Ok what would the correct probability be?
Pretend that you need 3 mutations in order to have a selective benefit, (1 or 2 mutations would be useless) if behe is wrong with his numbers, what would the correct probability be?
So mutation 1 would be useless meaning that the mutation would be neutral, so it will only get fixed by chance (genetic drift) the same is true with mutation 2,....... But the combination of 1+2+3 will result in a benefit that would be selected by natural selection
So please provide the correct calculations and show that it is possible to have 3 mutations that together would produce a benefit, but by themselves would be neutral. Acoording to Behe such an event can't happen.
For the record, observations support Behes affirmations.
When an organism (say a bacteria) requires 1 mutation to gain a selective benefit such mutation is likely to occure
When it requires 2 mutations it is unlikely but possible but only under optimal conditions and in microbes that reproduce fast.
When it requires 3 mutations or more, well that has never been observed
As stated, the probability can't be determined. It depends on much more than simply the mutations required.
Well, what is the probability of each mutation? What is the size of the population? What is the overall rate of mutation? Are you wanting a specific protein, or just one that work (for which there may be several).
Once again, the 'multiply probabilities' is only correct if you want a single protein to spontaneously appear with no precursors and with all three mutations all at once. Anything that allows sequential steps drastically increases the probability. Anything that allows fixation in a population drastically increases the probability.
Let me give another example. Suppose I had 20 dice and asked you to throw all 6's. The probability for each throw would be one in 6^20, which is about 4 quadrillion, So you would expect to make about this many throws before you are successful.
Now, suppose that I asked you to throw 20 dice and keep all the 6's that appear. Then throw the remainder and keep all the 6's that appear. And then throw the remainder and keep all the 6's that appear. Now, how many throws, on average, will it take to get all 20 dice to show 6's? I think it quite easy to see that the number of throws is less than 6*20=120.
I don't feel like computing the exact probability even for this one, but I think that it is easily seen that the 'multiply probabilities' calculation is simply wrong and badly so.
In your scenario, if it takes an average of, say, 1000 generation for each mutation to appear on average and if, once it appears it stays fixed, then for three mutations, you would need an average of less than 3000 generations. In other words, instead of multiplying the probabilities, you simply divide the value by the number of required probabilities and this is an *underestimate* for the actual probability.
A complete probability calculation, like I said, would be incredibly difficult. It would need to take into consideration population size, mutation rate, deal with both mutations producing the needed changes and other mutations changing away from them, notice that the order the mutations occur may not be fixed, etc.
But you are simply wrong that three neutral mutations would be effectively impossible.