[algogeeks] Re: Interesting Probability Question
On 3/9/07, Nat (Padmanabhan Natarajan) [EMAIL PROTECTED] wrote: 1. We cannot bound the triangle if we don't bound the space...thats the reason why I choose a unit square I think we don't really need to bound anything. I think the question as is phrased can only yield what I guess should be called _limiting_ probabilities. ;-) 2. It is true that there are a lot of points outside the triangle that you cannot choose but they all lie in a finite set of lines Again, I think this is incorrect. Please refer to my argument in my previous post. -- Regards, Rajiv Mathews --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~--~~~~--~~--~--~---
[algogeeks] Re: Interesting Probability Question
I am not sure if I am thinking right about this problem but here is my approach 1. Hull is not a quadrilateral if at least 3 points are collinear to begin with 2. Pick two random points, draw a line through them 3. Pick the third point, the probability of the third point lying on the line = 0 (bound - not exact) 4. Now join these three points to get a triangle, again the hull is not quadrilateral if it lies within this triangle. Now the probability is 1 - (area of triangle)/(area of Sq.) I think that we don't really lose anything by considering the plane as a unit square 5. The question now is what is the expected area of a random triangle inside a square. I would think it is half the size of the largest that could be inscribed. Any takers? Cheers, Nat On 3/6/07, Rajiv Mathews [EMAIL PROTECTED] wrote: I came across this interesting probability question recently. Consider an infinite two dimensional plane. 4 points are chosen at random on this plane. What is the probability that the convex hull of the 4 points will be a quadrilateral? -- Regards, Rajiv Mathews --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~--~~~~--~~--~--~---
[algogeeks] Re: Interesting Probability Question
why is it 1 - (area of triangle)/(area of Sq.)? why do we need a square since what would happen is that the 4th point can be anywhere in the space but the area of the triangle is bounded. The probability of choosing a point outside the triangle would be 1 (bound - not exact by reasoning as in (3)) and that would lead to a probability of 1. If the bound reasoning is correct then we would have a probability of 1 to get a quadrialetral when choosing random points in a plane. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~--~~~~--~~--~--~---
[algogeeks] Re: Interesting Probability Question
On 3/8/07, Karthik Singaram L [EMAIL PROTECTED] wrote: If the bound reasoning is correct then we would have a probability of 1 to get a quadrialetral when choosing random points in a plane. The bound reasoning is valid under the condition that size of plane tends to infinity. As I've posted above this still doesn't mean that the probability of getting a quadrilateral hull tends to 1. -- Regards, Rajiv Mathews --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~--~~~~--~~--~--~---
