I don’t think the chance of fixation formula given (2s, where (1+s) is the fitness) is correct. It seems to fail some basic sanity tests.
First of all, it’s possible to imagine a mutation that makes you have 10x the number of children of a regular person. This gives us s=9. Is the probability of fixation then… 18?
Second, it seems like this would have to depend on other factors, such as the base probability an individual has children in the first place. In a dangerous environment where (as a base rate) 99% of all individuals die before they can reproduce at all, the individual has 0.01(1+s) chance of surviving, which is an upper bound for the probability of fixation. In an environment where everyone is guaranteed to have at least one child (but some have more), the mutation should (if dominant) stick around no matter what.
First of all, it’s possible to imagine a mutation that makes you have 10x the number of children of a regular person. This gives us s=9. Is the probability of fixation then… 18?
Good spot. The P(fixation) = 2s formula is actually an approximation that holds only when s is small but positive and the population size is very large. A more precise formula (here’s a publicly-accessible paper deriving it) is P(fixation) = (1 - exp(-2s)) / (1 - exp(-4N**s)), where N is the population size. When s is small one can approximate the numerator by 2s and when s is positive and N is infinite the denominator becomes 1 — so when all of these conditions hold the formula reduces to 2s. In your example s is large and the P(fixation) = 2s approximation is no longer a good one.
I don’t think the chance of fixation formula given (2s, where (1+s) is the fitness) is correct. It seems to fail some basic sanity tests.
First of all, it’s possible to imagine a mutation that makes you have 10x the number of children of a regular person. This gives us s=9. Is the probability of fixation then… 18?
Second, it seems like this would have to depend on other factors, such as the base probability an individual has children in the first place. In a dangerous environment where (as a base rate) 99% of all individuals die before they can reproduce at all, the individual has 0.01(1+s) chance of surviving, which is an upper bound for the probability of fixation. In an environment where everyone is guaranteed to have at least one child (but some have more), the mutation should (if dominant) stick around no matter what.
Good spot. The P(fixation) = 2s formula is actually an approximation that holds only when s is small but positive and the population size is very large. A more precise formula (here’s a publicly-accessible paper deriving it) is P(fixation) = (1 - exp(-2s)) / (1 - exp(-4N**s)), where N is the population size. When s is small one can approximate the numerator by 2s and when s is positive and N is infinite the denominator becomes 1 — so when all of these conditions hold the formula reduces to 2s. In your example s is large and the P(fixation) = 2s approximation is no longer a good one.
Really? Can you give an example? Because I don’t see how that could be done.
Edit: Oh, right. Nevermind.
I think he means that in the sense of a theoretical possibility that shows that the math must be flawed. Not something likely to actually crop up.