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I was wondering how to apply Markov steady-state chains to a simple case in genetics, eye colour.

Assuming the probabilities of being brown, green or blue eyed, given your parents were a certain colour combination are:

Br + Br - Br 75%  , Gr 18.75% , Bl 6.25%
Gr + Br - Br 50%  , Gr 37.5%  , Bl 12.5%
Bl + Br - Br 50%  , Gr 0%     , Bl 50%
Gr + Gr - Br 0.5% , Gr 74.75% , Bl 24.75%
Gr + Bl - Br 0%   , Gr 50%    , Bl 50%
Bl + Bl - Br 0%   , Gr 1%     , Bl 99%

Corresponding matrix:

{{3/4, 1/2, 1/2, 0.005, 0, 0}, {0.1875, 0.375, 0, 0.7475, 1/2, 1/
  100}, {0.0625, 0.125, 1/2, 0.2475, 1/2, 99/100}}

I want to find the steady state for the eye colour of the population.

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If you've made any attempt at solving this yourself it would be a good idea to share that attempt. Providing data that's been formatted for Mathematica is also appreciated. –  Pickett Aug 28 '13 at 1:43
    
This question requires an initial gentype frequency. some assumption or declaration of probabilities of genotypes mating and then use of Mendelian genetic rules to develop transition matrix for genotypes (as states). The eye colors can be inferred from the genotype-phenotype relationships after determining stationary(steady state) distribution for given initial state. The matrix you have, though rows add to 1, is not a transition matrix, eg entry (1,1) is not transition from state 1 to state 1 etc. –  ubpdqn Aug 28 '13 at 4:20
    
I haven't made any attempts that led anywhere. I saw the table of values on twitter and immediately wondered what the steady-state would be. I was willing to assume that any one person can marry any other person on Earth to remove any regional differences there may be, but I was hoping that it was possible to somehow use the steady-state as the probability that person A is Br/Gr/Bl and same for person B and solve? I haven't studied Markov chains in that much detail and it seems like it is a lot more complicated than I had hoped. –  DannyBland Aug 28 '13 at 8:14
    
I thought about setting: P(Brown)=0.75*P(Brown)*P(Brown)*... etc. It would be assuming that there is equal likelihood of a person with brown eyes having a child with a person with green, blue or brown eyes (so there is no bias from a particular eye colour to a particular eye colour) and that eye colour is spread the same universally. It would seem to lead to a solution but I don't have a maths package and there are too many terms for Wolfram Alpha. –  DannyBland Sep 9 '13 at 17:03

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