6. Among the 10 vertices of a regular decagon, if any four points are chosen, the probability that the quadrilateral formed by these four points is a trapezoid is $\qquad$
6. $\frac{2}{7}$.

Let the regular decagon be $A_{1} A_{2} \cdots A_{10}$.
(1) With $A_{1} A_{2}$ as the base, there are 3 trapezoids: trapezoid $A_{1} A_{2} A_{3} A_{10}$, trapezoid $A_{1} A_{2} A_{4} A_{9}$, and trapezoid $A_{1} A_{2} A_{5} A_{8}$.

Similarly, there are 3 trapezoids each for $A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{5}, \cdots, A_{9} A_{10}$, and $A_{10} A_{1}$ as the base.
Thus, there are a total of 30 trapezoids.
(2) With $A_{1} A_{3}$ as the base, there are 2 trapezoids: trapezoid $A_{1} A_{3} A_{4} A_{10}$ and trapezoid $A_{1} A_{3} A_{5} A_{9}$.

Similarly, there are 2 trapezoids each for $A_{2} A_{4}, A_{3} A_{5}, A_{4} A_{6}, \cdots, A_{9} A_{1}, A_{10} A_{2}$ as the base.
Thus, there are a total of 20 trapezoids.
(3) With $A_{1} A_{4}$ as the base, there is only 1 trapezoid: trapezoid $A_{1} A_{4} A_{5} A_{10}$.

Similarly, there is 1 trapezoid each for $A_{2} A_{5}, A_{3} A_{6}, A_{4} A_{7}, \cdots, A_{9} A_{2}, A_{10} A_{3}$ as the base.
Thus, there are a total of 10 trapezoids.
In summary, the required probability is
$$
P=\frac{30+20+10}{\mathrm{C}_{10}^{4}}=\frac{2}{7} \text {. }
$$