Friday, 23 August 2013

Corrigendum to previous post

This post is a corrigendum to the previous post, which you can either read by scrolling down a bit or here.

This corrigendum is being issued as a result of blogger and general talker of sense, Michael Bench-Capon, who alerted me to an error in the model presented in the previous post.

The error comes from the fact that I suggested that CF and DG parents would mate as one third CF-CF, one third DG-DG and one third CF-DG. I was muddled and forgetting about men having to mate with women and in fact you have one quarter CF-CF, one quarter DG-DG and one half CF-DG. This is, as was explained to me, down to the same maths as that which explains why you get 1/4 heads-heads, 1/4 tails-tails and 1/2 heads-tails when you flip two coins.

Anyway, correcting this error threw something of a spanner in my model's works, because it meant that 1/8 of offspring emerging from DG and CF mating amongst themselves were gay, which made 11.5 out of 90 given the parameters in the previous post. Clearly, if a population with 10 gay people brings forth 11.5 (or more) gay people, then it isn't sustainable in the sense of maintaining the same levels, but has a growing gay population.

The aim of this exercise is to show that a population can maintain a steady rate of 10% gay people.

Given the new, correct, maths, I had to fiddle with my spreadsheet, but I found that 10 gay people could be sustained in a population of 100 if:
  • we have a split 10 CG, 34 CF, 34 DG, 22 DF (this means 22 dirty fighty people)
  • the 68 normally-attractive people produce 72 children amongst themselves - this is around 2.18 kids each, a fair bit less than the 2.25 in the first (erroneous) post
  • the 10 CG (gay) people only produce 2 kids, both of whom are made from reproductions with DG or CF 
  • the other 26 kids are a result of DF-CF and DF-DG reproductions.
 In detail, this means that you have:
  • 18 kids from CF-CF, all of whom are CF
  • 18 kids from DG-DG, all of whom are DG
  • 36 kids from CF-DG, of whom:
    • 9 are CG (gay)
    • 9 are CF
    • 9 are DG
    • 9 are DF.
  • One kid from CG-CF, of whom:
    • 0.5 will be CG (gay)
    • 0.5 will be CF
  • One kid from CG-DG, of whom:
    •  0.5. will be CG (gay)
    • 0.5 will be DG
  • 13 kids from  DF-CF, of whom:
    • 6.5 will be CF
    • 6.5 will be DF
  • 13 kids from DF-DG, of whom:
    • 6.5 will be DG
    • 6.5 will be DF.

Unless I've made another mistake, this gives you 100 kids, who are split 10 CG, 34 CF, 34 DG, 22 DF, the same as what we started with.

The most interesting thing to note here is that, given the correct maths, the gay people have to have far fewer kids and the dirty fighty people can get away with being significantly less unattractive than with the wrong maths.

And as real life uses correct maths, this shows that it's even more possible to sustain a level of 10% gay people within a population than I suggested in my first post.

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