
Re: Random Triangle Problem
Posted:
Jul 24, 1997 2:09 PM


On 22 Jul 1997 15:56:59 GMT, tony richards <tony.richards@rl.ac.uk> wrote:
>mathwft@math.canterbury.ac.nz (Bill Taylor) wrote: >>Mike Housky <mike@webworldinc.com> writes:
>>> > Derive a method of finding the chances of 3 random points placed on >>> > a plane that will yield an obtuse triangle.
>without loss of generality, we can assume point 1 is at the origin with probability =1 >so P1(anywhere)=1.0.
>Without loss of generality, point 2 can be on the x axis at either x>0 (probability 0.5) >or at x<0 (probabilty = 0.5). > SO P2(x>0)=0.5 and P2(x<0)=0.5
>Considering only the xcoordinate of point 3, x3
>You get an ACUTE triangle if x3 is BETWEEN x1 and x2
Only if the point is not within the circle of which the segment P1P2 is a diameter.
>I propose that it is equally likely that x3>x1 as x3<x1 (probability = 0.5 each) >and also that it is equally likely that x3>x2 as x3<x2 (probability=0.5 each) >So, the probability that x3 lies between x1 and x2 will be the >product of the probabilities 0.5 and 0.5, one factor drawn from each of the above pairs > giving P=0.25 for x3 between x1 and x2.
Even assuming for the sake of argument that there's a reasonable probability measure floating around here somewhere, this calculation seems unjustifiable: the events in question aren't independent. But the whole approach seems unworkable. If Pr(x3 < x1) = Pr(x3 < x2) = 1/2, then (assuming that x1 < x2) Pr(x1 < x3 < x2) = 0. The plane is just too big.
Brian M. Scott

