Finding the transition probably matrix

If I have an urn that contains six tags, three are red and three are green. Two tags are selected from the urn. If one tag is red and the other is green, then the selected tags are discarded and two blue tags are returned to the urn. Otherwise the selected tags are resumed to the urn. This process repeats until the urn contains only blue tags.

Then, I let $X_n$ denote the number of red tags in the urn after the nth draw, with $X_0=3$ and I'm supposed to find the transition probability matrix. I'm kind of lost as to hard to start this. Thanks for any and all help!

Solution 1:

See that $X_n\le3$, so that the set of states of the process is $\{0,1,2,3\}$ being $0$ an absorbing state. Now, since $X_{n+1}=X_{n}$ or $X_{n+1}=X_{n}-1$, the probabilities you're interested in are $P_{3,3}, P_{3,2}, P_{2,2}, P_{2,1}, P_{1,1}, P_{1,0}$ (the others are $0$ except for $P_{0,0}$ which is $1$). These are going to be the elements of the matrix you are looking for.

$P_{3,3}=P(X_{n}=3|X_{n-1}=3)= 2\frac{\binom{3}{2}{\binom{3}{0}}}{\binom{6}{2}}$ (that is: both of the tags you get are red or both green)

$P_{3,2}=P(X_{n}=2|X_{n-1}=3)= \frac{{\binom{3}{1}}{\binom{3}{1}}}{\binom{6}{2}}$ (one tag you get is red and one is green)

$P_{2,1}=P(X_{n}=1|X_{n-1}=2)= \frac{{\binom{2}{1}}{\binom{2}{1}}{\binom{2}{0}}} {\binom{6}{2}}$ (now you have two blue tags inside the urn)

$P_{2,2}=1-P_{2,1}$

I'll leave the others to you. Hope it's been of some help.