1 - ItÃ‚Â´s a basic concept in probability , the sum of the individuals probabilities must be equal 1.So, the sum of the probabilities of all positions in the genome must be 1.If the probability is equal in all positions and you have k positions then

k * p = 1.

Sigh.... The above statement is false. Deadlock, I really don't mind helping you understand some issues in probability theory. However, when you make these errors and then attack those helping you by saying things like the quote below, you really create a lot of unnecessary hostility.

I suspect statements like the one below even go against forum rules.

ItÃ‚Â´s amazing your lack of knowledge about probability.But you dont need to believe me.Please, go to a university near your home and ask a teacher of statistics and after that tell me what is his answer.

And when you make such statements and then make basic errors, you just look foolish. Could we tone down this rhetoric just a bit? Everyone makes mistakes. Certainly, I make mistakes (just ask my wife). However, if I make a basic error and it is pointed out, at least I hope I will have the courage to admit it.

1 - ItÃ‚Â´s a basic concept in probability , the sum of the individuals probabilities must be equal 1.So, the sum of the probabilities of all positions in the genome must be 1.If the probability is equal in all positions and you have k positions then

k * p = 1.

So where is the error in the above statement? The sum of probabilities = 1 only when you consider all possible states that the system can be in (like you did in the coin flip example). It does not work when you apply it to the probability at each position. Consider an example where the probability of a mutation is 0.1 and you have three positions on the genome.

does k * p = 1?

does 3 * (0.1) = 1?

of course not

With a genome size of 3 billion and 4 kinds of base pairs, the number of possible states is

4^(3,000,000,000) which is a very very large number (my calculator says infinity but it should be something like 10^2,000,000,000). So if you could calculate the probability of each one of those states (whatever that would mean), it should sum to 1.

in the mutation case p = 10^-8, so you say that the probability of mutation does not happen in 3 positions would be :

p = ( 1 - 10^-8 )^3

p = ( ( 10^8 - 1 )/10^8) ^3

p = ( ( 100000000 - 1 )/10^8 )^3

p = ( ( 999999999/100000000)^3

p = 1.

In other words you are saying that is practically certain that 3 mutations will not happen.

Even me, a creationist, isnÃ‚Â´t so pessimistic about evolution

Ok, now apply that equation to the size of the genome

p(zero mutations) = (1-10^-8)^(3,000,000,000)

=(.99999999)^(3,000,000,000)

=9.3*10^-14 (if my little calculator got all those digits right)

So there is an extremely small chance of having no mutations

p(of at least 1 mutation) =1 - p(zero mutations)

= 0.99999999999999 (or something like that)

So the probability that your child will have at least one mutation is very very high.

On average a person has over 100 new mutations each generation. Luckily, most all of those are neutral. Some small fraction are harmful and if you have one of those you will have fewer offspring and the gene will vanish from the gene pool. Some smaller fraction are beneficial and if you have one of those, you will have more offspring (relative to your peers) and the gene can multiply through the gene pool.

Beneficial simply means that those with that new gene have a slightly higher chance of producing fertile offspring. What we can not calculate is how many different possible mutations would be beneficial.

After a few generations there will be more of the beneficial genes relative to the harmful ones. Since we do not know how many mutations might be beneficial (without trying out every possible combination of mutations with every possible combination of genome in the population), we can not calculate the probability that any particular mutation would be beneficial.

I guess a very very rough analogy would be the game of poker where you already hold a hand of cards with a few redundant cards (7 card stud). If you changed one of the cards at random, what would be the chance that it would give you a winning hand? To answer that, we would need to know the rest of your cards and know the cards of the other players. One does not want to calculate the probability of a particular set of cards. Rather, one wants to calculate the probability that the card you drew gave you a better hand - remembering that there are several different ways to build a winning hand.

James