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. 

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 $\frac{1}{b+1}$.
