1. In the Cartesian coordinate system, there are eight points:
$$
A_{i}(i, 0), B_{i}(0, i)(i=1,2,3,4),
$$

Divide them into two groups, each containing four points, to form convex quadrilaterals $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. Then

the probability is $\qquad$
$-1 \cdot \frac{5}{12}$.
When and only when two points are taken from each of $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$ and $\left\{B_{1}, B_{2}, B_{3}, B_{4}\right\}$ to form a pair, two convex quadrilaterals are formed. There are $\left(\frac{4 \times 3}{2}\right)^{2}=36$ ways to do this.

Consider the case where $S_{\text {quadrilateral } B C D}=S_{\text {quadrilateral } A^{\prime} B^{\prime} C^{\prime} D^{\prime}}$. In this case, the first group is
$$
\begin{array}{l}
\left(A_{1}, A_{2}, B_{3}, B_{4}\right),\left(A_{1}, A_{3}, B_{2}, B_{4}\right), \\
\left(A_{1}, A_{4}, B_{2}, B_{3}\right),\left(A_{2}, A_{3}, B_{1}, B_{4}\right), \\
\left(A_{2}, A_{4}, B_{1}, B_{3}\right),\left(A_{3}, A_{4}, B_{1}, B_{2}\right) .
\end{array}
$$

Therefore, the required probability is $\frac{1}{2}\left(1-\frac{6}{36}\right)=\frac{5}{12}$.