6. $\frac{16}{81}$.

It is easy to see that there are a total of $3^{4}=81$ different ordered quadruples $(a, b, c, d)$.

For a good quadruple $(a, b, c, d)$, we know that $a \neq d, b \neq c$, and with only three numbers to choose from, there must be two identical numbers.

Assume without loss of generality that $a=b$, then $c \neq d$. Thus, $b, c, d$ are all distinct, and $b \neq 0$ (if $b=0$, then one of $c+d, a+c, b+d$ must also be 0). Therefore, we have
$$
\begin{array}{l}
(a, b, c, d) \\
=(1,1,0,-1),(1,1,-1,0), \\
(-1,-1,0,1),(-1,-1,1,0) .
\end{array}
$$

Hence, there are a total of $4 \times 4=16$ good quadruples.
Therefore, the required probability is $\frac{16}{81}$.
6. Let $a, b, c, d \in\{-1,0,1\}$. If the ordered array $(a, b, c, d)$ satisfies that $a+b$, $c+d$, $a+c$, and $b+d$ are all distinct, then $(a, b, c, d)$ is called a "good array". The probability that an ordered array $(a, b, c, d)$ is a good array among all possible ordered arrays $(a, b, c, d)$ is $\qquad$