10. (20 points) Let $a, b, c, d$ be real numbers, and call $\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$ a $2 \times 2$ number table, and refer to $a, b, c, d$ as the elements of the number table. If $a+b, c+d, a+c, b+d$ are all distinct, then the $2 \times 2$ number table is called a "good number table". If the set of number tables where each element is equal to $-1, 0$ or 1 is denoted as $M$, and a number table is randomly selected from $M$, find the probability that it is a good number table.
10. Since each element in the number table $M$ has three different choices, there are a total of $3^{4}=81$ different number tables in $M$.
It is easy to see that, in a good number table, $a \neq d, b \neq c$.
Also, $a, b, c, d \in\{-1,0,1\}$, where two numbers are the same (let's assume $a=b$).
Clearly, $c \neq d$. Otherwise, $a+c=b+d$, which is a contradiction.
Therefore, $a, c, d$ are all distinct, i.e.,
$$
\{a, c, d\}=\{-1,0,1\} \text {. }
$$

If $a=0$, then $a+b=0=c+d$, which is a contradiction.
Thus, $a \neq 0$.
It is easy to see that
$$
\begin{array}{l}
(a, b, c, d) \\
=(1,1,0,-1),(1,1,-1,0), \\
\quad(-1,-1,0,1),(-1,-1,1,0)
\end{array}
$$

satisfy the conditions.
By symmetry, there are $4 \times 4=16$ good number tables in $M$.

Therefore, the probability of randomly selecting a good number table from $M$ is $\frac{16}{81}$.