a loaded die? Practice 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????
 A: 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.

