# Probability to draw a black ball in a set of black and white balls with mixed replacement conditions

When a black ball is drawn, it is not replaced in the set, but white balls are replaced.

I have thought of this, with the notations:

• $b$, $w$ the initial number of black and white balls
• $x_i = (b - i)/(b + w - i)$

The probability of drawing a black ball $Pb(n)$ after n draws:

This sum seems infinite with n, even if some terms are null since $x_{i \ge b}=0$

Except $b=1$:
$Pb(n) = (1-x_0)^nx_0$

For $b=2$:
$Pb(n)= x_0(1-x_1)^n + x_0x_1\sum\limits_{i+j=n-1} (1-x_0)^i(1-x_1)^j$

Is there a known solution to this problem?

Let the initial number of white balls be $w$ and black balls be $b$. The question describes a Markov chain whose states are indexed by the possible numbers of black balls $i \in \{0, 1, 2, \ldots, b\}.$ The transition probabilities are

The first describes drawing a white ball, in which case $i$ does not change, and the second describes drawing a black ball, in which case $i$ is reduced by $1$.

From now on let us drop the explicit subscript "$w$," taking this value as fixed throughout. The eigenvalues of the transition matrix $\mathbb{P}$ are

corresponding to the matrix $\mathbb{Q}$ given by

whose inverse is

That is,

Consequently the distribution after $n$ transitions out of the state $b$ is given by the vector of probabilities

That is, the chance there are $i$ black balls left after $n$ draws is

For example, starting with any number of white balls and $b=2$ black balls, the probability distribution after $n \ge 1$ draws is

The curves in this figure track the probabilities of the states $i=0$ (blue), $i=1$ (red), and $i=2$ (gold) as a function of the number of draws $n$ when $w=5$; that is, the urn begins with two black balls and five white balls.

The state $i=0$ (running out of black balls) is an absorbing state: in the limit as $n$ grows without bound, the probability of this state approaches unity (but never exactly reaches it).