4. In a square $A B C D$ with an area of 1, a point $P$ is chosen at random. Then the probability that the areas of $\triangle P A B$, $\triangle P B C$, $\triangle P C D$, and $\triangle P D A$ are all greater than $\frac{1}{6}$ is $\qquad$ .
4. $\frac{1}{9}$.

As shown in Figure 5, establish a rectangular coordinate system with $A$ as the origin and $AB$ as the $x$-axis.

Let $P(x, y), 0 < x < 1, 0 < y < 1$. Then, the conditions are:
$$
\left\{\begin{array}{l}
\frac{1}{2} x > \frac{1}{6}, \\
\frac{1}{2} y > \frac{1}{6}, \\
\frac{1}{2}(1 - x) > \frac{1}{6}, \\
\frac{1}{2}(1 - y) > \frac{1}{6},
\end{array} \text{ which is } \left\{\begin{array}{l}
\frac{1}{3} < x < \frac{2}{3}, \\
\frac{1}{3} < y < \frac{2}{3}
\end{array} .\right.\right.
$$

Therefore, the point $P$ that satisfies the given conditions must be within the square region enclosed by the lines $x=\frac{1}{3}, x=\frac{2}{3}$ and $y=\frac{1}{3}, y=\frac{2}{3}$.
Thus, the required probability is $\frac{\left(\frac{1}{3}\right)^{2}}{1^{2}}=\frac{1}{9}$.