Further Question on " What is the probability that the center of the circle is contained within the triangle? " Q: Consider the triangle formed by randomly distributing three points on a circle. What is the probability of the center of the circle be contained within the triangle?
This question was raised by joriki in Jan 17 '13 at 21:47 and had been answered with: $1/4$
Based on this answer, I wish to know if there have a total numbers of random triangles in a circle or it is just tends to infinity.
 A: Since a circle shows symmetry, we can fix one point on the circumference. Next, we take the 2nd point on the circle anywhere else except the same point and diametrically opposite point. Now, the points which can be the third point form the arc which is symmetrically opposite to the arc joining our two points.
Thus, for a given two points, the probability is the arc length divided by circumference. Since the expected distance between our two points is $1/4$ times the circumference i.e. midway between first point and diametrically opposite point, the answer is $1/4$.
A: There is another way to identify a probability of $1/4$, by geometry alone.
Consider any three points $A,B,C$ such that no two are on a common diameter.  We can pair this arrangement with another, equally probability-dense arrangement by swapping $A$ for the diametrically opposite point $A'$, and we can do the same with $B$, $C$ or any combination of the three points.
Then we have eight equally probability-dense arrangements, four of which are congruent with the other four ($180°$ rotations).  Of the four noncongruent arrangements one will have only acute angles when the triangle is drawn and the other three will have an obtuse angle (which would be exchanged for an acute angle, leaving both other angles acute, by swapping the appropriate vertex).  Ergo one outbid four arrangements has the center inside the triangle.
