Urn Problem - black and white balls, infinite trials.

Question

Imagine that there are 10 black balls and 20 white balls in an urn. Two balls are removed at random from the urn. The second ball removed is recolored, such that it matches the color of the first ball, then we put the balls back into the urn and repeat this process indefinitely. What is the probability that all the balls in the urn turn black?

My approach is to model this problem similar to gambler's ruin problem. We start with 20 white balls and we should find out the probability that we reach the absorbing state of 0. We have three possibilities after every trial, to stay in the same state(no change in the number of white balls of urn after the trial), one white ball less(we go to the state i-1), on white ball more(we go to the state i+1). P varies at each trial and only depends on the current state. Is this method feasible? or should I look for variants of Polya urns?

Answer 1

It is a pure gambler's ruin problem. While there are balls of each colour in the urn, with probability $\frac{1}{2}$ the number of black decreases by $1$, and with probability $\frac{1}{2}$ it increases by $1$. The ties are irrelevant.

Answer 2

Yes, As long as there are white balls and black balls, the probability of increasing whites $W/(B+W) * B/(B+W-1)$ is equal to probability of increasing blacks $B/(B+W) * W/(B+W-1)$, so the process is as likely to go toward W as to go toward B. But once we reach a state of only one color the probabilities are 1 (for no change) and 0 (for change). The expected length of the chain of events toward all W is much shorter than that to all B. This is where the difference appears. In the end, the probability of all B is much smaller than all W, so $P[all B]<<0.50$.