Probability of red ball i before any black ball Assume we have $r$ red balls and $b$ black balls in a box and we remove one ball at a time without replacement. Red balls are labeled from $1$ to $r$. We want to calculate the probability a particular ball $i$ is chosen before any black ball is chosen. 
I know how to solve the problem using conditioning on the first ball taken, and also by counting the number of permutations of b+1 objects (1 corresponds to the red ball) where the red ball is before any black ball. In both cases we find that:
Prob(red ball i is chosen before any black ball)=$\frac{1}{b+1}$
Can we solve the above problem by counting the permutations of all the $r+b$ objects where red ball $i$ is before any black balls in order to find the probability of interest? 
I have tried the above but I find a dependence on $r$, the number of red balls which does seem to agree with the $\frac{1}{b+1}$ result we took above. 
In other words, the question can become: do we care at all about the rest of the red balls (except red ball i) or not? If we count them, our sample space changes. 
 A: You can work it out (the long way) by counting how many red balls show up before the red ball labeled $i$. Say there are $k$ red balls before the red ball $i$, then we choose which $k$ out of the remaining $r-1$ red balls in $\binom{r-1}{k}$ ways, arrange the $k$ balls in $k!$ ways, and arrange the balls after the red ball labeled $i$ in $(r+b - k - 1)!$ ways.
Therefore, the number of ways of arranging them is $\sum_{k=0}^{r-1} \binom{r-1}{k}k!(r+b-k-1)!$
This simplifies to $(r-1)!\sum_{k=0}^{r-1} \frac{(r+b-k-1)!}{(r-k-1)!} = b!(r-1)!\sum_{k=0}^{r-1}\binom{r+b-k-1}{b}$
Relabeling the sum, gives $b!(r-1)! \sum_{i=b}^{r+b-1} \binom{i}{b}$, where the sum is a well known identity solving to $\binom{r+b}{b+1}$. Therefore, there are $b!(r-1)!\binom{r+b}{b+1} = \frac{(r+b)!}{b+1}$ ways for the red ball labeled $i$ to appear before any black ball. Since there are $(r+b)!$ arrangements total, the probability is $\frac{1}{b+1}$.
A: If you only care about ball $i$ and the black balls, you can ignore all the rest.  Imagine painting them all green-what does that change here?  You certainly could solve it by counting all permutations of the $r+b$ objects and finding how many have ball $i$ before any black balls, but that is a lot more work than what you already did.  You could say there are $(b+r-1)!$ permutations where ball $i$ is first, and it precedes all the black balls in all of them.  There are $(b+r-1)!$ where ball $i$ is second and it preceeds all the black balls in $\frac {r-1}{b+r-1}$ of them, and so on.
