These are all basic probability questions. You can get this in the link that Kega provided in a couple posts up if you are interested.

If the location is specific, please tell me what is it ?

Not quite sure of your question. For the calculation, I made the assumption that the distribution was uniform. That is, that every point the probability of a mutation was the same p=10^-8

1. Three people each have a deck of cards (52 cards) and each person draws one card at random from their deck. What is the chance that at least one person will draw an ace of spades? (k=3, p=1/52)

As previously mentioned, the equation is just

P(of at least one)=1-(1-p)^k

P(of at least one)=1-(1-1/52)^3

2 - What is the Chance that only the first person draws an ace os spades ?

That is easy, assuming they each have an independent deck like in the first question.

Probability of the first person getting an ace of spades is p1=1/52

Probability of the second person not getting an ace of spades is p2=(1-1/52)=51/52

Probability of the third person not getting an ace is spades p3 = (1-1/52)=51/52

So the probability of A and B and C is just

p(first person drawing an ace of spades and second two not drawing it)=

p1*p2*p3 = (1/52)*(51/52)*(51/52)

notice that p1+p2+p3 does not equal 1. Also, as I noted before with your coin flip example, if you looked at all possible permutations (there are 8 permutations in this 3 draw case) then the probabilities of all permutations must sum to 1.

However, there is no reason to presume that k(number of samples)*p=1

That would be like saying that if I watch 10 people over a given day, the probability that one will get in an accident must be 1/10. But if I have 20 people the probability must be 1/20th (assuming a uniform distribution). Of course, that is just a bit silly. The distribution can be uniform but they can take on any probability between 0 and 1. Each of the 10 people could have a 1/1000 chance of getting in an accident.

So the probability of a mutation at any single position (p), is completely unrelated to the number of positions in the genome (k). Of course, we are ignoring mutations other than single nucleotide mutations (e.g., gene duplications or deletions) and this is critical if you want to peform a full comparison of two different genomes (e.g., chimp versus human), one must be clear whether one is using just single nucleotide mutations or all forms of mutations. And this is the cause of a lot of confusion when someone in the press says we are 99% similar (and why this thread started).

So why do you have so much difficulty admitting you made a minor error? For me, winning means learning. If I make an error and learn from it, I win. Is there some problem with admitting to an "evolutionist" that you may have made any kind of error? Even a simple algebra question? Could we even come to an agreement that if we discover that we made an error, we will admit it? I will agree to that.

I don't understand having any kind of discussion with you, when you feel the goal is to never admit even a minor error no matter how the evidence stacks up. Just seems very odd to me.

Can you address this point please? If you simply made a simple error, admit it and my respect for you will increase. If I was wrong, then just tell me what the right answer is since you have been a bit shy regarding your answer.

Explain me, how ( 0.99999999 )^(3,000,000,000 ) could be equal = 9.3*10^-14 ?

correct answer is (99999999/10^8)^(3*10^9) = ? , put it in the Excel and see the answer it Ã‚Â´ll give you.

The thing is worse than I thought.The problem begins in algebra.

Thanks, James