Drawing balls from urn with conditional replacement

Question

An urn contains $n + m$ balls, of which $n$ are red and $m$ are black. On each step one ball is taken from the urn. If it is red, it is returned to the urn, if it is black, a red ball is instead returned to the urn. How to find the expected value of red balls in the urn after $k$ drawing operations?

Answer 1

If you draw a red ball then nothing changes, if you draw a black ball then this ball is removed and a red ball is added to the urn. Let's first calculate the expected number of black balls.

At the beginning, before we draw $(k=0)$ the number of black balls is

$m_0=m$.

After the first draw they number is reduced by $1$ multiplied by the probability to draw a black sphere

$m_1=m-1\dfrac{m}{m+n}$.

For the second draw the same applies but now we have to account for $m_1$ black balls

$m_2=m_1-1\dfrac{m_1}{m+n}$.

So we can write in general that after $k$ draws the number of black balls is

$m_k=m_{k-1}-\dfrac{m_{k-1}}{m+n}=m_{k-1}\left(1-\dfrac{1}{m+n}\right)$,

and as the number of total spheres $m+n$ is constant we can write

$m_k=m\left(1-\dfrac{1}{m+n}\right)^k$.

For the expected number of red spheres after $k$ draws we get

$n_k=m+n-m_k=m+n-m\left(1-\dfrac{1}{m+n}\right)^k=n+m\left(1-\left(1-\dfrac{1}{m+n}\right)^k\right)$

Example for the expected number of red balls for the first 100 draws for m=n=10

Example for the expected number of red balls for the first 100 draws for m=10,n=1