# Normalisation (probability)

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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

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

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

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

where ${\displaystyle c_{\mathrm {i} }}$ is a coefficient of probability for outcome i. Assuming the dice are symmetrical we assume all values of ${\displaystyle c_{\mathrm {i} }}$ 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

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

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

## 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/ns to the integer n: the normalizing factor is then the value of the Riemann zeta function

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