## Probability to get a specific triangle

 Quote by kai_sikorski It's the problem I pointed out to you before. In my diagram when B goes past π/2, you're changing the 'good' region from (π,π + B) to (2π-B,2π) but that's not right. Even when B goes past π/2, the 'good' region is still (π,π + B).
Oh I see it now. When the angle goes past pi/2 I can NOT just flip my orientation and consider the complementary acute angle. The acute region is always the angle in the positive direction ranging from zero to pi between the first and second points. So now I can take the average of a random angle between 0 and pi to get pi/2 as the average acute region, leaving 3pi/2 as the obtuse region.

Fun problem.
 Thank you Kai. I got it backwards. The probability of an obtuse angle is 0.75, acute is 0.25. I did find an easier way. There is an obtuse angle iff the arc length between any two points is greater than one half of the circumference. So the problem is equivalent to picking two points from the interval [0,1], thus subdividing the interval into 3 sub-intervals, and finding the probability that any sub-interval has length >1/2. The result is as Kai found.
 Hi guys. Thanks very much for the thoughts! A variety of ideas is already presented here, so I don't have much to add besides this little experiment when I simulated 1,000,000 events (circle was simplified to 1,296,000 points) and evaluated the largest angle of every triangle: The largest angle is a decisive factor for knowing the type of a triangle. I found it interesting, because this interval of 90-180 degrees (meaning an obtuse triangle) is three times as big as interval of 60-90 degrees (meaning an acute triangle) while the ratio of P(obtuse) and P(acute) is also exactly three. So I wanted to see how the likelihood of those angles is distributed.

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