math - Approximation Algorithm- using expectation -

note: don't have understand approximation algorithms answer this

hello.
need prove algorithm approximation using expectation.
algorithm takes x_i \in {0,1,2} such i\in 1,...n+2 , there constants c_i \in 0,1,2 such i\in 1,...,n , find assignment variables such x_i +x_(i+1)+x_(i+2) != 0 \mod(3) i such number of indexes such x_i +x_(i+2) = c_i \mod(3) maximized.
following:
choose x_1 , x_2 \in 0,1,2 independently , uniformly @ random ,
each i\in 3,...,n+2 given values of x_(i-2),x_(i-1) assign x_i 1 of 2 values in {b\in 0,1,2 | x_(i-1)+x_(i-2)+b != 0 \mod(3)} uniformly @ random.
the assignment each x_i independent x_j such j<i-2.

need prove algorithm gives 1/3 approximation problem described, using expectation(meaning proving x random variable assign question, e[x]=1/3)
struggling defining such x , calculating this, keep getting 2\3 instead of 1\3.
can calculation?

you can prove x_i uniformly distributed on {0, 1, 2} , x_i pairwise independent induction. base case (n=2) immediate, , induction step follows fact you're given x_i being independent x_j (with j < i-2), , simple enumeration of cases.

then result follows immediately, since p(x_i + x_{i+2} = c_i) 1/3, , linearity of expectation, e[x] = n/3.

clarifying last statement: let v_i random variable that's 1 if x_i + x_{i+2} = c_i , 0 otherwise. x = sum(v_i i=1..n), , e[x] = sum(e[v_i] i=1..n) linearity of expectation. e[v_i] = p(x_i + x_{i+2} = c_i) = 1/3.