Draw and Add Balls

Problem

Suppose you have $x$ red balls and $y$ black balls. Each round, you randomly draw a ball, put it back and add $k$ balls with the same color. What’s the probability that the $n$-th draw is red.

Solution 1

We can work on Expectation:

$$\mathbb{E}[\#\text{red balls in }n\text{-th round}]=\mathbb{E}[\#\text{red balls in }(n-1)\text{-th round}]+\mathbb{E}[\#\text{red balls added after}(n-1)\text{-th draw}].$$

Therefor we have

$$\frac{\mathbb{E}[N_2^{\text{red}}]}{\mathbb{E}[N_2^{\text{red}}]+\mathbb{E}[N_2^{\text{black}}]}=\frac{x+xk/(x+y)}{x+y+k}=\frac{x}{x+y}.$$

Then we can induce that $P=\frac{x}{x+y}$ for any $n$ and $k$.

Solution 2

We can utilize symmetry and relabeling. Instead of labeling balls as red and black, we can see each of them representing a color, and $x$ colors belong to a “red” family, $y$ colors belong to a “black” family. Then there are $x+y$ colors, and each time you add $k$ balls of the same color.

Then by symmetry, we know that the probability of drawing any color is $P^{\prime }=\frac{1}{x+y}$. Then we know that the probability of drawing a ball in “red” family is $P=x\cdot P^{\prime }=\frac{x}{x+y}$, regardless of $k$ and $n$.