Testing the Central Limit Theorem with the Shapiro-Wilk test on dice rolling simulations Rolling a single dice repeatedly will result in a uniform distribution. But if we roll multiple dice the sum would be the Normal distribution due the Central Limit Theorem (CLT).
To verify this, simulated multiple dice rolls number of times to get a distribution and then used the Shapiro-Wilk test to calculate p-value.
What seems strange is that if we increase the number of rolls, the histogram becomes smoother and apparently closer to normal by p- value tends to decline until it stays below 0.05 for any number of repetitions.
Both Python and R gives the same results. Below are two pieces of code in Python. One plots p value vs. the number of rolls. The other repeats the experiment 1000 times and calculates the percentage of results supporting the null hypothesis with p > 0.05.
E.g., with 20 dice, the percentage for 100 rolls is about 94; 700 rolls - 75%, 1500 rolls - 11%, 2000 rolls - 0%.
How can we explain this phenomenon?
import numpy as np
import matplotlib.pyplot as plt
from scipy import stats

def shapiro_p(n_dice, n_throws):

    rng = np.random.default_rng()
    throwing = rng.integers(1, high=6, size=(n_throws, n_dice), endpoint=True)
    X = np.sum(throwing, axis=1) / n_dice
    shapiro_test = stats.shapiro(X)

    return shapiro_test.pvalue

n_dice = int(input('Input the number of dice '))
n_max = int(input('Input the number of rolls '))

x = [n for n in range(3, n_max + 1)]
y = [shapiro_p(n_dice, n) for n in range(3, n_max + 1)]

figure_title = 'Number of dice: {0}'.format(n_dice)

plt.plot(x, y)
plt.hlines(0.05, 3, n_max + 1, colors='red')
plt.xlabel('Number of rolls')
plt.ylabel('p')
plt.title(figure_title)
plt.grid(True)
plt.show()

The second piece of code:
from scipy import stats
import random

dice = [i for i in range(1, 7)]

def shapiro_p(n_dice, n_rolls):
    sampl = [sum(random.choices(dice, k=n_dice)) for j in range(n_rolls)]
    shapiro_test = stats.shapiro(sampl)
    return shapiro_test.pvalue

n_dice = int(input('Number of dice '))
n_rolls = int(input('Number of rolls in each test '))
# n_tests =  int(input('Number of tests '))
n_tests = 1000

null_h = 0
for n in range(n_tests):
    if shapiro_p(n_dice, n_rolls) >= 0.05:
        null_h += 1
print('Percentage of tests supporting the null hypothesis: ', 100*null_h/n_tests)


The CLT is an asymptotic result valid for $n\rightarrow \infty$. Namely, it says for iid random variables $X_i$ with mean $\mu$ and variance $\sigma^2<\infty$ with the following definition:

$$ S_n = \frac{1}{\sigma\sqrt{n}}\sum_{i=1}^{n}(X_i-\mu) =
> \frac{\sqrt{n}(\bar{X}_n-\mu)}{\sigma} $$ Then $S_{n} \xrightarrow{d}
 Z\sim \operatorname{N}(0, 1)$ as $n\rightarrow \infty$.

The CLT says nothing about finite $n$ but the Berry-Esseen theorem gives bounds on the error of the approximation.
The distribution of the (appropriately scaled) sum is never exactly normally distributed at finite $n$, except when the $X_i$ have a normal distribution to begin with. For example: The support of the sum of $m$ d6 dice throws is $[m, 6m]$ whereas the normal distribution has support $(-\infty, \infty)$. So we'll expect the $p$-value of the Shapiro-Wilk test to increase with increasing number of dice, because the sum (or scaled sum) will approximate the normal distribution better.
The second effect that you observe is that the power of the Shapiro-Wilk test to reject deviations from a normal distribution increases with increasing number of rolls. Hence, with a fixed $m$ (number of dice), the $p$-value of the Shapiro-Wilk test will get smaller with increasing number of rolls. That's what you see in the graph resulting from the first piece of your Python code.
