----- Original Message -----

From: asimovd@aol.com

To: math-fun@mailman.xmission.com

Sent: Wednesday, December 11, 2002 12:41 AM

Subject: Re: [math-fun] Thought up while drawing polygons ...

I found a math discussion site on the web where someone calling themself Poliwrath
posed and solved the question of what is the probability that if A,B,C,D are random points in a square, then AB and CD intersect?

Poliwrath's clever reasoning is as follows:  If X,Y,Z are any three random points in the unit square, let T be the expected area of the triangle XYZ.  (This just happens to be 11/144.)  Hence the probability of (A lying in BCD) = T, and same for B lying in ACD, C lying in ABD and D lying in ABD.  Since these 4 events are disjoint (!), the probability that one of them occurs is 4*T, and so the probability that none of them occurs, i.e. the probability that the convex hull of A,B,C,D is a quadrilateral, is 1-4T.
[snip]

 

I don't see that the events are disjoint. Consider A and B to be fixed and put C down somewhere. Now suppose D

is about as far from line AB as C is. Let D move directly toward AB. As it gets closer to ABC, C has a decreasing

chance of being inside ABD. Is there possibly some compensating effect? Even if there is, can you just say the

events are disjoint or does it have to be proved? Confusing.

 

Steve Gray