Order of drawing colored balls from a bag We're looking for an upper bound on (or a method to compute) probabilities of the following type:  Suppose I put 12 yellow balls, 18 red balls, 7 white balls, and 2 green balls in a bag.  Then I start withdrawing random balls one by one. What is the probability that I draw at least one red ball and at least one yellow ball before I draw the first green ball?

More generally, I have a set $A$ of $m$ colors, a set $B \subset A$ of colors, a color $c \in A \setminus B$ and a set of $n$ balls with colors from $A$ where the number of balls of each color is known. Whats the probability that I draw at least one ball of each color in $B$ before I draw a ball of color $c$?

Peter Shor points out that this is related to the coupon collector's problem (thanks!), the main difference being that we want to know the probability of a particular color being last (not the number of steps required to collect every coupon).  The literature we've searched on the coupon collector's problem (in the last few hours) doesn't seem to address this question.  (A minor difference is that we're sampling without replacement, though a solution for sampling with replacement would be welcome too.)  Ideally we'd like a closed-form upper bound (an asymptotic upper bound would probably do), but a polynomial-time algorithm to compute an upper bound would be useful too.

We seek this bound to help us analyze a randomized geometric algorithm.
First observation: the cases with and without replacement have exactly the same answer. (Once you have drawn a red ball, say, you don't care about future draws of red balls: all you care about now is the relative order of future green and yellow balls).

A good way to attack the case with replacement is by embedding in continuous time. 

Imagine that the balls arrive according to independent Poisson processes with appropriate rates: rate 12 for yellow balls, rate 18 for red balls, rate 2 for green balls. Then whenever a ball arrives, it has probability 12/32 of being yellow, 18/32 of being red and 2/32 of being green, as desired - I'm ignoring the white balls since they're irrevelant. So the sequence of colours observed is exactly as it would be from a sequence of independent draws with replacement. 

Now you want to know the following: does the first event in the green process occur after the first events in the yellow and red processes? This is easy enough to calculate, since the first event in a Poisson process of rate $r$ has exponential distribution with rate $r$, and the three processes are independent. 

So, integrating over the time of the first event in the green process, you get the following expression for your probability:

$\int_0^\infty 2e^{-2x} (1-e^{-12x})(1-e^{-18x})dx$

which is straightforward to evaluate and generalises nicely.