Simulation and modeling - Old Questions

The testing for uniformity can be achieved through different frequency test. These tests use the Kolmogorov-Smirnov or the chi- square test to compare the distribution of the set of numbers generated to a uniform distribution.

The Poker Test is the test for independence based on the frequency with which certain digits are repeated with in a series of numbers. This test not only tests for the randomness of the sequence of numbers, but also the digits comprising of each of the numbers. The expected value of each of the combination of digits in a number is compared with the observed value by means of the chi-square test for independence. The acceptance is done if the observed value of chi-square sums for all the possible combinations of digits is less than the acceptable value for the given degree of freedom at the specified confidence interval.

Example:

Poker test for three digit numbers

In each case, a pair of like digits appears in the number that was generated. In three-digit numbers there are only three possibilities, as follows:

    1. The individual numbers can all be different.

    2. The individual numbers can all be the same.

    3. There can be one pair of like digits.

The probability associated with each of these possibilities is given by the following

    P (three different digits) = (0.9)*(0.8) = 0.72

    P (three like digits) = (0.1)*(0.1) = 0.01

    P (exactly one pair) = 1 - 0.72 - 0.01 = 0.27

E.g.

Suppose a sequence of 1000 three-digit numbers has been generated and an analysis indicates that 680 have three different digits, 289 contain exactly one pair of like digits, and 31 three like digits. Based on the poker test, we are going to check are these number independent? Let α = 0.05..

Since, 47.65 > X²_0.05,2 = 5.99 (tabulated value), the independence of the numbers is rejected on the basis of this test.