# 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.

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Is $D2$ a two sided "dice", i.e a coin and $D6$ a regular dice? – milcak Feb 3 '11 at 5:32
Related, but not exactly dupe: math.stackexchange.com/questions/4632/… – Aryabhata Feb 3 '11 at 5:43
@milcak yes, that is standard notation for dice of different sizes. Sorry if that wasn't obvious. – dimo414 Feb 3 '11 at 5:51

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.

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Could you please elaborate on how you get from the probabilities of the different dice to the expression on the right? I cannot follow that reasoning. – dimo414 Feb 3 '11 at 6:07
@dimo: It is adding up terms in geometric progression. – Aryabhata Feb 3 '11 at 6:09
Sorry, I'm afraid this is all a bit over head. I'll have to look into this all further. static.funnyjunk.com/pictures/twelve_plane_of_torment0.jpg – dimo414 Feb 3 '11 at 7:30
@Moron: what will it do to expand out $\frac{1}{(x-1)^2}$ in this case? For $x\gt8$ in this case, the probability is 0 anyway... – Qiang Li Feb 3 '11 at 18:26
@QIang: You mean $k \gt 8$? YOu don't have to expand out the whole thing, just upto $x^k$. – Aryabhata Feb 3 '11 at 18:44

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

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Sorry, maybe I wasn't clear. I'm trying to figure out a general function that works on different sided dice in the same way F(s,n) in the Wikipedia article works. – dimo414 Feb 3 '11 at 5:40
@Moron: Moron has it right for that. See his answer. – Ross Millikan Feb 3 '11 at 5:46

Suppose we have dice $Da$ and $Db$, with $a \le b$. Then there are three cases:

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