# Dynamics of an urn, with additional replacement of one colour and removal of the opposite colour

PREMISE

Suppose we have an urn with $$N$$ balls, with $$m$$ white balls and $$N-m$$ black balls. At every iteration, a ball is chosen at random. If a white ball is chosen, we add an extra white ball and remove a black ball, so that there are always $$N$$ balls. If a black ball is chosen, then we simply put it back in the urn. The process terminates when $$m=N$$, since it can no longer continue with the same rules.

QUESTIONS

1. What is the expected value of the number of white balls after $$k$$ iterations? Alternatively, how many iterations does it take on average to reach $$m=N$$?
2. Is this a well known probability distribution? Maybe a variant of the hypergeometric distribution?

WHAT I KNOW

In a particular simulation, if we have $$s$$ successes out of $$k$$ iterations, then the probability of this event is $$Pr(S=s) = \frac{(m+s)!/(m-1)!}{N^k} \times Pr(k-s \ \text{failures})$$

always. The term in the expression representing the probability of successes is always the same/independent of the order of successes and failures.

The probability of having a failure at any iteration is only dependent on the number of previous successes; order matters here.

• I suspect this can been seen as a kind of coupon collector's problem run backwards in time, particularly if $m=1$. To actually solve it, you want to sum the expected numbers of draws until you draw the next white ball (which in each case has a geometric distribution with a parameter depending on $N$ and $m$). The sum, i.e. the answer to question 1, will be $N$ times a harmonic number. Dec 7, 2022 at 1:21

As @Henry noted, the number of draws at each $$m$$ is geometrically distributed:

$$P(X = x)=\frac{m}{N}\Big(1-\frac{m}{N}\Big)^{x-1}$$

which has a mean of $$\frac{N}{m}$$

The expected number of draws until $$m=N$$ is

$$\sum_{i=m_0}^{N-1}\frac{N}{i}=N\cdot(H_{N-1}-H_{m_0-1})$$

where $$m_0$$ is the initial number of white balls, $$H_n$$ is the $$n^{th}$$ harmonic number, and $$H_0=0$$.

This is the same as the expected number of draws for the coupon collector's problem for collecting the second through the $$(N-m+1)^{th}$$ coupon out of $$N$$ coupons.