Probability of four randomly choosen points A, B, C, D on a circle making line segments that cross each other?

Question

Line segment AB crossing line CD.

Answer 1

Dear curvature,

If the density for choosing the points on the circle is uniform, then the probability that the two chords intersect is 1/3.

To see this, let v be the smaller of the two angles formed at the circle's center O between A and B, i.e., v is the angle AOB, where 0 <= v <= pi. Then the probability that C will fall on the shorter arc between A and B is v / (2 pi), and the probability that C will fall on the longer arc between A and B is 1 - v / (2 pi). We have the same two probabilities for the location of D with respect to the shorter and longer arcs between A and B.

Now we observe that chords AB and CD will intersect if one of C or D falls on the shorter arc between A and B, and the other falls on the longer arc between A and B. Thus, the probability of C being on the shorter arc and D on the longer is
[v / (2 pi)] [1 - v / (2 pi)]. By symmetry, the probability of D being on the shorter arc and C on the longer is identical. This let's us write the probability of the two chords intersecting, given the angle v.

P(AB intersects CD | v ) = 2 [v / (2 pi)] [1 - v / (2 pi)]
= 2 [v / (2 pi)] [(2 pi - v) / (2 pi)]
= 2 v (2 pi - v) / (4 pi^2)
= v (2 pi - v) / (2 pi^2) .

With v having uniform density of 1 / pi for 0 <= v <= pi, we can integrate out v to get the unconditional probability that the two chords intersect.

P(AB intersects CD) = int[0,pi] P(AB intersects CB | v) (1 / pi) dv
= int[0,pi] v (2 pi - v) / (2 pi^2) (1 / pi) dv
= [1 / (2 pi^3)] int[0,pi] v (2 pi - v) dv
= [1 / (2 pi^3)] int[0,pi] (2 pi v - v^2) dv
= [1 / (2 pi^3)] [2 pi v^2 / 2 - v^3 / 3] | [0,pi]
= [1 / (2 pi^3)] [pi^3 - pi^3 / 3]
= [1 / (2 pi^3)] [2 pi^3 / 3]
= 1/3 .

Answer 2

If points A and B are next to each other the line segments do not cross, but if they have C between them on one side and D between them on the other the line segments do cross.

Put the four points on the circle. There are two ways A and B can be adjacent and only one way they are not. Therefore the probability that the line segments AB and CD cross each other is 1/(1 + 2) = 1/3.

Answer 3

A brute tension sampling of triangles formed with vertices (x,y), chosen uniformly from [-a million,a million], yet rejected while radius to beginning exceeded a million, for one hundred,000 samples gave answer .40038. while vertices have been chosed with uniformly random radius from [0,a million] and theta from [0, 2pi], answer for one hundred,000 strengthen into .46441. So i've got self belief this determination concerns.

Answer 4

100% if you connect A to C or B or D whatever it takes to make to intersect the other two.

Answer 5

there is a 50 50 chance.good luck!!!

Answer 6

P(AOC < AOB) = 0.5
P(AOB < AOD) = 0.5
P(AOC > AOB) = 0.5
P(AOB > AOD) = 0.5
P(AOC < AOB) ∩ P(AOB < AOD) U P(AOC > AOB) ∩ P(AOB > AOD) = 0.5*0.5 + 0.5*0.5 = 0.5