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While trying to answer this question I realized that the probability for two points to lie on the same side of the line joining two other points is directly related to the probability for four points to form a convex quadrilateral. Since results are known for the latter for various distributions but I couldn't find any results for the former, I thought it might be useful to record the connection and transfer the known results for easy reference in the form of an answer here.

So let some distribution on the plane be given, for instance a uniform distribution over some region. If we know the probability that four points form a convex quadrilateral, how can we obtain the probability that two points lie on the same side of the line joining two other points (where all points are independently drawn from the given distribution)?

(Note that I'm asking for two particular points to lie on the same side of the line joining two particular points, not for any two out of four to lie on the same side of the line joing the other two.)

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1 Answer 1

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Given four points, the number of lines joining pairs of points that have the other two points on the same side is the number of edges of the convex hull of the four points. This is $4$ if the points form a convex quadrilateral and $3$ otherwise. Thus if the probability for the points to form a convex quadrilateral is $p$, the expected number of lines that have the other two points on the same side is $p\cdot4+(1-p)\cdot3=3+p$. Since all six lines have the same probability $q$ of having the other two points on the same side, this is $6$ times that probability, which is therefore $(3+p)/6=\frac12+\frac p6$.

The results for $p$ for uniform distributions over various regions given at MathWorld thus translate to the following results for $q$ (with "pentagon" and "hexagon" referring to regular polygons):

$$ \begin{array}{c|rl|rl} \text{shape}&p&&q&\\\hline \text{triangle}&\frac23&\approx0.66667&\frac{11}{18}&\approx0.61111\\\hline \text{parallelogram}&\frac{25}{36}&\approx0.69444&\frac{133}{216}&\approx0.61574\\\hline \text{pentagon}&\frac2{45}(18-\sqrt5)&\approx0.70062&\frac1{270}(171-2\sqrt5)&\approx0.61677\\\hline \text{hexagon}&\frac{683}{972}&\approx0.70267&\frac{3599}{5832}&\approx0.61711\\\hline \text{ellipse}&1-\frac{35}{12\pi^2}&\approx0.70448&\frac23-\frac{35}{72\pi^2}&\approx0.61741 \end{array} $$

Here's code to check the results for the parallelogram and ellipse numerically.

Since the value of $p$ for a uniform distribution over any open convex region with finite area is known to lie between the values for a triangle and an ellipse, the value of $q$ is analogously constrained. The length of the interval for $p$ is divided by $6$, so $q$ is known to lie in the narrow range

$$0.61111\lt q\lt0.61742\;.$$

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