In trying to solve the classic problem —
"Find the probability that three randomly chosen points on a circle provides an acute triangle"
I came across this page that seems to have a good explanation.
However, I do not understand how they came up with the probability as
$\int_{0}^{\pi }{\frac{1}{\pi }\cdot \frac{\theta }{2\pi }\cdot d\theta }$
I understood that the probability is the length of sector of the circle between the two dotted lines divided by the total circumference, but do not see how/why the integral is needed.
Best Answer
I answered a more general form of this problem here. Think of the integral as the sum of all the angles you can have, and the division as averaging it out. Just as $\frac{1}{n}\sum_{i=1}^n f(i)$ is an average, so is $\frac{1}{\pi}\int_0^\pi f(\theta)\mathrm d\theta$. So, in this case, you want to take the average of all the $\theta$ you can have, but there are an infinite amount of these, so how do you take an average? By this integral, of course!
Once you have an average $\theta$, you need to recognize that the length of the arc the third point can occupy in order to form the triangle is exactly $\theta$, the length of the minor arc between the first two points. The probability of landing in this arc is $\theta/2\pi$, and thus your integral.