Expected number of trials of pairs 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!
 A: *

*Let $n = 2m$, so  $m$ boxes with $2$ balls each, totalling $3$ red and $2m-3$ black


*The red balls can either be $1-1-1$ in $3$ separate boxes or $2-1$ in two boxes


*Placing the $1-1-1$ pattern, the first red ball can go anywhere, the second and third must go in different boxes, thus $Pr = \large\frac{2m-2}{2m-1}\cdot\frac{2m-4}{2m-2} = \frac{2m-4}{2m-1}$, and the other pattern's Pr would be the complement $= \frac3{2m-1}$


*In the $1-1-1$ pattern, the three red balls can be considered as separators that, on an average, divide the boxes into four equal parts, thus the first "good" box is expected to  be seen after $\large\frac{m-3}{4}\;$ boxes, at box $\large\frac{m+1}4$


*Similarly, in the $2-1$ case, the first good box is expected at  $\large\frac{m+1}2$,


*Thus the overall expected value weighted by probabilities will be $\left(\Large\frac{2m-4}{2m-1}\cdot\frac{m+1}4\right) +\left(\Large \frac3{2m-1}\cdot \frac{m+1}2\right) = \Large\frac{(m+1)^2}{4m-2}= \frac{(n+2)^2}{8(n-1)}$
A: 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.
A: There are two scenarios to consider:

*

*The first red ball is drawn out at the same time as a black ball.


*Two red balls are drawn together, then later the third red ball is drawn together with a black ball.
The probability of #1 occurring on the $k$th trial is the probability that we draw $k - 1$ pairs of black balls, followed by one black and one red ball.
The probability of #2 is a bit trickier, but you can look at the probability that it happens on the $j$th trial conditional on the pair of red balls being drawn in the $k$th draw, with the latter event being related to the probability of #1.
Then you just need to collapse all of that down to get the probability that the red-black pair is drawn on the $k$th draw, and then use that to get an expectation.
