Statistics dice question $6$ people, $A, B, C, D, E, F$ sit in a circle to play a dice game.

Rule of game: as you roll a die and get a number which is a multiple of $3$, give the die to the person on the right(counterclockwise). If the number is not a multiple of $3$, give it to the person on the left.
What is the probability that $B$ has the die after five trials starting from $A$?

My approach:
The probability that the die would go right is $\frac 13$ and left is $\frac 23$. Am I supposed to find all the possible outcomes and add the probabilities?
For example, it could go $A-B-C-D-C-B$. So $\frac 13\times \frac 13 \times \frac 13 \times \frac 23 \times \frac 23$.
Do I find all the paths that it could take and add all these probabilities? Or is there another way to solve this??

Let A=0 B=1 C=2 D=3 E=4 F=5
You are at A (ie 1), you have 5 chances and each chance can be +1 with Probability X(=1/3) or -1 with Probability Y(=2/3) such that
v + w + x + y + z = 1
Since RHS is positive, therefore number of positive term on LHS > no of negative term, hence solution is of the form
(v + w + x) + (y + z) = 1 where first bracket is of positive terms and second bracket of negative terms. This can be arranged in L = 5C3 ways.
Therefore Prob is [(1/3)^3 x (2/3)^2] x L = 10 x (1/27) x (4/9) = 0.164
(As per the suggestion from comments)
Also 
v + w + x + y + z = -5
or 
v' + w' + x' + y' + z' = 5 just change signed to make calculation intitutive.
Now same analysis as above, only possible if 
solution of for (v' + w' + x' + y' + z' ) = 5 ie all terms positive, hence I = 5C5 = 1, therefore Probability is (2/3)^5 * 5C5 = (32/243) = 0.131
Therefore net probability is 0.131 + 0.164 = 0.295
P.S Feel free to ask or correct me!
