Suppose we have $n$ blue balls and $m$ red balls inside an urn. In each step, we pick a random ball from the urn, look its color and turn the ball back to the urn with another ball of the same color. If we do this process $k$ times, what's the probability of $k+1$-th ball being blue?
- For $k=0$: $$\dfrac{n}{m+n}$$
- For $k=1$: We have (blue, blue) or (red, blue): $$(\dfrac{n}{m+n}\times\dfrac{n+1}{m+n+1})+(\dfrac{m}{m+n}\times\dfrac{n}{m+n+1})=\dfrac{n(m+n+1)}{(m+n)(m+n+1)}=\dfrac{n}{m+n}$$
- For $k=2$: We have (blue, blue, blue) or (blue, red, blue) or (red, blue, blue) or (red, red, blue): $$(\dfrac{n}{m+n}\times\dfrac{n+1}{m+n+1}\times\dfrac{n+2}{m+n+2})+2(\dfrac{n}{m+n}\times\dfrac{m}{m+n+1}\times\dfrac{n+1}{m+n+2})+(\dfrac{m}{m+n}\times\dfrac{m+1}{m+n+1}\times\dfrac{n}{m+n+2})=\dfrac{n(n^2+m^2+3m+3n++2mn+2)}{(m+n)(n^2+m^2+3m+3n++2mn+2)}=\dfrac{n}{n+m}$$
$B_{i}:= i\text{-th ball is blue}$ & $B_{i}^c=R_{i}:= i\text{-th ball is red}$
$$\mathbb{P}(B_{p+1})=\mathbb{P}(B_{p+1}\mid B_p)\mathbb{P}(B_{p})+\mathbb{P}(B_{p+1}\mid B_p^c)\mathbb{P}(B_p^c)$$
$$\dfrac{n}{m+n}=\dfrac{n'}{m+n+p}\implies n'=\dfrac{n(m+n+p)}{m+n}$$ $$\mathbb{P}(B_{p+1}\mid B_p)=\dfrac{\dfrac{n(m+n+p)}{m+n}+1}{m+n+p+1}, \mathbb{P}(B_{p+1}\mid B_p^c)=\dfrac{\dfrac{n(m+n+p)}{m+n}}{m+n+p+1}$$
$\mathbb{P}(B_{p+1}\mid B_p)=(\dfrac{\dfrac{n(m+n+p)}{m+n}+1}{m+n+p+1}\times\dfrac{n}{m+n})+(\dfrac{\dfrac{n(m+n+p)}{m+n}}{m+n+p+1}\times\dfrac{m}{m+n})$ $=\dfrac{n}{(m+n)^2(m+n+p+1)}\times\left[n(m+n+p)+m+n+m(m+n+p)\right]$ $=\dfrac{n}{(m+n)^2(m+n+p+1)}\times\left[(m+n+p)(m+n)+(m+n)\right]$ $=\dfrac{n}{(m+n)^2(m+n+p+1)}\times\left[(m+n+p+1)(m+n)\right]$
$=\dfrac{n}{(m+n)}$
Consider the case there are $m+n$ balls, all of which have different colors. If you apply the rule above, then by symmetry, the probability of getting any color is $\dfrac 1{m+n}.$ Just pick $n$ of those colors and call them blue.