3 types of non-infinite coupons in urn with halting, no replacement An urn has 6 coupons in it.


*

*3 red, 2 blue, 1 green


in order to win a prize I must collect a complete set of all the coupons of a particular colour.  So all three reds, two blues or just 1 green constitute a winning set. I continue to draw until I have completed a set but must halt once I do.  Coupons are not replaced, I draw one at a time and each coupon has an equal chance of being picked.
Example draws: RRR, RRBR, RG, G, BRRB, RRBG, etc

What is the probability of completing each set?
I already used brute force and found probabilities:
Red = 54/360 ; Green = 96/360 ; Blue = 210/360
The denominator is 6!/2! because I only checked the first 4 draws, the last two are irrelevant.

So my question is: How do I do this using Maths?  I'm looking for a general solution that can be applied to larger numbers of coupons and more colours.
Edit: Maybe involves Hypergeometric distribution, unsure how to apply.
 A: You could try to use absorbing Markov chains.
For this example there will be 9 states (the first 6 listed below are transient, the last three are absorbing). We can enumerate the states as follows:


*

*Initial state, no coupons observed

*1 Red

*1 Blue

*2 Red

*1 Red, 1 Blue

*2 Red, 1 Blue

*Complete set of Red

*Complete set of Blue

*Complete set of Green


The transition matrix is then
$ P = \begin{bmatrix}
0 & 3/6 & 2/6 & 0 & 0 & 0 & 0 & 0 & 1/6 \\
0 & 0 & 0 & 2/5 & 2/5 & 0 & 0 & 0 & 1/5 \\
0 & 0 & 0 & 0 & 3/5 & 0 & 0 & 1/5 & 1/5 \\
0 & 0 & 0 & 0 & 0 & 2/4 & 1/4 & 0 & 1/4 \\
0 & 0 & 0 & 0 & 0 & 2/4 & 0 & 1/4 & 1/4 \\
0 & 0 & 0 & 0 & 0 & 0 & 1/3 & 1/3 & 1/3 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}
$
With the transition matrix in hand you can use the formulas given in the wiki to calculate various different properties of the urn model. Here is a Matlab script which computes the probabilities you are interested in.
Q = [0,3/6,2/6,0,0,0;
     0,0,0,2/5,2/5,0;
     0,0,0,0,3/5,0;
     0,0,0,0,0,2/4;
     0,0,0,0,0,2/4;
     0,0,0,0,0,0;];

R = [0,0,1/6;
      0,0,1/5;
      0,1/5,1/5;
      1/4,0,1/4;
      0,1/4,1/4;
      1/3,1/3,1/3;];

N = inv(eye(6) - Q);
B = N*R;
absorption_probabilities = B(1,:)

%If only interested in absorption probabilities we can avoid explicitly
%forming (I-Q)^(-1)
%e1 = [1;0;0;0;0;0];
%absorption_probabilities = R'*(((eye(6)-Q)')\e1);

In order to generalize this, you would need to find a convenient way to enumerate the states and transition probabilities for an arbitrary number of colors/coupons.
A: Say you have $k$ colours, and $n_i$ coupons of colour $i$. Denote the set of colours by $I$. The probability that an entire subset $S\subseteq I\setminus\{i\}$ of other colours is completed before colour $i$ is completed is
$$
\frac{n_i}{n_i+\sum_{j\in S}n_j}
$$
(the probability that the last coupon of all the colours in $S\cup\{i\}$ that's drawn has colour $i$).
Thus, by inclusion-exclusion, the probability to complete colour $i$ after $m$ draws without completing any of the other colours is
$$
\sum_{S\subseteq I\setminus\{i\}}(-1)^{|S|}\frac{n_i}{n_i+\sum_{j\in S}n_j}\;.
$$
In your case, $k=3$, $n_1=3$, $n_2=2$, $n_3=1$, so the probability to first complete red is
$$
\frac33-\frac34-\frac35+\frac36=\frac3{20}\;,
$$
the probability to first complete green is
$$
\frac22-\frac23-\frac25+\frac26=\frac4{15}\;,
$$
and the probability to first complete blue is
$$
\frac11-\frac13-\frac14+\frac16=\frac7{12}\;,
$$
in agreement with your results.
