Computing long run proportion with Markov chain Two-thirds of all trucks have a car behind them but just only one-fifth of the cars have a truck behind them. What proportion of vehicles are cars?
This problem from Markov chain chapter. But it seems like the states here are not easy to find. Maybe it's because you don't even need a markov chain ? 
If there are $y$ trucks then $(2/3)y$ cars are behind them and $(1/5)(2/3)(y)$ of those cars have another truck behind them and $(2/3)(1/5)(2/3)(y)$ of those trucks have a car behind them.
How do I solve this sum  though? Looks difficult and I don't see it
 A: I think the Markov Chain you want to consider looks like this:
We have two states, "truck" and "car", and we can think of each turn as moving back along the line of vehicles by one. The transition probabilities would be
\begin{align*}
p_{cc} &= 4/5 \\
p_{ct} &= 1/5 \\
p_{tc} &= 2/3 \\
p_{tt} &= 1/3,
\end{align*}
and if $C$ is the number of times $c$ is visited and $T$ is the number of times $t$ is visited, then the answer would be $\frac{\mathbb{E}C}{\mathbb{E}C+\mathbb{E}T}$. The problem is solvable but I'm not exactly sure what the initial probabilities would be. Is there an equal chance of starting with a car or a truck, or is there no specification at all?
A: To get anywhere, you need to make the assumption that there are no other vehicles besides cars and trucks.  The Markovity arises because the probability that the currently passing vehicle is a car or truck depends on (and presumably only on) whether the previous one was a car or truck.   You can model this as a two-state Markov chain with transition matrix
$$
P=\pmatrix{\frac{1}{3}&\frac{2}{3}\\
         \frac{1}{5}&\frac{4}{5}}\ ,
$$
and which is in state $1$ if the current passing vehicle is a truck, or state $2$ if it's a car.  The equations for the equilibrium distribution $\ \pi\ $ are
\begin{align}
-\frac{2\pi_1 }{3}+ \frac{\pi_2}{5}&=0\\
\pi_1+\pi_2 &= 1\ ,
\end{align}
which have solution $\ \pi_1=\frac{3}{13}, \pi_2=\frac{10}{13}\ $.
