I was wondering if anyone had an idea of how to approach this problem. Although I have written a simulation, I would like to see if I can get to a mathematical expression instead.
We have three types of balls, Red (R), Blue (B), and Green (G). We will also assume that R, B, G also indicate the number of balls which are red, blue, and green, respectively for this problem.
These balls are placed in an urn. Two balls are pulled from the urn at a time. If the balls are red and blue (R,B) or blue and red (B,R) then we annotate the draw count. We wish to determine the distribution of the number of draws between pulling (R,B) or (B,R).
For example: Let R = 3, B = 2, G = 3
Draw 1: R,B <-- D = 1
Draw 2: G,G
Draw 3: R,G
Draw 4: B,R <-- D = 4
Therefore, the time between draws is 1-0=1 and 4-1=3. So the times between the draws are 1 and 3.
This example does not capture all possible states but I put it here to illustrate what I am trying to get at.
Thank you for your help!
It doesn't matter what the previous draw was, since the draws are independent. You simply want the distribution of the number of draws until the first red-blue draw. This is a geometrically-distributed random variable with probability of success $$p={RB\over {R+B+G\choose2}}={2RB\over N(N-1)}$$ where $N=R+B+G$. The probability that it takes $k$ draws until the first success is $$(1-p)^{k-1}p$$ since we must have $k-1$ failures, followed by a success.