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. Among the 24 equally likely ways that A,B,C,D can be assigned to the corners of any given quadrilateral, just 1/3 of them involve AB interesecting CD (since wherever A is, B can be adjacent with prob. = 2/3, or opposite, with prob. = 1/3). This shows that P = [the probability AB and CD intersect] is (1-4T)/3. Assuming T = 11/144, this gives P = 25/108. (I don't know the calculation giving T = 11/144, but both this figure and the 25/108 are well approximated by my computer simulations, so I believe them.) --Dan
----- 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
----- 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. Among the 24 equally likely ways that A,B,C,D can be assigned to the corners of any given quadrilateral, just 1/3 of them involve AB interesecting CD (since wherever A is, B can be adjacent with prob. = 2/3, or opposite, with prob. = 1/3). This shows that P = [the probability AB and CD intersect] is (1-4T)/3. Assuming T = 11/144, this gives P = 25/108. (I don't know the calculation giving T = 11/144, but both this figure and the 25/108 are well approximated by my computer simulations, so I believe them.) --Dan If this reasoning is true (see my other post, sent about 1 minute ago), it's interesting that it seems to be powerless in the 3-D case of determining the chance that 6 ordered connected segments form a knot. Does anyone have any ideas on that?
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asimovd＠aol.com -
Steve Gray