Picking balls from a bag until one colour is missing A bag contains balls of three different colours namely $A，B，C$. The number of each types of balls is $n$.
Assume you pick 3 balls randomly from the bag every round, without replacement. And you stop if you have picked up all balls of one colour.  What's the probability distribution of number of the rounds you spend?
I think we can identify this with the same question except you pick one ball a time，but I still cannot handle it.
 A: My answer is not as detailed as Marko's and it does not give the probability of rounds but probability of draws before a ball of one colour to become extinct first.  But a simple enumeration of the combination of draws could lead to the below and all X's are the number of draws that yield the combination. Now let us denote each of them with sub-scripts.
$AAA -X_1$
$AAB -X_2$
$AAC -X_3$
$ABB -X_4$
$ACC -X_5$
$ABC -X_6$
$BBB -X_7$
$BBC -X_8$
$BCC -X_9$
$CCC -X_10$
Let us analyse Ball A
$3X_1+2X_2+2X_3+X_4+X_5+X_6 = n$
If you notice it is same for B and C too.
Let us analyze for Ball B
$3X_7+2X_4+2X_8+X_2+X_6+X_9 = n$
So is for Ball C
Summation of the solution of all three will make sure that one of those balls will get to run out the first.
Now use generating function to find the solution for any one of the equations. And the coefficient of $X^n$ is the number of ways you can draw all n balls of any type first.
Let us assume $n=20$
Thus
$(1+x^3+x^6+x^9+x^{12}+x^{15}+x^{18}+x^{21})(1+x^2+x^4+x^6+x^8+x^{10}+x^{12}+x^{14}+x^{16}+x^{18}+x^{20})^2(\sum_{i=0}^{20}x^i)^3$
The coefficient of $x^{20}$ in the above generating function is $6602$.
Total number of ways all such draws could result in a sum of 60 
$3(X_1+X_2+X_3+X_4+X_5+X_6 +X_7+X_8+X_9+X_{10})= 60$
$(X_1+X_2+X_3+X_4+X_5+X_6 +X_7+X_8+X_9+X_{10})= 20$
and the coefficient of $x^{20}$ in the below generation function
$(\sum_{i=0}^{20}x^i)^{10}$ is $10015005$.
Thus the required probability = $\frac{6602\times 3}{10015005}$
different n, you will have to go through this procedure to obtain the probability.
I hope I have given you a simple solution.
Goodluck.
