \frac{x^2(x^2-1)(x^6 - 1)}{12(x-1)^2}$$
You need to find the coefficient of $x^k$ in this to get the probability that the sum is $k$.
You can use binomial theorem to expand out $\frac{1}{(x-1)^2}$ in the form $\sum_{n=0}^{\infty} a_n x^n$
You can generalized it to any number of dice with varying sides.
I will leave the formula to you.
A: For D2+D6, you can only get 2 or 8 in one case each, probability 1/12.  Other totals 3-7 you can attain two ways (for example 6=5+1 or 4+2), probability 2/12=1/6.  A good check is the total over all the values is 1. The  check is 2*(1/12)+(7-3+1)*(1/6)=1
A: Suppose we have dice $Da$ and $Db$, with $a \le b$. Then there are three cases:  


*

*If $2 \le n \le a$, the probability of throwing $n$ is $\frac{n-1}{ab}$.

*If $a+1 \le n \le b$, the probability of throwing $n$ is $\frac{1}{b}$.

*If $b+1 \le n \le a+b$, the probability of throwing $n$ is $\frac{a+b+1-n}{ab}$.

