conditional probability markov chain A man has three shirts. Each day he chooses one to wear at random from those that he was not wearing the previous day. The shirts are labelled by 1, 2 and 3 and X$_n$ is the label of the shirt he wears on day $n$.
What is the probability that the next day he will wear shirt 2 given that he wears shirt 1 today?
I look at this and the answer seems obvious, namely P(X$_{n+1}$ =2|X$_n$ =1) = $\frac{1}{2}$.
However the method presented uses the definition of conditional probability and is given as
P(X$_{n+1}$=2|X$_n$=1) = $\frac{P(X_{n+1}=2 \cap X_n=1)}{P(X_n=1)}$ = $\frac{1/3\times1/2}{1/3}=\frac{1}{2}$
What does $X_{n+1}=2 \cap X_n=1$ look like and why does it equal $\frac{1}{3}\times \frac{1}{2}$? Any feedback on this appreciated.
 A: A more solid explanation comes from the study of stationary distributions in Markov chains. Allow me a brief divagation. 
Say we have a discrete Markov chain $X=\{X_n\}$ with space of possible states $S$ and transition matrix $P=(p_{xy})_{x,y\in S}$. A vector $\pi$ with non-negative entries is said to be a stationary distribution of $X$ if $\sum_{x\in S}\pi(x)=1$ and $\pi P = \pi$, that is if it satisfies $\pi(y) = \sum_{x\in S}p_{xy}\pi(x)$ for every $y\in S$
Let $\pi$ be a stationary distribution, then for every $n\in\mathbb N$
$$\pi P^n = (\pi P)P^{n-1} = \pi P^{n-1}=\ldots=\pi P = \pi$$
Thus, if $X$ has initial distribution $\pi$, for every $n\in\mathbb N$ we have that
$$\mathcal P(X_n=y)=\sum_{x\in X}\pi(x)p_{xy}^{(n)} =\sum_{x\in X}\pi(x)p_{xy}=\pi(y)$$
On the other hand, if the distribution of $X_n$ is independent of the time $n$, then the initial distribution $\pi_0$ is such that 
$$\mathcal \pi_0(y)= P(X_0=y) =\mathcal P(X_1=y)=\sum_{x\in S}\pi_0(x)p_{xy}$$ 
thus, $\pi_0$ is an stationary distribution.
In summary, the distribution of $X_n$ is independent of $n$ iff the initial distribution is a stationary distribution.
Now, let's get back to the man-shirt problem. As you've pointed out, it has transition matrix 
$$P=\begin{pmatrix}
0 &\frac12&\frac12\\
\frac12&0&\frac12\\
\frac12&\frac12&0
\end{pmatrix}
$$
We're interested in finding $\pi$, its stationary distribution. That is, we solve a simple system of equations
$$\begin{align}
\pi(1) & = & 0 & +\frac12\pi(2) &+\frac12\pi(3)\\
\pi(2) & = &\frac12\pi(1)&+0&\frac12\pi(3)\\
\pi(3) & = &\frac12\pi(1)&+\frac12\pi(2)&+0\\
1 & = & \frac12\pi(1)&+\frac12\pi(2)&+\frac12\pi(3)
\end{align}
$$
which has solution $\pi=(\frac13,\frac13,\frac13)$. But the original problem states that the first day, the man chooses a shirt randomly. That is, the initial distribution is the stationary distribution. So $\mathcal P(X_n=y)=\frac13$ for $y=1,2,3$ and for every $n\in\mathbb N$.
Now, getting back to your question, what does $\mathcal P(X_{n+1}=2, X_n=1)$ mean? It is the probability that the man wore the 1st shirt the $n$-th day AND he wore the 2nd shirt the next day. The probability of choosing the 1st shirt the $n$-th day is $\mathcal P(X_n=1) =\frac13$, and the probability that then he wore the 2nd shirt is choosing it at random from 2 posible shirts, that is $\frac12$. Hence $\mathcal P(X_{n+1}=2, X_n=1)=\frac13\times\frac12$
Notice that this reasoning (of computing $\mathcal P(X_{n+1}|X_n)=\frac{\mathcal P(X_{n+1},X_n)}{\mathcal P(X_n)}$) is possible thanks to the fact that our initial distribution is the stationary distribution. Had the initial distribution been $\pi_0=(\frac12,\frac18,\frac38)$, we would have to make a much more complicated argument.
