Points on the Same Semicircle

Let us generalize this problem:

Say we have to choose $n$ points on the circumference such the lie within the same semicircle.

We can try fixing any of the $n$ points and then multiply by $n$ as total no. of fixings could be $n$ .

All of remaining $n-1$ points must lie within a semicircle around the fixed point,

the probability for a given point to lie within a semicircle $= \frac{1}{2}$ ,

Hence, for $n-1$ points , probability = $\frac{1}{2^{n-1}}$ .

Finally, multiplying by $n$ ,

Required probability = $\frac{n}{2^{n-1}}$

Here $n = 6$ , hence $P = \boxed{\frac{3}{16}}$

Suppose that there are $n$ points to be placed on the circle. Let $p_j$ be the probability that all $n$ points lie in the same semicircle, with the $j$ th point the "first" point occurring in that semicircle (going around the circle in an anticlockwise direction). This is just the probability that the other $n-1$ points lie in that semicircle defined with the $j$ th point at its start (going anticlockwise), and hence $p_j \; = \; 2^{1-n}$ It is clear that the desired probability is just (since the probabilities $p_1,p_2,\ldots,p_n$ represent disjoint events) $p_1 + p_2 + \cdots + p_n \; =\; n2^{1-n}$ which equals $\tfrac{3}{16}$ when $n=6$ , gving the answer $3+16=19$ .

Let $S= \{P_1, P_2, \cdots, P_6\}$ be a random selection of six points on the circumference of a circle. Let $O$ be the center of this circle. For convention we take $\angle P_1OP_1$ to be $0$ . Then any other point $P_i$ on the circumference will be uniquely described by the angle $\angle P_iOP_1$ . We then assign a coordinate $\theta_i$ to each of the $P_i$ s, which measures the angle $\angle P_iOP_1$ .

Note: For convention, we take the anticlockwise angles to be positive, and clockwise angles to be negative.

Note that each of the $\theta_i$ 's are either positive, or $0$ , or non-negative, but are pairwise distinct. Now, note that if $\theta_2>0$ , the necessary and sufficient condition for the six points to lie in the same semicircle is $\theta_3, \theta_4, \theta_5, \theta_6>0$ . Since the probability that each $\theta_i$ is positive is equal to the probability that each $\theta_i$ is negative (from symmetry), the probability of each $\theta_i$ being positive is $\frac{1}{2}$ . Thus, this case gives the probabiity $\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} \times \frac{1}{2}= \left ( \frac{1}{2} \right ) ^6= \frac{1}{64}$ Similarly, if $\theta_2<0$ , we get the same probability. Summing them up, we get the probability $\frac{1}{64} + \frac{1}{64}= \frac{1}{32}$ However, note that this is not the answer. We are taking the point $P_1$ as the measuring point. To count the total probability, we need to count the probability by taking each of the other $P_i$ s as the measuring points. By symmetry, this will just increase the answer $6$ times, so our desired answer is $6 \times \frac{1}{32}= \frac{3}{16}$ Which gives the sum $3+16=\boxed{19}$ .

Consider a simple circle with n random points on its circumference. Lets take any reference point. We can do this in C(n,1) ways. Now we are left with n-1 points. Let us, without loss of generality consider that these n-1 points lie on the semi-circle in clockwise sense to out reference points. Each point has a probability of 1/2 to lie in the clockwise circle or not, yielding us a probability 1/2^(n-1). Since we have a choice of C(n,1) reference points, our required answer is C(n,1)/2^(n-1) We shall substitute n=6 for this question.

I should come back and do this in Latex. But the proof is somewhat straightforward. For large n, given 2n points on a circle, so that any semicircle contains n points, there are (2n)^6 ways 6 points can fall on the circumference. Given an arc containing x points, and given that 2 points fall on its endpoints, there are x^4 ways for 4 points to fall on it, and thus (2n)(6)(5)x^4 ways for 6 points to fall on an arc containing x points. We add up all the ways 6 points can fall on an arc containing 0 to n points, and divide it by (2n)^6.

(2n)^(-6) (2n)(6)(5) ∫ (x = 0 to n) x^4 dx = 3/16

We can use this method for the general case of k points, the probability being k 2^(1-k) for all points falling on a semicircle.