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 probabilites 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 Probabilty Distributions

Discrete probability distributions are used througout 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 (Dice1,Dice2)	occurances (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 occuring 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.