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??
 A: This is a basic Markov model problem; we can model the die as a state in the Markov model---i.e., the model is in state $x$ when player $x$ has the die. The transition matrix has entries on the +1 and -1 diagonals only, indicating right and left transitions between players, and they are based on the die roll probabilities, which are $\frac{1}{3}$ and $\frac{2}{3}$. We thus have a transition matrix $\mathbf{M}$:
$$
\mathbf{M} = \begin{pmatrix}
  0 & 1/3 & 0 & 0 & 0 & 2/3 \\
  2/3 & 0 & 1/3 & 0 & 0 & 0 \\
  0 & 2/3 & 0 & 1/3 & 0 & 0 \\
  0 & 0 & 2/3 & 0 & 1/3 & 0 \\
  0 & 0 & 0 & 2/3 & 0 & 1/3 \\
  1/3 & 0 & 0 & 0 & 2/3 & 0
\end{pmatrix}.
$$
The initial state is just a vector, $\mathbf{\pi}_0$, that indicates that the first player has a 100% probability of holding the die at the beginning:
$$
\mathbf{\pi}_0 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.
$$
The probability distribution of who has the die after $n$ rolls is simply $\mathbf{M}^n \cdot \mathbf{\pi}_0$ where the exponent on the matrix indicates that the matrix $\mathbf{M}$ should be matrix-multiplied by itself $n$ times.
$$
\mathbf{M}^5 \cdot \mathbf{\pi}_0 = \mathbf{M} \cdot \mathbf{M} \cdot \mathbf{M} \cdot \mathbf{M} \cdot \mathbf{M} \cdot \mathbf{\pi}_0 = \begin{pmatrix} 0 \\ 1/3 \\ 0 \\ 10/27 \\ 0 \\ 8/27 \end{pmatrix},
$$
so the player directly on the left of player 1 (the way I have setup the matrix) has a $1/3$ chance of having the die after 5 rolls, the person on the right of the first player has a 8/27 chance, and the person halfway between them has a 10/27 chance. The other players cannot possibly have the die after an odd number of rolls, which should be obvious.
Note that heropup is correct concerning the ordering problem; so I can't say definitively whether person $B$ is to the left or right, but this should show you how to solve the problem generally.
A: Has anyone noticed that the probability depends on whether $B$ sits to the left or the right of $A$, which is not specified?  This is because the probability of passing left is not equal to the probability of passing right.
Suppose for the time being that $B$ sits to the right of $A$.
If $p = 1/3$ is the probability of getting a multiple of 3 (i.e., a person rolls a 3 or a 6), and therefore is the probability of passing to the right (counterclockwise), then let $$X \sim \operatorname{Binomial}(n = 5, p)$$ be a binomial random variable that counts the random number of counterclockwise passes.  There are two cases:  either we observe $X = 3$ counterclockwise passes and $n-X = 5-3 = 2$ clockwise passes, in any order, or we observe $X = 0$ counterclockwise and $n-X = 5$ clockwise passes for the die to also arrive at $B$.  Then the desired probability is simply $$\Pr[X = 3] + \Pr[X = 0] = \binom{5}{3}(1/3)^3 (2/3)^2 + \binom{5}{0}(1/3)^0(2/3)^5 = \frac{8}{27}.$$
Now suppose $B$ sits to the left of $A$.
Then the desired probability, under the same notation, is now $$\Pr[X = 2] + \Pr[X = 5] = \binom{5}{2}(1/3)^2 (2/3)^3 + \binom{5}{5}(1/3)^5 = \frac{1}{3}.$$
Consequently, one cannot give a single answer to the question unless we are told on which side $B$ sits relative to $A$.
A: 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!
