Normalisation (probability)

In mathematical probability equations, which are used in nearly all branches of science, a normalization constant (or function) is often used to ensure that the sum of all probabilities totals one, or

$\sum {P_{\mathrm {i} }}=1$

Probability distributions can be divided into two main groups: discrete probability distributions and continuous probability distributions.

Discrete Probability Distributions

Discrete probability distributions are used throughout gaming theory. Consider the simple example of rolling a pair of six-sided dice. Summing up the total roll of the dice yields the following possibilities:

Total (i)	Possible outcomes (Die1,Die2)	Occurrences (ni)
2	(1,1)	1
3	(1,2), (2,1)	2
4	(1,3), (3,1), (2,2)	3
5	(1,4), (4,1), (2,3), (3,2)	4
6	(1,5), (5,1), (2,4), (4,2), (3,3)	5
7	(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)	6
8	(2,6), (6,2), (5,3), (3,5), (4,4)	5
9	(3,6), (6,3), (4,5), (5,4)	4
10	(4,6), (6,4), (5,5)	3
11	(5,6), (6,5)	2
12	(6,6)	1

Since the probability of any particular outcome is proportional to the number of ways it can occur

$\sum {P_{\mathrm {i} }}=\sum {c_{\mathrm {i} }n_{\mathrm {i} }}=\sum {Nn_{\mathrm {i} }}=1$

where $c_{\mathrm {i} }$ is a coefficient of probability for outcome i. Assuming the dice are symmetrical we assume all values of $c_{\mathrm {i} }$ are equal and their sum equals 1.

Solving for N yields 1/36, the number of possible outcomes, so that the probability of total = i occurring are

$P_{\mathrm {i} }=\left({\frac {1}{36}}\right)n_{\mathrm {i} }$, and the sum of all probabilities is one

$\sum {P_{\mathrm {i} }}=\left({\frac {1}{36}}\right)\left(1+2+3+4+5+6+5+4+3+2+1\right)={\frac {36}{36}}=1$

Continuous probability distributions

In most scientific equations, probability functions are continuous functions, and the probability coefficients are sometimes functions rather than constants. For example, the zeta distribution with parameter s assigns probability proportional to 1/n^s to the integer n: the normalizing factor is then the value of the Riemann zeta function

$\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}.$