I suggested that Larry take the Behe challenge.
UPDATE: And Larry took the challenge!
I asked Professor Moran in the combox:
"Hi Larry,
Thanks for taking "the Behe challenge." I'll be curious to see what Behe's response is. Meanwhile, you wrote:
The probability that the other required mutation will occur is roughly 125/25 Mb = 5 × 10‐6, or about five in a million.
I'm curious. Does that mean that if a third mutation was required, then the probability would be 5x10-6 x 5x10-6? And if a fourth mutation was required, would the probability be 5x10-6x5x10-6x5x10-6?"
If the correct answer is "yes," then the probability of getting all four required mutations would be 125 x 10-18, or 1.25 x 10-16, which is not too far off from what Behe estimated the probability was for malaria to develop chloroquine resistance, and also the probability he estimated for evolving one new protein binding site.