Solution:

Answer: $\frac{1816}{6561}$

By Lucas' Theorem we're looking at
$$\prod_{i=1}^{4} \binom{a_{i}}{b_{i}}$$
where the $a_{i}$ and $b_{i}$ are the digits of $a$ and $b$ in base $3$. If any $a_{i}<b_{i}$, then the product is zero modulo $3$.

Otherwise, the potential residues are $\binom{2}{0}=1$, $\binom{2}{1}=2$, $\binom{2}{2}=1$, $\binom{1}{0}=1$, $\binom{1}{1}=1$, $\binom{0}{0}=1$.

So each term in the product has a $\frac{1}{3}$ chance of being zero; given that everything is nonzero, each term has a $\frac{1}{6}$ chance of being $2$ and a $\frac{5}{6}$ chance of being $1$. The probability that an even number of terms are $1$ given that none are zero is then given by the roots of unity filter
$$\frac{\left(\frac{5}{6}+\frac{1}{6} \cdot(1)\right)^{4}+\left(\frac{5}{6}+\frac{1}{6} \cdot(-1)\right)^{4}}{2}=\frac{81+16}{162}=\frac{97}{162}$$
Thus the expected value is
$$\left(\frac{2}{3}\right)^{4}\left(2-\frac{97}{162}\right)=\frac{1816}{6561}$$