I suggested that Larry take the Behe challenge.
UPDATE: And Larry took the challenge!
I asked Professor Moran in the combox:
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?"