I have $2n$ balls labeled $1, 2, \ldots, 2n$, and two boxes, Box $1$ and Box $2$. I take $n$ of the balls at random without replacement and place them in the Box $1$. I take the remaining $n$ balls and place them in the Box $2$. I take the two smallest-numbered balls from each box. What is the probability that I end up with balls $1, 2, 3$ and $4$?
I'm trying this for small values of $n$ but am struggling to see the pattern. It is the probability $ 1-P(\text{any 3 of balls 1,2,3,4 in the same box})-P(\text{all balls 1,2,3,4 in same box}) .$ I'm having trouble finding the formula form here. Thanks!
Ok this part is more complicated. I have 3n balls and 3 boxes. I take n of the balls at random w/out replacement and place them in Box 1. I take n of the remaining 2n balls at random w/out replacement and put them in Box 2. I take the remaining n balls and put them in Box 3. I then take the smallest-numbered ball from each box and then the smallest-numbered ball among the balls remaining. What is the probability that I end up with balls 1,2,3,4?
So this time, it is okay to have two of Balls 1,2,3,4 in the one of boxes because he will take the smallest and come back for the next smallest. So I'm thinking that it is $\binom{4}{3}$ * ($\binom{3}{1}$\binom{3n-3}{n-1}$)/$\binom{3n}{n}$ *($\binom{2}{1}$\binom{2n-2}{n-1}$)/$\binom{2n}{n}$
I'm not sure if I need that first factor of $\binom{4}{3}$ though. Thanks!