Semicircle Probability

I'll look at the general problem of finding the probability that $n$ points chosen at random on the circumference of a circle lie on a common semicircle.

First choose any one of the $n$ points. Now for each of the other $(n - 1)$ points the probability that they are within $\pi$ radians going clockwise of the chosen point is $\frac{1}{2},$ and so the probability that all the remaining $(n - 1)$ points lie on the (clockwise) semicircle that has the initially chosen point as the 'leftmost' point is $(\frac{1}{2})^{n-1}.$

Since the choice of the initial point was arbitrary, this result holds for each of the $n$ points on the circle, and since these 'events' are mutually disjoint, (i.e., we can't have two different points serving as the leftmost point on the clockwise semicircle on which all $n$ points lie), we can simply sum these $n$ probabilities to find that the probability that all points lie on the same semicircle is $n \times (\frac{1}{2})^{n-1}.$

In this case we have $n = 3,$ which gives us the solution $3 \times (\frac{1}{2})^{2} = \frac{3}{4} = \boxed{0.75}.$

First, fix a point on the circumference of the circle, call it $P$ now, draw a diameter through the circles center $C$ and call the other end point of the diameter $Q$ .

Now, place a second point $S$ anywhere on the circle and draw a diameter with $S$ as one end point and $T$ as the other.

Label angle $\angle SCP$ as $\theta$ . Hence $\angle TCQ$ also equals $\theta$ .

Now, the third point can't be on minor arc $TQ$ . So the probability that it's not on that arc is

$\dfrac{2\pi-\theta}{2\pi}=1-\dfrac{\theta}{2\pi}$ .

Note that theta varies between $0\leq \theta <\pi$

Using probability distribution function, we have

$\dfrac{1}{\pi} \displaystyle \int_0^{\pi} 1-\dfrac{\theta}{2\pi}=\frac{3}{4}$

A nice geometric argument is this....let x,y,z be the proportions of the circle lying between the three points.

If you plot $(x,y,z$ in 3-space you get a point uniformly distributed in the region $x+y+z=1$ with $x,y,z\geq0$ which geometrically is an equilateral triangle with vertices at i, j and k.

The three points do not lie on the same semicircle if $x,y,z<\frac{1}{2}$ which is the equilateral triangle with vertices $\frac{i}{2},\frac{j}{2},\frac{k}{2}$ which has area one quarter of the original triangle. (Easy to see in a diagram, it's just the middle quarter if you split original triangle by joining midpoints of adjacent sides)

Let there be $x$ points on a semicircle.

Now we have to find out the probability to have all the points on a semicircle which will be equal to 1 - {probability of having no three points in same semicircle}

Probability of selecting 1 point on the semicircle = 1/2

Probability of selecting 2 points on another semicircle = $x(x-1)/2x(2x-1)$

This selection can be done in two ways so the probability of selecting three points that do not lie on a semicircle = $2/2 * x(x-1)/2x(2x-1) = x-1/2x-1$

There fore the probability of having all the 3 points on a semicircle = (1- (x-1/2x-1) = 3x-1/4x-1 Now as x tends to infinity the limit comes to be 0.75.

Fix the first point and choose the second point and then find the probability that the third is on a semi-circle.

Extreme Case I: The first two points are opposite each other. In this case the probability that the third point is on a semi-circle with the first two points is 1.

Extreme Case II: The first two points are the same point. In this case the probability that the third point is on a semi-circle with the first two points is $\frac{1}{2}$ .

Assuming the probability is linearly distributed, the overall probability is the mean of 1 and $\frac{1}{2}$ , which is $\frac{3}{4}$ .

Consider the circle having a circumference of $2$ , then let's put the first to point two points having a distance $k$ between them with $k$ between $0$ and $1$ . Then for $k$ chosen randomly between $0$ and $1$ the probability that the other point is in the same semicircle is $(1+k)/2=3/4$ .

This is the complementary problem to the following: break a stick in two places - what is the probability that you can form a triangle with the three pieces that result?