What is the probability of taking out a 6-advantage of black marbles? Suppose you have only 1 bag with 6 white marbles and 5 black ones. A person takes out a marble at a time, and then get it into the bag again so the probability will remain the same (6/11 for white, 5/11 for black) for next times. Question is: What is the probability of the scenario where there is a -first time- 6-advantage of black marbles over the number of white marbles? My teacher meant: I require you tell me the probability of getting a score of ( 6 black - 0 white ) + probability of (7 black - 1 white (excluding the scenario where there is 6 - 0 black advantage and then 1 marble of each color appear because 6 - 0 is one of our desired scenarios and it was considered already ) + ( 8 black - 2 white (excluding all the combinations of the previously calculated 7-1 and 6-0 desired scenarios) ) + ( 9 black - 3 white (excluding the already mentioned 8-2,7-1,6-0 scenarios)) + (10 - 4 excluding 9-3,8-2,7-1,6-0) + (11 - 5 excluding 10-4,9-3,8-2,7-1,6-0) etc... For me it is very difficult to formalize this problem, even the teacher mentioned might need a limit expression and maybe calculus!?? Please help :S Thanks :D
 A: Suppose you draw $n$ balls and $b$ are black.  You clearly need $6 \le b \le n$ to satisfy your condition.  
If $b \ge \frac{n+6}{2}$ then you will have at least six more black balls than white in the final position and there will be ${n \choose b}$ patterns of possible draws to get there.
If $b \lt \frac{n+6}{2} $ then you will have not have this as the final position, but you might have had six more black balls than white at an earlier stage. You can use the random walk reflection principle to show there will be ${n \choose b-6}$ patterns of possible draws to have had six more black balls than white at an earlier stage.
So you want $$\lim_{n \to \infty}\left(\sum_{b: 6 \le b \lt \frac{n+6}{2}}   {n \choose b-6} \left(\frac{5}{11} \right)^b\left(\frac{6}{11} \right)^{n-b} + \sum_{b: \frac{n+6}{2} \le b \le n}   {n \choose b} \left(\frac{5}{11} \right)^b\left(\frac{6}{11} \right)^{n-b} \right)$$
As $n$ increases, the first sum tends to $\left(\frac{5/11}{6/11}\right)^6 = \left(\frac{5}{6}\right)^6 $ about $0.334898$, and the second sum tends to zero.
