3 ms·

You chose N in such a way to make your hypothesis true.

susam 7y ago
The variable 'n' has to be sufficiently large. The larger 'n' is, the more it agrees with Benford's law. For example, with n = 10 (as opposed to n = 10000 in my original code), we get results that disagree with Benford's law quite a bit: 0 => 0.000 1 => 0.301 2 => 0.200 3 => 0.100 4 => 0.100 5 => 0.100 6 => 0.100 7 => 0.000 8 => 0.100 9 => 0.000 In fact, the results for digits 7 and 9 are in complete disagreement with Benford's law. That's not surprising because there is no integer n between 0 and 10 such that 2^n has 7 or 9 as its first digit. Therefore, no matter how many trials we perform, the frequency of the first digit as 7 or 9 would always be 0. The first power of 2 for which we get 7 as the leading digit is 2^46. Similarly, the first power of 2 for which we get 9 as the leading digit is 2^53. So unless the powers are much larger than these numbers, we won't get frequencies for 7 and 9 that agree with Benford's law. With n = 100, the experimental results become closer to Benford's law: 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 But with n = 1000, the results become even closer to Benford's law: 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
- susam 7y ago
  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.
  - dmurray 7y ago
    You keep choosing N to be an integer power of 10. Make it a power of 2 or 7 or pretty much any number, and you'll see different results.
    - susam 7y ago
      I do not see different results: import random N = 8192 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)) Output: 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 I still see results that agree with Benford's law.
      - logicallee 7y ago
        I did something similar (except you're a better coder, haha), but when I looked up how to get a random number in Python the first thing I saw was how to get one between 0 and 1 so I went with it. Here is my comment (very similar in spirit to yours): https://news.ycombinator.com/item?id=22341416 https://news.ycombinator.com/item?id=22341416 Any comments?
        susam 7y ago
        Replied on your comment thread.

The variable 'n' has to be sufficiently large. The larger 'n' is, the more it agrees with Benford's law. For example, with n = 10 (as opposed to n = 10000 in my original code), we get results that disagree with Benford's law quite a bit: 0 => 0.000 1 => 0.301 2 => 0.200 3 => 0.100 4 => 0.100 5 => 0.100 6 => 0.100 7 => 0.000 8 => 0.100 9 => 0.000 In fact, the results for digits 7 and 9 are in complete disagreement with Benford's law. That's not surprising because there is no integer n between 0 and 10 such that 2^n has 7 or 9 as its first digit. Therefore, no matter how many trials we perform, the frequency of the first digit as 7 or 9 would always be 0. The first power of 2 for which we get 7 as the leading digit is 2^46. Similarly, the first power of 2 for which we get 9 as the leading digit is 2^53. So unless the powers are much larger than these numbers, we won't get frequencies for 7 and 9 that agree with Benford's law. With n = 100, the experimental results become closer to Benford's law: 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 But with n = 1000, the results become even closer to Benford's law: 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

susam 7y ago
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.
- dmurray 7y ago
  You keep choosing N to be an integer power of 10. Make it a power of 2 or 7 or pretty much any number, and you'll see different results.
  - susam 7y ago
    I do not see different results: import random N = 8192 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)) Output: 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 I still see results that agree with Benford's law.
    - logicallee 7y ago
      I did something similar (except you're a better coder, haha), but when I looked up how to get a random number in Python the first thing I saw was how to get one between 0 and 1 so I went with it. Here is my comment (very similar in spirit to yours): https://news.ycombinator.com/item?id=22341416 https://news.ycombinator.com/item?id=22341416 Any comments?
      - susam 7y ago
        Replied on your comment thread.