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 X, for 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.