I have a binomial response variable of 18 observations of which 9 yes and 9 no. What is the chance that I find a variable which has a perfect separation of no's and yes's? So getting 9 times yes and then 9 times no in that order (or the exact reverse).
After some juggling with multiplying fractions I derived this fraction: (9!*9!)/18! = 1/48620 chance of 9*yes before 9*no.
But if this is true, I'm very much surprised this is the inverse of the combination rule: n!/((n-k)!*k!)
Can anyone confirm my approach and this finding? Thanks a million.
Your approach is almost correct. The correct probability is $2\cdot \frac{9!\cdot 9!}{18!}$, in other words double the value of your probability. The reason is that the block of no's or the block of yes's can come first.
In other to see why your observation is true, you can count the number of solutions in a different way: First, pick the nine spots where no's should go and then fill in the remaining as yes's. There are $\binom{18}{9}$ ways to do this and 2 give valid orderings, thus the probability is $\frac{2}{\binom{18}{9}}$.