As it is such a small MRP, it is possible to solve it quickly and analytically using simultaneous equations based on the Bellman equation:

$$v(s) = \sum_{r,s'} p(r,s'|s)(r + v(s'))$$

and substituting in each state in turn. Using the variables $a,b,c,d,e$ to represent $v(A), v(B), v(C), v(D), v(E)$ makes this easier to read:

$a = \frac{1}{2}(0 + b)$

$b = \frac{1}{2}(a+c)$

$c = \frac{1}{2}(b+d)$

$d = \frac{1}{2}(c+e)$

$e = \frac{1}{2}(d+1)$

It takes multiple substitutions to resolve any individual value, at least to start with. But it is possible to resolve these terms by repeated substitution of how many times $a$ they are, and then proceeding to the next equation getting each variable:

$a = \frac{1}{2}(0 + b) \therefore b = 2a$

$b = \frac{1}{2}(a+c) \rightarrow 2a = \frac{1}{2}(a + c) \therefore c = 3a$

$c = \frac{1}{2}(b+d) \rightarrow 3a = \frac{1}{2}(2a + d) \therefore d = 4a$

$d = \frac{1}{2}(c+e) \rightarrow 4a = \frac{1}{2}(3a + e) \therefore e = 5a$

$e = \frac{1}{2}(d+1) \rightarrow 5a = \frac{1}{2}(4a+1) \therefore a = \frac{1}{6}$

That last one resolves everything, and gives the expected answer. You can also see a very strong pattern here, which should intuitively hold for any length of one dimensional random walk with equal $p=0.5$ - there is a proof for that more general case that would void the need to work the substitutions in detail for each length, but that is more complex.

One other simple intuition: Each state value in turn is the *mean* of the two state values either side of it, without exception. That means taking any three values next to each other, they should be co-linear; the two higher and lower values with the mean between them. As this holds on all values with overlap, plotting the values in order should place all points all on the same line. That in turn means that there is a simple linear relationship between position in the random walk and the state value.