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](https://tio.run/##fVFNb4MwDL3nV1g9EbqqkHCq1B2nHXufdqB8DCYIUQha8uuZnUI1OAwJzHt5frZj7W0zKDnPtRl6MLkqMbS9HoxdEGNlVUOTd/Wty1UVufQFPL5OYMTXYeAXBviUcAUXR17ACSUcjuBShIREQAJRSpAHfY76p/pcBu5OnCOREytXBE7EIRct/fPEVHYyCrq8v5c5OIu92AvksbNY7h57CgVjKkWHNEkSpsT6x@phMrehVfZ96roR6cAZaKFVNPtXFak0TOYo/XEbEXXud9iJ3fkOO7k732ASFMvUdEf0kfwsSRh4T7wn3i@8pgb@WUmBsaBKWmyFq0BilBuh3ApXweq8CFFJV/T954rEsnyXbYfE7rPdmABtDePURzpyGRpmHF4hCY6aHD80ltLYn5afHK5XEA9rKrrZ1RF3yJg2SESHN5MXth0UDDXYarQj/LS2geykKQEayrgc@CLfOp1xxRADDsHn@Rc) 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 <s>more effort than I want to spend on this right now</s> left as an exercise.
r~~
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2022-11-21T02:55:16Z (over 1 year ago)