Probability that only balls of the same color are left in an urn An urn contains x red balls and y green balls. The balls are withdrawn one at a time until only those of the same color are left. What is the probability that the remaining balls are green?
I worked with the assumption that you draw balls until only one is left, and this last ball is green. To find the total number of events in which I could arrange all balls in the urn,I can specify that this is equal to (x + y)! / x!y!. Similarly, for the case in which only one ball remains, I would have (x + y -1)! / x! (y-1)! events. Then, the probability, after working out the factorials, is x/(x+y). 
However, I'm having trouble with the question asked. x/(x+y), as I understand, is the probability for an specific scenario of the probability that is being asked.
That "overall" probability, as I'm thinking, should encompass the scenario in which I have only one green ball left, the scenario in which two green balls are left, and so on till I get to the scenario where I draw all red balls from one go and I have all green balls still inside the urn. 
I thought I could extend the previous analysis, so that the number of events in which I have n green balls remaining (where 1 < n <= y) could be calculated as (x+y-n)!*n! / x!(y-n)!, but this is just my assumption.
I have seen similar questions in this page, but I see that the results focus on the probability of the last ball being green (as in this case).I was hoping someone could help me understand how I should approach this problem.
 A: You can continue until only one ball is left, as this assumes nothing about the order that all the others are removed. It would be a different problem to find the probability that the second-last would be red and the last green.
A: Imagine arranging the balls in a line, and you sequentially draw the balls along this line until you have drawn the $x^{\rm th}$ red ball.  Therefore the outcome of interest, i.e. the remaining balls are green, occurs if and only if the $x^{\rm th}$ red ball is not the final ball.  So the final ball must be green.
Clearly, there are $\binom{x+y}{x}$ ways to choose an ordering of the $x$ red balls.  Of these, there are $\binom{x+y-1}{y-1}$ ways to order the balls such that the final ball is green.  Thus the desired probability is $$\frac{\binom{x+y-1}{y-1}}{\binom{x+y}{x}} = \frac{y}{x+y}.$$
A: Like you said, let's assume that there are $x$ red balls and $y$ green balls.
In order for the first set of monochromatic balls(say $n$ balls) to be green, the colour of the last removed ball($n+1^{th}$ from last) must be red. Otherwise the first monochromatic set would have been $n+1$ green balls.
Now that being said, the probability that the first monochromatic set has n green balls is $\frac{\frac {(x+y-n-1)!}{(x-1)!(y-n)!}}{\frac{(x+y)!}{x!y!}}$. Hence the probability that the first monochromatic set is green is 
$$\frac{\sum_{i=1}^y{\frac {(x+y-i-1)!}{(x-1)!(y-i)!}}}{\frac{(x+y)!}{x!y!}}$$
$$=\frac{x.y!}{(x+y)!}\sum_{i=1}^y{\frac {(x+y-i-1)!}{(y-i)!}}$$
