I thought I could try this really easily in Python.

Here's what I did:

Open IDLE and type:

   from random import seed
   from random import random
   seed(1)

You can then try typing random() and get values:

   >>>random()
   0.13436424411240122

   >>>random()
   0.8474337369372327

   >>>random()
   0.763774618976614

I actually forgot that random() doesn't just return an integer, like in c where if you type rand() you get a random integer. But these numbers looked great to me. I could easily imagine the first one 0.13436424411240122 just being 13436424411240122 and so on. I thought it's a great way to get the first digit of a random numbers with an unlimited size, and for this result I thought this is a good way to go ahead.

So with the above basis, I want to look at the distribution of the first nonzero digit after the decimal point.

So next put the following function in. (basically it just gets a random as above, but keeps multiplying by 10 as long as the first digit of the string conversion is a "0". this returns the first non-zero digit at the left.)

    def getone():
       floaty = random()
       while(str(floaty)[0] == '0'):
          floaty *= 10
       return int(str(floaty)[0])

This now gets us the leftmost digit, check it out:

   >>>getone()
   3
   >>>getone()
   2
   >>>getone()
   4

   >>>getone()
   2

All right. Now let's just do this a few million times and keep track of buckets. Make some buckets:

   buckets = [0,0,0,0,0,0,0,0,0,0] # buckets for: 0,1,2,3,4,5,6,7,8,9

This next line will run for up to a few minutes:

   for tries in range(10000000):
      buckets[getone()] += 1

Now print the results:

   for result in range(10):
      print (result, ":", buckets[result])

This gets me....

   0 : 0
   1 : 1112927
   2 : 1110821
   3 : 1109741
   4 : 1110793
   5 : 1111604
   6 : 1112178
   7 : 1111272
   8 : 1110402
   9 : 1110262

Clearly Benford's law does NOT apply here.

Why not?