Probability of hitting state i I have just started learning Markov chain and I have no idea how to solve this question
An ant walks along the edges of a cube, starting from the vertex marked 0. Upon reaching a vertex, the ant continues to edges incident to this vertex, with equal probability for each.
There are 2 spiders at two vertices marked α and β waiting for the ant. What is the probability that the ant will reach to the spider vertex α? What about the vertex β?
 A: This problem can be simplified to the point of admitting an easy solution.  Use this as a guide when working through the Markov Chain calculations to check your work.

Let $p_s$ be the chance of ending up at $\alpha$ when starting at vertex $s.$  We need to find $p_0.$ Since inevitably the caterpillar will wind up glued (prove this!), $1-p_s$ is its chance of ending up at $\beta.$
From the symmetries of the cube notice that
$$p_0 = 1-p_3;\quad p_1=p_5=1-p_2=1-p_4.$$
Since $p_\alpha=1$ and $p_\beta=0,$ that leaves us needing find just two quantities; say, $p_0$ and $p_1.$
Only three moves are possible from $0,$ each with equal probability to states $1,3,$ and $5.$ Therefore (state this rigorously in terms of conditional probability!)
$$p_0 = \left(p_1+p_3+p_5\right)/3 = \left(p_1+1-p_0+p_1\right)/3 = (1-p_0+2p_1)/3,$$
permitting us to express $p_1$ in terms of $p_0,$
$$p_1 = (4p_0-1)/2.$$
From state $1$ there are three equiprobable moves to states $0, 2,$ and $\alpha,$ whence
$$p_1 = (p_0+p_2+p_\alpha)/3 = (p_0 + 1-p_1 + 1)/3.$$
In conjunction with the antecedent equation this gives the unique solution
$$p_0 = 4/7.$$
The full solution can now be directly computed from the foregoing as
$$(p_0,p_1,p_2,p_3,p_4,p_5,p_\alpha,p_\beta) = (8, 9, 5, 6, 5, 9, 14, 0)/14.$$
