Judging unfairness. A common way to judge whether faces on a die are equally likely
is to use a chi-squared goodness-of-fit test.
Suppose you roll a fair die 600 times and count the numbers
of 1s, 2s, ..., 6s observed. We can simulate this by using R:
set.seed(116) # for reproducibility
d = sample(1:6, 600, rep=T)
x = tabulate(d); x
[1] 103 116 88 98 91 104
As you say, the expected counts are $E_i = 100, i = 1,\dots,6.$
The chi-squared statistic is the sum
$$Q = \sum_{i=1}^6 \frac{(X_i-E_i)^2}{E_i}
= \frac{(103 - 100)^2}{100} + \cdots + \frac{(104-100)^2}{100} = 5.1.$$
sum((x - 100)^2/100)
[1] 5.1
If the die is 'behaving' fairly, then the $X_i$s will tend to be relatively near to the expected value $100$ and the terms of $Q$ will be relatively small. If the die is biased, then $Q$ will tend to be large. The question
is how large can $Q$ be before we begin to suspect the die is unfair.
Under the null hypothesis that the die is fair, $Q \stackrel{aprx}{\sim}
\mathsf{Chisq}(\nu = 6-1=5),$ the chi-squared distribution with five degrees of freedom. If we cut 5% of the probability from the upper tail of
$\mathsf{Chisq}(5),$ we get the 5% critical value $c = 11.07.$
So if
$Q \ge c = 11.07$ we say that is evidence at the 5% level of significance
that the die is not fair. We conclude that our data above give no evidence of unfairness.
qchisq(.95, 5)
[1] 11.0705
Thus, traditionally, we do not look at the behavior of any one of the six
counts to decide whether a die is fair, but we look at the overall 'profile' of all six counts as measured by the chi-squared GOF statistic $Q.$
Required sample size. Another part of your question has to do with the number of rolls of the
die it takes in order to detect unfairness. Suppose that the die
is unfair in such a way that the respective faces have probabilities
$(2/18, 3/18, 3/18,$ $3/18, 3/18, 4/18).$ [Maybe someone put a small lead
weight into the die on side 1, so the 1 occurs too seldom and 6 too often.] Are $n = 600$ rolls of the die enough to have a good chance of
detecting this degree of unfairness?
Let's look at two sessions of 600 rolls of such an unfair die as simulated in R:
set.seed(1234)
d = sample(1:6, 600, rep=T, p=c(2,3,3,3,3,4)/18)
x = tabulate(d); x
[1] 72 105 103 93 106 121
sum((x-100)^2/100)
[1] 13.44
d = sample(1:6, 600, rep=T, p=c(2,3,3,3,3,4)/18)
x = tabulate(d); x
[1] 74 79 107 96 101 143
sum((x-100)^2/100)
[1] 30.32
It happens that both sessions gave values of $Q$ that are large enough to detect
unfairness.
In R, the procedure chisq.test automates the testing, giving a P-value
smaller than 0.05 if the the null hypothesis should be rejected. For the
first example above, we have x = c(72, 105, 103, 93, 106, 121). [Unless the contrary is stated, chisq.test assumes the null hypothesis is that
all faces are equally likely.]
chisq.test(x)
Chi-squared test for given probabilities
data: x
X-squared = 13.44, df = 5, p-value = 0.01959
Thus, it is possible to do a simulation for $m = 100,000$ 600-roll experiments with a specific kind of unfair die in order to see how often
unfairness is detected. The answer is: almost always. So $n =600$ is
plenty of rolls to detect this level of unfairness.
set.seed(2020)
unf = c(2,3,3,3,3,5)/18
pv = replicate(10^5, chisq.test(tabulate(
sample(1:6,600,rep=T,p=unf)))$p.val)
mean(pv <= .05)
[1] 0.99986
However, if I try to get by with only $n = 150$ rolls of this biased die
I will detect biasedness only about 3/4 of the time.
set.seed(2020)
unf = c(2,3,3,3,3,5)/18
pv = replicate(10^5, chisq.test(tabulate(
sample(1:6,150,rep=T,p=unf)))$p.val)
mean(pv <= .05)
[1] 0.74368
And with only $n = 60$ rolls, we would detect unfairness only about $1/3$
of the time [simulation not shown].
We say that the power of the test is about $.99, .75,$ and $.33$ for $n = 600, 150,$ and $60,$ respectively. There are formulas for determining the power
of the chi-squared GOF test for various scenarios of unfairness. [If you
are interested in the technical details you can look at this Q&A or try googling power chi-squared test.]
