Ken Kriesel <[EMAIL PROTECTED]> writes:
> George asks, what about the bit second from the left?
> In total, 26 0 bits vs only 12 1 bits. (0 26 of 38 times or ~68.4%;
> 26/12=2.16666...)
> Excluding p=2 and p= 3, to look only at interior bits, 25 0 bits vs
> only 11 1 bits; 0 25 of 36 times, ~69.4%; 25/11=2.2727...)
> Now why would that be, that a 0 bit is more than twice as likely as a 1 bit?
> (Note that Mp#38, M6972593, has the less likely 1 bit in the second most
> significant position; searching biased for these probabilities would have
> delayed finding it. But the 4 previous Mersenne primes all had a 0 bit
> in that position.)
The law of leading digits predicts that, for decimal numbers,
log10(2) ~= 30.1% will have leading digit 1. More precisely,
the fractional parts of the base-10 logarithms are assumed
to be uniformly distributed. This distribution is invariant
under many transformations, such as converting a physical constant
from miles/hour to meters/second.
For binary numbers, this predicts the leading bit is 1 with probability
log2(2) = 1, a not surprising result. The two leading bits are
10 with probability log2(3) - log2(2) = ln(3/2)/ln(2) ~= 58.5%.
This prediction is smaller than the observed 68.4%, but above 50%.
Peter Montgomery
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