The probability of ending up with red balls In a bag there are $a$ red balls and $b$ blue balls ($a,b>0$). You conduct the following process.

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*Randomly fetch a ball from the bag, record the colour as the current colour. And throw it away. Go to 2.


*If there's no ball left, break. Otherwise, randomly fetch a ball from the bag. If the colour coincides with current colour, throw it away, and repeat 2; otherwise, put back the ball, and go back to 1.
Qs:

*

*What's the probability that the last ball fetched is red, in terms of $a,b$?
(I seem to lack the necessary combinatorial tools to tackle it. Yeah I deem it to be combinatorial. Initially thought of constructing some kind of martingale based on the balls fetched and apply optional stopping, but didn't work out...)


*Does a closed form solution exist?
(I have absolutely no idea other than imagination about a numerical implementation.)


*Is it generalisable to multicoloured balls (more than two) cases?
(This one just for fun.)

The solution for the 2-color case is surprisingly:
If $a,b > 0$ then the probability to fetch either color last is $\frac12.$
This can be proved via strong induction on $a+b$. 
Induction start: For $a=b\;(=1)$ the probability is $\frac12$ for symmetry reasons, as the algorithm doesn't handle any color special.
Induction step: Assume the probability to fetch a red ball last is $\frac12$ $\forall a,b > 0, a+b < n$ for some $n > 2$. Consider any $a,b > 0, a+b = n$.
Let's first consider that happens when we fetch a red ball in step 1 at the beginning! In that case, in step 2 there are 2 possibilities: 
a) all the remaining $a-1$ red balls are fetched before finally a blue ball is fetched (and put back), or 
b) there is still at least one red ball in the bag at end of step 2, when we return to step 1.
For a) to happen, the first ball fetched in step 2 must be red (probability $\frac{a-1}{a+b-1}$), and the second ball fetched must be red (probability $\frac{a-2}{a+b-2}$, if first fetched ball was red), a.s.o until the $(a-1)$-st ball must be red (probability $\frac{1}{b+1}$, if all previously fetched balls were red). Overall, the probability that case a) happens is
$$p_{red,a}=\frac{a-1}{a+b-1}\frac{a-2}{a+b-2}\ldots\frac{1}{b+1}=\frac{(a-1)!\,b!}{(a+b-1)!}$$
Since a) and b) are complementary events, we have
$$p_{red,b}=1-p_{red,a}.$$
If a) happens, we have no longer any red balls when we return to step 1), so the probability to fetch a red ball last is $0$. For case b), we have still $b$ blue balls and some number $0 < a' \le a-1$ of red balls, so by induction hypothesis the probability to fetch a red ball last is $\frac12$. That means the probability to fetch a red ball last (provided a red ball was fetched first) is
$$p_{red}=0p_{red,a} + \frac12p_{red,b}=\frac12 \left(1-\frac{(a-1)!\,b!}{(a+b-1)!}\right).$$
If we consider the case of fetching a blue ball first in step 1, we get a similar partition into to possibilities as before
a) all other blue balls are fetched in step 2 before we return to step 1, or
b) there is still at least one blue ball in the bag at end of step 2, when we return to step 1.
Case a) happens with probability
$$p_{blue,a}=\frac{(b-1)!\,a!}{(a+b-1)!}$$
(same derivation as above) and again
$$p_{blue,b}=1-p_{blue,a} = 1-\frac{(b-1)!\,a!}{(a+b-1)!}$$
Now, if case a) happens we are guaranteed to fetch a red ball last; in case b) we are back at the induction hypothesis. This leads to the following formula for the probability that we fetch a red ball last, assuming we fetched a blue ball first:
$$p_{blue}=1p_{blue,a}+\frac12\left(1-p_{blue,a}\right) = \frac12 + \frac12p_{blue,a}= \frac12+\frac12\frac{(b-1)!\,a!}{(a+b-1)!}$$
Now, the probability to fetch a red ball or a blue ball first in step 1 are $\frac{a}{a+b}$ and $\frac{b}{a+b}$, resp. That means we can finally calculate the probability of a red ball being fatched last:
$$
\begin{align}
p & = \frac{a}{a+b}p_{red} + \frac{b}{a+b}p_{blue} \\
  & = \frac{a}{a+b}\frac12 \left(1-\frac{(a-1)!\,b!}{(a+b-1)!}\right) + \frac{b}{a+b}\left(\frac12+\frac12\frac{(b-1)!\,a!}{(a+b-1)!}\right)\\
  &  = \frac12\frac{a}{a+b} - \frac12\frac{a!\,b!}{(a+b)!} + \frac12\frac{b}{a+b} + \frac12\frac{a!\,b!}{(a+b)!}\\
  & = \frac12\frac{a}{a+b} + \frac12\frac{b}{a+b}\\
  & =  \frac12.
\end{align}
$$

I guess there is a much easier solution that uses some equivalent formulation of the algorithm to prove the result in a much shorter way, but I can't think of any at the moment.
