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| quote: | Originally posted by DrUg_Tit0
Well..Since the gene is dominant, the probability varies. If they're 65 and 65, it's 75%. If they're 66 and 65, it's 100%, and so is if they're 66 and 66. Additionally, if they're 66 and 55, it's a 100%, 65 and 55 have a 50% chance, while 55 and 55 have a 0%.
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To cut it short, what I need would be the following probabilities (P(56|65,66) means "if a 65 parent and 66 parent gets a kid, then the probability of that kid being a 56 is P(56|65,66)"):
P(66|66,66)
P(65|66,66)
P(56|66,66)
P(66|66,65)
P(65|66,65)
P(56|66,65)
P(66|66,56)
P(65|66,56)
P(56|66,56)
P(66|66,55)
P(65|66,55)
P(56|66,55)
P(66|65,65)
P(65|65,65)
P(56|65,65)
P(66|65,56)
P(65|65,56)
P(56|65,56)
P(66|65,55)
P(65|65,55)
P(56|65,55)
P(66|56,56)
P(65|56,56)
P(56|56,56)
P(66|56,55)
P(65|56,55)
P(56|56,55)
P(66|55,55)
P(65|55,55)
P(56|55,55)
The last three should sum to 0.002, according to the requirement that one out of 500 kids born to normal parents would have the gene anomaly.
| quote: | Originally posted by DrUg_Tit0
However this makes it even more complicated since you'd also have to keep track of the recessive genes. If there's just 1000 people starting, I suppose you can say that each of the 2 6-fingered people has just one gene. But on larger populations, I'd say that it's best to say that 1/3 of them have a 66 and 2/3 have a 65.
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I don't follow you here. Which gene is it that you see as recessive? Would it be somehow captured by the detailed list of probabilities listed above?
| quote: | Originally posted by DrUg_Tit0
Well, if it's just the average couple that manages to raise 2 kids, I'd say it just means that the population is neither growing nor declining. Some might have more and some less. My guess is that on average, 6-fingered people have 10% more.
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This I also something I don't understand. When you say that 6-fingered people have 10% more kids, do you then mean than each member of *the entire group* of 6-fingered people on the average have 10% more kids than people with five fingers? I could agree with that.
| quote: | Originally posted by DrUg_Tit0
Now there's another problem here I'd say. If there would be 50% of the 6 fingered ppl in the population, then to maintain population equilibrium they'd have to have about 2.0952 children per couple, while 5-fingered ones would have some 1.9046 children. Mixed couples would then I suppose have 2 children exactly. But the thing is, if there's, say 10% of 6 fingered people in a group, the numbers are different, because he's closer to 2.1 children, while the rest is closer to 2.0. Ultimately, when 6 fingered people prevail, they'll have 2.0 children while the last 5 fingered couples will have about 1.81. So that's kinda one more dynamic parametar here.
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Well, the entire thing seems highly unrealistic if everyone procreates. And it definitely clashes with the intuition of 6-digit individuals somehow having an advantage in getting to the mating stage. I would think of the reproduction mechanism in the following manner:
while (not enough new individuals to constitute a generation)
- sample two individuals from the old population, that has not already been sampled. Each individual with 6-digits on at least one hand has a 10% higher chance of being picked than someone with only 5 digits.
- generate x new individuals from these sampled individuals (the children of the couple)
Come to think of it, maybe the 10% doesn't hold for an 66-individual?
| quote: | Originally posted by DrUg_Tit0
I don't really think the author actually thought about all these aspects, so we should make it as simple as possible. Survive does seem to substitute for procreate, and I think we should just have them be absolutely monogamous and living equal lifespans. I don't think changing that would make a big difference, aside from making the problem even more complicated. I'd say just substitute survive for procreate. If they all didn't have a 100% survival rate, with just 2 children per couple they'd die out pretty soon. So my guess is that the author wanted to say procreate. |
I agree here. Also having equal lifetimes is a must if the question about "how many generations" is to make any sense.
| quote: | Originally posted by DrUg_Tit0
Now if you actually go and do this, I'd really like to see how you did it |
It won't be very deep - and I have a bad habit of not commenting my code, but of course I can post it (if I get the probabilities, that is).
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