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Q&A

What is the probability that the convex hull of $n$ randomly distributed points has $l$ of the points on its boundary?

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Consider a square in which $n$ points are uniformly randomly distributed. Now consider the convex hull of these points. The "length" of the convex hull is defined as the number of points in the perimeter of the convex hull i.e. $n \\; -$ the number of points strictly contained by the convex hull.

I conjecture that the probability that the convex hull has "length" $l$ is the same for all $3 \le l \le n$ as the points are randomly arranged in the grid. However, I’m unable to come up with a proof for this claim.

Is my conjecture correct, and if not, what is the correct probability distribution?

Note that the points do not lie on any grid - there are infinite possibilities for the arrangement of points and (almost surely) there will be no identical points.

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Would "population" be easier to understand than "length"? (1 comment)
Are two points allowed to be identical? Otherwise the case of four points on a 2×2 grid is an obvious... (9 comments)

1 answer

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Your conjecture is incorrect.

Consider the case of four points. Place three of these points randomly; these points form a triangle, and each pair of points defines a line. Consider the three half-planes bounded by each of these lines and containing the centroid of the triangle. If and only if a fourth point is contained in exactly two of these half-planes, all four points are on the boundary of their convex hull. (It's impossible to be excluded from all three half-planes, so there are three possibilities: if the point is included in only one half-plane, it along with the two points defining the line corresponding to that half plane form a triangle that contains the remaining point. If the point is included in all three half-planes, it is in the interior of the triangle. In the remaining case, by inspection of the four points, none of them are contained in the triangle defined by the other three.)

A Monte Carlo simulation shows that the probability of a fourth random point being contained in exactly two of the three half-planes is somewhat larger than 50%; it appears to be around 69%.

Finding the exact probability of this is more effort than I want to spend on this right now left as an exercise.

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