Three dice were rolled. The first two were rolled 25 times, and the third was rolled 50 times. For each die find a number that ranks it relative to the others in terms of how fair it is (or how much its distribution of scores differs from chance). Circle which die you think is Most Fair, Somewhat fair, and least fair. Make sure your numbers support your ranking. There are no ties (i.e., only one die is Most Fair, etc.).

Die 1: Rolling a 5-sided die. The # of times each result appears.
"1" 6 times
"2" 5 times
"3" 4 times
"4" 4 times
"5" 6 times

Die #2
"1" 3 times
"2" 5 times
"3" 10 times
"4" 4 times
"5" 3 times

Die #3
"1" 11 times
"2" 10 times
"3" 9 times
"4" 9 times
"5" 11 times

What is your formula????

  • 1
    $\begingroup$ Consider chi-squared 'goodness-of-fit` statistic. Difference in sample sizes is a potential difficulty, Suggest you look in basic statistics text, search this site for examples; maybe Wikipedia article is helpful. $\endgroup$ – BruceET Sep 9 '16 at 16:42
  • $\begingroup$ The title is a bit obscure. Would it make sense to title this question "rank three dice for fairness, using observed data"? $\endgroup$ – David K Sep 10 '16 at 2:04

Chi-squared Goodness-of-Fit test. Suppose there are $k$ categories, for which the null hypothesis specifies probabilities $p = (p_1, p_2, \dots, p_k),$ with $\sum_{i=1}^k p_i = 1.$ Also suppose you have $n$ observations with counts $X = (X_1, X_2, \dots, X_n),$ for the respective categories, with $\sum_{i=1}^k x_i = n.$ Then, under the null hypothesis, the expected count for the $i$th category is $E_i = np_i.$ and the goodness-of-fit statistic is $$Q = \sum_{i=1}^k \frac{(X_i - E_i)^2}{E_i}.$$ If all counts $X_i$ exactly match the corresponding $E_i,$ then $Q = 0,$ and larger values of $Q$ indicate departure from a perfect fit. Under the null hypothesis $Q$ is approximately distributed as $Chisq(\nu),$ with $\nu = n - 1$ degrees of freedom. The approximation is reasonably good provided that all $E_i \ge 5$ and often useful if all $E_i \ge 3.$ If $Q$ exceeds the 95th percentile of $Chisq(\nu),$ called the critical value of the test, then the null hypothesis is rejected at the 5% level of significance.

Application to first die. In your problem the experiment with the first die has $k = 6,\,$ $\nu = 5,\,$ and $n=25.$ Also under the null hypothesis that the die is fair, $p_i \equiv 1/6$ and $E_i \equiv 25/6 = 4.25.$ (Notice that the $E_i$ need not be integers. Also, notice that in this problem we are working near the boundary of accuracy of conformity to a chi-squared distribution.)

For $n = 25$ observations $X = (6, 5, 4, 4, 5),$ we have $Q = 2.13.$ The P-value of the test is the probability of getting $Q \ge 2.13,$ assuming that the null hypothesis is true.

Related computations in R statistical software are shown below. The critical value can be found in printed tables of the chi-squared distribution. (Generally, software is required to find the P-value.)

X = c(6, 5, 4, 4, 5); E = 25/4
Q = sum((X-E)^2/E);  Q
## 2.13
crit = qchisq(.95, 5); crit
## 11.0705
p.val = 1 - pchisq(Q, 5); p.val
## 0.8308804

Judging fairness. You could do similar computations for the other two dice. The die with the smallest $Q$ (or the largest P-value) can be taken as the 'fairest' die. I have used R for a compact display, but hand computation of $Q$ is easy enough on a calculator.

However, you should be aware that you cannot get a really good idea of fairness with so few rolls because of the randomness of the experiment. With so few rolls it is possible to get a small $Q$ (and no rejection of the null hypothesis) even if a die is badly out of balance. You have a little better information for the third die (tossed 50 times).

Figure for first die. The figure below shows the PDF of $Chisq(5);$ the vertical purple line is at $Q = 2.13$ and the vertical red line is at the critical value 11.07. The area under the curve to the right of the purple line is the P-value 0.8309, and the area under the curve to the right of the red line is 0.05.

enter image description here


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