4. As shown in Figure 1, given that the two medians $A D$ and $B E$ of $\triangle A B C$ intersect at point $G$, we can obtain 8 figures:
$\triangle A B D$, $\triangle A C D$,
$\triangle B A E$, $\triangle B C E$,
$\triangle G A B$, $\triangle G A E$,
$\triangle G B D$, quadrilateral $C E G D$.
If two figures are randomly selected from these, the probability that the areas of these two figures are equal is . $\qquad$
4. $\frac{2}{7}$.

From 8 figures, choosing any 2 figures has $\frac{8 \times 7}{2}=28$ ways, among which there are three cases where the areas are equal:
(1) Triangles with an area of $\frac{1}{2} S_{\triangle A B C}$ are 4 $(\triangle A B D, \triangle A C D, \triangle B A E, \triangle B C E)$, resulting in $\frac{4 \times 3}{2}=6$ pairs of figures with equal area;
(2) Triangles with an area of $\frac{1}{6} S_{\triangle A B C}$ are 2 $(\triangle G A E, \triangle G B D)$, resulting in 1 pair of figures with equal area;
(3) Figures with an area of $\frac{1}{3} S_{\triangle A B C}$ are 2 $(\triangle G A B$, quadrilateral $C E G D$), resulting in 1 pair of figures with equal area.

In total, there are 8 pairs of figures with equal area. Therefore, the probability of picking two figures with equal area is $\frac{8}{28}=\frac{2}{7}$.