Suppose I have x red and y blue balls. At each timestep I draw a ball with probability $$P(\text{red ball}) = (x/(x+y))^z, P(\text{blue ball}) = 1-P(\text{red ball})$$ where z is fixed. Each ball is returned to the urn, but only after b timesteps. Therefore, the sum x+y in the denominator stays constant.
I am interested in the expected value of the probabilities or the number of balls. Any hints to existing, similar problems (and their solutions!) or how to solve this one is welcome.
