Find probability that last drawn ball is white. Urn contains $w$ white and $b$ black balls. We draw balls with the following procedure:


*

*We draw ball and put it aside.

*We draw another ball. If its color is different than last drawn ball then we put it back to urn and go to state 1. Otherwise we put this ball aside and go to state 2.

 A: As was pointed out in a comment the probability in question is $\frac12$ if at least one ball of each color is present.
First of all we prove the following combinatorial lemma:
Let $m$ and $n$ be integer numbers such that $n\ge0$, $m>0$. Then:
$$
\sum_{k=0}^n \frac{n!}{k!}\frac{(m+k)!}{(m+n)!}\frac{m}{m+k}=1.\tag1
$$
Obviously the lemma holds for $n=0$ and any $m>0$. We demonstrate now that if the lemma is valid for $n$ it holds for $n+1$ as well:
$$\begin{align}
\sum_{k=0}^{n+1} \frac{(n+1)!}{k!}\frac{(m+k)!}{(m+n+1)!}\frac{m}{m+k}
&=\frac{m}{m+n+1}+\frac{n+1}{m+n+1}\sum_{k=0}^n\frac{n!}{k!}\frac{(m+k)!}{(m+n)!}\frac{m}{m+k}\\
&\stackrel{I.H}=\frac{m}{m+n+1}+\frac{n+1}{m+n+1}=1.
\end{align}
$$
Let $P(w,b)$ be the probability of the last drawn ball being white if we start from the state 1 with $w$ white and $b$ black balls. 
Observe that to return to state 1 (with a different number of balls) we need to draw first $k$ balls of the same color and then a ball of the other color. In the case of a series of black balls the probability of such event is:
$$
\frac{b}{w+b}\frac{b-1}{w+b-1}\cdots\frac{b-k+1}{w+b-k+1}\frac{w}{w+b-k}=
\frac{b!}{(b-k)!}\frac{(w+b-k)!}{(w+b)!}\frac{w}{w+b-k}
$$
and similar expression for the series of white balls. Thus, we have 
for $w,b>0$:
$$\begin{align}
P(w,b)
&=\sum_{k=1}^b\frac{b!}{(b-k)!}\frac{(w+b-k)!}{(w+b)!}\frac{w}{w+b-k}P(w,b-k)\\
&+\sum_{k=1}^{w}\frac{w!}{(w-k)!}\frac{(w+b-k)!}{(w+b)!}\frac{b}{w+b-k}P(w-k,b),
\end{align}
$$
with $P(w,0)=1$, $P(0,b)=0$.
We claim: $$\forall w,b>0: P(w,b)=\frac12.\tag2$$ 
For $P(1,1)$ the claim is obviously valid. Assume it is valid for all $P(w,b-k)$ with $0<k<b$ and $P(w-k,b)$ with $0<k<w$.
We have:
$$\begin{align}
P(w,b)
&\stackrel{I.H.}=\frac12\sum_{k=1}^b\frac{b!}{(b-k)!}\frac{(w+b-k)!}{(w+b)!}\frac{w}{w+b-k}
+\frac{b!}2\frac{w!}{(w+b)!}\\
&+\frac12\sum_{k=1}^{w}\frac{w!}{(w-k)!}\frac{(w+b-k)!}{(w+b)!}\frac{b}{w+b-k}
-\frac{w!}2\frac{b!}{(w+b)!}\\
&\stackrel{(1)}=\frac12\left(1-\frac{b}{w+b}\right)+\frac12\left(1-\frac{w}{w+b}\right)=\frac12.
\end{align}
$$
Thus, the claim (2) is proved.
A: Let denote by:
$a_n$ - from urn we thrown $n$ balls and last drawn ball was white.
$b_n$ - from urn we thrown $n$ balls and last drawn ball was black.
$A$ - white ball was drawn
$B$ - black ball was drawn
then
$$P(a_n) = P(A|a_{n-1})P(a_{n-1})+P(AA|b_{n-1})P(b_{n-1})+P(AB|a_{n-1})P(a_{n-1})$$
$$P(a_n) = P(a_n)P(a_{n-1})+P(a_n)P(a_{n-1})(1-P(a_{n-1}))+P(a_n)(1-P(a_n))P(a_{n-1})$$
$$P(a_n)(1-P(a_{n-1}))(2P(a_{n-1})-1) = 0$$
this give a solution
$$P(a_n) = 0 \text{ for } w=0$$
$$P(a_n) = 1 \text{ for } b=0$$
$$P(a_n) = \frac{1}{2} \text{ otherwise }$$
