Normalisation (probability): Difference between revisions
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In mathematical probability equations, which are used in nearly all branches of science, a '''normalization''' constant is often used to ensure that the sum of all probabilites totals one | 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 | ||
<math> | <math> \sum{P_\mathrm{i}} = 1 </math> | ||
Probability distributions can be divided into two main groups, discrete probability distributions and continuous probability distributions. | Probability distributions can be divided into two main groups, discrete probability distributions and continuous probability distributions. | ||
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</table> | </table> | ||
Since the probability of any particular outcome is proportional to the number of ways it can occur | |||
<math> \sum{P_\mathrm{i}} = \sum{c_\mathrm{i}n_\mathrm{i}} = \sum{Nn_\mathrm{i}} = 1 </math> | |||
where <math> c_\mathrm{i} </math> is a coefficient of probability for outcome i. Assuming the dice are symmetrical we assume all values of <math> c_\mathrm{i} </math> 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 | |||
<math> P_\mathrm{i} = \left(\frac{1}{36}\right)n_\mathrm{i} |
Revision as of 15:07, 9 October 2007
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
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
where is a coefficient of probability for outcome i. Assuming the dice are symmetrical we assume all values of 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
<math> P_\mathrm{i} = \left(\frac{1}{36}\right)n_\mathrm{i}