> I'm not sure what the law of leading digits is, but I read this as talking
> only about base 10 numbers... so the leading digit 1 is compared to
> leading digit 2, 3, 4, ..., 9. Right? So for numbers 2^n (in Base 10),
> [or is it 2^p?] there are a lot more leading ones than one would "expect"
> naievely (you would expect 1/9 to start with "1", I imagine).
As Jud said, this is called Benford's law. It was originally discoved in
the late 1800's by "Some Guy," who was never noticed. It was then
rediscovered by Benford in the 1930's. I think that he worked for AT&T.
Both guys discovered this law by looking at wear-and-tear on Logarithm books.
This law states that for a given base a, and a given leading digit b (0<b<a-1),
the probability that the leading digit is b is log_a(1+1/b).
What Peter was doing when he was talking about the second digit was (aparently)
converting it to base 4, and working witht the usual Benfords law.
>> The law of leading digits predicts that, for decimal numbers,
>>log10(2) ~= 30.1% will have leading digit 1.
>Uh, won't they *ALL* have a leading digit of one? Anything with a leading
>digit of ZERO can be totally discounted....
Read the rest of his email, and you will be enlightened...
>> The law of leading digits predicts that, for decimal numbers,
>>log10(2) ~= 30.1% will have leading digit 1.
>It depends on what set of numbers you are considering.
That's the point of Benford's law, it is supposed to be relatively independent
of the set of numbers.
Note that in the set of mersenne prime exponents (so far), the leading
digit 1 (in decimal), turns up 10 times as opposed to the 4.2 times
expected by equal leading digit distribution...
-Lucas Wiman
P.S. I seem to recall this being mentioned (benford's law), shortly after
I joine the list. May have to go into the FAQ.
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