I've been confused with this question for awhile:
Suppose there are $n$ balls in a bag and $n$ is even; $3$ are red and $n-3$ are black. At each trial, $2$ balls are taken from the bag without replacement. What is the expected number of trials until a pair with exactly $1$ red ball and $1$ black ball is taken from the bag?
I've tried different sums like $\sum_{n=1}^{\frac{x}{2}}\frac{x-3}{x-1}\cdot n$, which I thought might work because it uses the probability of a red ball being paired with a black ball, but I realized it doesn't account for the changing probability, since the balls aren't replaced. When I tried to account for these changes by subtracting 2 after the balls are taken out, the expected number still didn't come out correct.
Any hints or help would be greatly appreciated, I've just been so stuck and can't come up with a good probability to work with this question!
Best Answer
Let $n=2m$ and for $k=1,2,\dots,m$ let $A_k$ denote the event that at the $k$-th draw for the first time a red and black ball are drawn.
Then:$$P\left(A_{k}\right)=\frac{2\left[\binom{2m-2k}{2}+k-1\right]}{\binom{2m}{3}}$$ Here the denominator equals the number of ways that we can place the $3$ red balls in a row of $2m=n$ balls.
Thinking from left to right the numerator is the number of ways of placing $3$ red balls in this row such that event $A_k$ occurs.
Then for the expectation we find:$$\sum_{k=1}^mkP(A_k)=$$$$\binom{2m}3^{-1}\sum_{k=1}^m2k\left[\binom{2m-2k}{2}+k-1\right]=\frac{\left(m+1\right)^{2}}{2\left(2m-1\right)}=\frac{\left(n+2\right)^{2}}{8\left(n-1\right)}$$ Not a nice job to work this summation out but it can be done.
It would not surprise me if someone provides a more elegant solution.