Thanks for the helpful comments on my example. I am satisfied this is good enough for the purpose of discussion. Back to it …
Let’s look again at Demski’s equation:
SC(E ) = I(E) - K(E) ≥ I(E) - |D|
And plugging in the values from the example
SC(E) ≥ 499 - 352 = 147 bits
FIRST, we know that Shannon Information describes a “bandwidth” needed for communications. "352 bits of Kolmogorov Information is the compressed length of D. This gives us “147 bits” of some quantity that is undefined in mathematical theory. Dembski calling it “Specified Complexity” does not make it meaningful. There is no meaning or interpretation for the difference of Shannon and Kolmogorov information. None. This is hot nonsense.
I got the same response when I posted this to FB:
In other words, Dembski is comparing unitless apples to unitless oranges. Matthew has more good comments in that discussion.
Second, we can’t actually get Shannon Information in this example because (A) we don’t have a probability distribution for E, or (B) we have a sample size of N=1. In the case of the discrete uniform distribution Dembski knows what the SI will be because he assumes the length of the sequences. He never estimates the probability of E from any data, he just assumes a longer (shorter) sequence when he needs a smaller (larger) probability. (Footnote #1)
Not only is Dembski comparing apples to oranges, but he doesn’t actually have any apples.
That should probably be the end of the story. I have a few additional notes and comments, but I will post them separately.
Footnotes
#1: I did the same thing in my example by taking an arbitrary precision of 150 digits. My first attempt had 50 digits of precision, giving
SC(E) ≥ 166.1 - 352 = -185.9 bits
Technically information cannot be negative and we should say “0 bits” if we get a negative number. But since we don’t actually have Information, I guess anything goes? That didn’t fit the original example which had non-zero SC, so I arbitrarily added another 100 digits to E to get a positive value. Problem solved.