What is the probability that a quadrilateral is convex?

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.

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Please specify what you mean by randomly chosen points in the plane. (It's not an idle question.) –  cardinal Apr 30 '11 at 20:49

Since any three points will create a convex polygon (a triangle) you just have to worry about where the fourth point is. So you can place the first three vertices anywhere, then the condition for your path to be a poygon will just be that the fourth vertex has to be in the half plane delimited by the line from $x_1$ to $x_3$ which doesn't contain $x_2$. So you could say here that the probability is 50% (it depends on how you define your probability space as pointed our by @cardinal).

Now if you want to make it a convex polygon, you have to restrict this even more. The fourth point has to be in that half plane and in the region delimited by the lines $\overline{x_2x_1}$ and $\overline{x_2x_3}$. Now to make this a probability starts to be harder, you need to come up with a decent probability space for this. Most likely, by comparing areas, you'll get a probability which is ratio of space occupied by the half-cone of angle $x_1x_2x_3$ (if you count the area of the triangle $x_1x_2x_3$ to have a probability of 0.

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On your first paragraph, it will also be a simple polygon if $x_4$ is inside the triangle of the other three points, though it will not be convex. Similarly if $x_4$ is in the region beyond $x_2$ delimited by the lines $\overline{x_1 x_2}$ and $\overline{x_3 x_2}.$ –  Henry Apr 30 '11 at 22:35
That's right Henry, thank you for pointing this out. It means then that the probability is much higher than 50% to get a simple polygon, we also need to add the portion of the plane given by that angle. –  David Kohler May 1 '11 at 16:05

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.

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