Instead of performing trials in an experiment, another way to look at this problem is to simply analyze the frequency distribution of the leading significant di

Instead of performing trials in an experiment, another way to look at this problem is to simply analyze the frequency distribution of the leading significant digit for a function that grows exponentially. Like before, let us choose 2^n as that function. Here is the code: import random N = 10 counts = [0] * 10 for n in range(N): first_digit = int(str(2**n)[0]) counts[first_digit] += 1 for i, count in enumerate(counts): print('{} => {:.3f}'.format(i, counts[i] / N)) For N = 10, i.e., 0 < n < 10, we get: 0 => 0.000 1 => 0.300 2 => 0.200 3 => 0.100 4 => 0.100 5 => 0.100 6 => 0.100 7 => 0.000 8 => 0.100 9 => 0.000 For N = 100, i.e., 0 < n < 100, we get: 0 => 0.000 1 => 0.300 2 => 0.170 3 => 0.130 4 => 0.100 5 => 0.070 6 => 0.070 7 => 0.060 8 => 0.050 9 => 0.050 For N = 1000, i.e., 0 < n < 1000, we get: 0 => 0.000 1 => 0.301 2 => 0.176 3 => 0.125 4 => 0.097 5 => 0.079 6 => 0.069 7 => 0.056 8 => 0.052 9 => 0.045 For N = 10000, i.e., 0 < n < 10000, we get: 0 => 0.000 1 => 0.301 2 => 0.176 3 => 0.125 4 => 0.097 5 => 0.079 6 => 0.067 7 => 0.058 8 => 0.051 9 => 0.046 These results are consistent with the experimental results in my previous comment. As we can see, the larger N is, the more the frequencies of the leading digit obey Benford's law. Similar results can be reproduced with many other exponentially increasing functions too, e.g., 3^n, 5^n, Fibonacci numbers, Lucas numbers, and even factorials that grow faster than exponential functions.