TITLE: What is the probability that a quadrilateral is convex?
QUESTION [7 upvotes]: Given $4$ distinct randomly chosen points $x_1$, $x_2$, $x_3$, and $x_4$ in the plane such that the polygonal path from $x_1$ to $x_2$ to $x_3$ to $x_4$ to $x_1$ describes a non-self-intersecting quadrilateral, what is the probability that the quadrilateral is convex?
This question is prompted by the observation that, although arrowheads, darts and chevrons are not rare, the corresponding quadrilateral is seldom mentioned/encountered in the study of elementary geometry.

REPLY [3 votes]: Assuming the distribution of your points are identical, independant and uniform in some convex body $K$, this is Sylvester's four-points problem.
The probability $p_4$, that the $4$ points are the vertices of a convex quadrilateral, depends on the convex body $K$. For example on the link pointed  above you will find:

if $K$ is a disk, $p_4=1-35/(12\pi^2)$,
if $K$ is a triangle,  $p_4=2/3$,
if $K$ is a square,  $p_4=25/36$,

and more results.