I have seen a lot on here about the stationary distribution for a birth-death chain that goes to infinity, such as the following [random walk][1]. [1]: https://i.stack.imgur.com/I4xp7.png
However, I am confused on finding the stationary distribution for the case with a finite end (with no queue), ie the random walk above but the domain is [0, 3] not $[0, \infty]$.
For example, consider you have 3 people. Let $X_t$ be the number of people wearing hats at time t. If not wearing a hat, they independently put on a hat with rate $\lambda$ (~exp). If wearing a hat, they take off their hat with rate $\mu$ (~exp). What is the stationary distribution of $X_t$?
I have seen for the general N, the stationary distribution is $$ \pi_k = \pi_0 \prod_{i=1}^{k}\frac{\lambda}{i \mu} = \pi_0 \frac{(\lambda / \mu)^k}{k!} $$ for k = 0, 1,..., N,
with
$$ \pi_0 = (\sum_{k=0}^N \prod_{i=1}^k \frac{\lambda}{i \mu})^{-1} = (\sum_{k=0}^N \frac{(\lambda / \mu)^k}{k!})^{-1} $$
But I'm not really sure where to go with the stationary distribution from there. Is it just a simple plug and chug from here on out? For example, in the hats example above, do I just manually calculate $\pi_0, \pi_1, \pi_2, \pi_3$ to find $\pi$?