Markov Chains and absorption probabilites

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The discussion focuses on calculating absorption probabilities and expected times to absorption for a single-celled organism with N particles, specifically when N=3. The absorbing states are identified as i=0 and i=3, but the user struggles to determine the probabilities of transitioning between states. They propose that the probabilities for reaching state 1 from state 2 could be calculated based on the combinations of daughter cells formed from independent draws of particles. The conversation highlights confusion regarding the correct approach to these calculations. Overall, the user seeks guidance on how to proceed with the problem.
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Homework Statement


Could someone please help me with this question?

A single-celled organism contains N particles, some of which are of type A, the others of
type B . The cell is said to be in state i , where 0<=i<=N if it contains exactly i particles
of type A. Daughter cells are formed by cell division, but fi rst each particle replicates itself;
the daughter cell inherits N particles chosen at random from the 2i particles of type A
and 2N-2i of type B in the parent cell.

Find the absorption probabilities and expected times to absorption for the case N = 3.


Homework Equations





The Attempt at a Solution


So far i have that the absorbing states are i=0 and i=3 but can't work out the probabilities or times to absorption. Don't know where to start now that i have my absorbing states
 
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From state i, the (A,B) numbers of the daughter are determined by N independent "draws", with probabilities P(A) = i/N and P(B) = 1-P(A) in each draw.
 
So is the probability of say reaching state 1 from state 2 1/5 from the number of possible combinations of the daughter cell or am i going about this the wrong way?
 
macca1994 said:
So is the probability of say reaching state 1 from state 2 1/5 from the number of possible combinations of the daughter cell or am i going about this the wrong way?

I have said all I intend to say on the subject.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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