Let's say you have an urn full of balls. Each ball has one or more colors on it. I'm trying to figure out, given that you draw E balls from an urn without replacement, what is the probability that you will remove all of the balls of one or more colors? I think I have it solved for 2 colors, but beyond that, double-counting becomes problematic, and I'm at a bit of a loss as to the proper way to incorporate it.
I've started by thinking about this problem with two colors, red and green, and no mixed color balls. What is the probability of removing all of the red balls, R, given E draws? Well, we can get that from the density function of the hypergeometric function. For shorthand, dh(a,b,c,d) is the probability of drawing a balls of a color with b balls of that color in the urn, c balls not of that color, and d draws.
The probability of drawing out all of the red balls is dh(R, R, G, E). The probability of drawing out all of the green balls is dh(G,G,R,E). The probability of drawing out all of either the red OR the green (since we can't do both with 2 colors) is dh(R, R, G, E)+dh(G,G,R,E).
Great. This leads to an easy general expression for single color balls. Let's sum over all colors, with i being the number of balls with color i. T is the total number of balls.
p(removing all of 1 or more) = sum(dh(i, i, T-i, E))
Now let's say there are M mixed balls - both red and green. Let's go back to the probability of drawing out all balls with red on it.
Here, we have dh(R+M, R+M, G, E). And we can just flip the Rs and Gs to get the same expression for G.
Can we just sum these two to get the answer to the probability of drawing out all balls with either red or green on them? Do we need to worry about double-counting? Not usually in the case of two colors only. One can only draw out either all of the red or all of the green balls, unless you are removing all of the balls. In which case double counting does become a problem.
How can I derive a general term that corrects for double counting with C number of colors, mixtures having any number of colors on them from 2...C, and E draws. I'm guessing that the 1 color per ball case is a special case. I'm just not seeing the correction term. Thoughts? Just pointers in the right direction would be appreciated.