Urn with non-uniform probabilities An urn contains N-1 red and 1 green ball. Each ball has an associated weight. If each ball is drawn (without replacement) with a probability proportional to how much its weight contributes to the urn, what is the expected number of attempts required to get the green ball?
Example: 2 red balls of weight 0.3 and 0.4, and green ball of weight 0.3.  
In first attempt, Pr(red1)=0.3, Pr(red2)=0.4, Pr(green)=0.3.  
Say, the red ball with weight 0.3 is chosen in the first attempt.
Then, in the next attempt, Pr(red2)=0.4/(0.4+0.3) and Pr(green)=0.3/(0.4+0.3). It becomes relatively difficult to keep track of the probabilities if there are more red balls.
 A: The same model is used by poker players to estimate the probability of finishing in each place in a tournament given the stack sizes. It is called the Independent Chip Model or ICM. You can download my program ICM Explorer which can calculate the finishing probabilities for up to $10$ balls/players.
Although there doesn't seem to be a simple expression for the probability of finishing in the $k$th place, it's actually quite easy to answer your question. The expected number of red balls you draw before the green ball is the sum of the probabilities that you draw each red ball before the green ball. That's the same as a "last longer" bet in a poker tournament. According to this model, you can ignore all of the other players: consider the first time that you draw the green ball or the $i$th red ball. The conditional probability that you draw the red ball of weight $r_i$ before the green ball of weight $g$ is $r_i/(r_i+g)$, so the expected number of red balls you draw before the green ball is 
$$\sum_i \frac{r_i}{r_i+g}$$
and the expected number of attempts necessary to find the green ball is $1$ more than this.
