Question

I want to test to see if a die is fair. Why shouldn’t I just do one test to see if there are too many ones, one test to see if there are too many twos, one test to see if there are too many threes, etc.?

Answer 1

A die has 6 faces. The goal is to check if the die is fair. For this, the die needs to be rolled a number of times. A good rule of hand is to roll at least 5 times the total number of faces. Hence, for a die with 6 faces, at least 6 * 5 = 30 times the die should be rolled.

Now the number of times each of the faces comes up can be counted. Let n1, n2, n3, n4, n5, or n6 be the number of times the rolls bring 1, 2, 3, 4, 5, or 6 up, respectively. Let n be the total number of rolls done. We have:

n = n1 + n2 + n3 + n4 + n5 + n6

If the die were fair, the expected number of times each face should come up will be n/6.

To check for the validity of the null, we have to perform a chi-square test:

The degrees of freedom will be df = n - 1 = 6 - 1 = 5. For a given significance level and degrees of freedom, we get a critical value. If the chi-square statistic as computed by the formula above is greater than the critical value, we can assume the die to be unfair. Else, the die is fair.

This method is ideal.

If however, each number is checked for individually, it brings the issue of 0 degrees of freedom. Also, there will be multiple hypotheses being tested therefore, resulting in a requirement of correction for multiple hypothesis which can result in a lower significance level.