# Probabilities for n Dice

Suppose we roll n regular balanced six-sided dice.  If X is the sum of the
n values which appear, what are the probabilities associated for each value
of Xfor the possible values X n,  ... , 6n?

In the case n =1, these probabilities are all 1/6.  For two dice , it is easiset to
consider  a table of possible outcomes:

 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Next,  we can consider the sums associated with these outcomes:

 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 7 8 9 10 11 12

Since there are 36 outcomes, each equally likely, we see that the probabilities are:

 X count P(X) 2 1 1/36 3 2 2/36 4 3 3/36 5 4 4/36 6 5 5/36 7 6 6/36 8 5 5/36 9 4 4/36 10 3 3/36 11 2 2/36 12 1 1/36

If there are more than two dice, it is difficult to make tables such as these.  The number
of outcomes will be 6n, so to calculate the probabilities it is sufficient to count the
number of times each  sum occurs among the 6n possible outcomes.  This is easily done
by considering the generating function for the number of times each sum appears:

f(x) = (x + x2 + x3 + x4 + x 5+x6)n
For example

(x + x2 + x3 + x4 + x 5+x6)2= x 2+2x3+3x4+4x5+5x6+6x7+5x8+4x9+3x10+2x11+x12

The coefficient of xn  is the number of times the count occurs.