If three points are chosen at random on a circle's edge, what is the probability that the triangle contains the circle's center? 
If I created a triangle with 3 random points on the outside edge of a circle, then what’s the probibility that the triangle contains the centerpoint of the circle?

Please answer in as many ways as possible. I’m only in 8th grade. You can use calculus, because my math teacher said that’s how he would solve it; however, I only know a little bit of calculus, so I would also like alternatives.
Also, I'm sorry if my question was confusing. I was having trouble with the wording.
 A: The center will be included if the three points are not all within the same semicircle.  This question shows the chance they are within a semicircle is $\frac 34$ so the chance the center is inside your triangle is $\frac 14$
A: The triangle will fail to contain the center if and only if the three points are on the same side of some diameter of the circle.  For simplicity, let the circle have radius 1.  Now, if the first two points make an angle $\theta$, then you can see that the third point needs to be chosen on the larger of the two arcs between the first two points, and such that it make an angle at least $\pi-\theta$ with each of the first two.  This forces it to lie along an arc of length $\theta$ from among $2\pi$ of possibilities.  The initial $theta$ was chosen from among $\pi$ possibilities.  Thus, the probability is the average of $\frac{\theta}{2\pi}$ for $\theta$ uniform between $0$ and $\pi$. When $\theta$ is $0$, this is $0$ chance.  When $\theta$ approaches $\pi$ this approaches a $1/2$ chance, and it is an average of everything in between.  So the answer is 1/4.  (You can also do the corresponding integral and see that the answer is 1/4.)
