Points $A$,$B$ and $C$ are randomly chosen from a circle, What is the probability that all the points are in one quadrant ($\frac{1}{4}$ circle)?

My try

Using this answer about semicircle, I tried to do the same for quadrant and I think the answer should be $\frac{3}{16}$. Is this correct? What are the other ways to find the answer?


More generally, let $n\in\Bbb N$ and $\alpha\in[0,\pi]$. Then we can compute the probability that $n$ random points $x_1,\ldots, x_n$ independently and uniformly distributed on a circle are in a common arc subtended by $\alpha$ as follows:

All points being in one $\alpha$-arc is equivalent to there being exactly one $k$, $1\le k\le n$, such that the $\alpha$-arc starting at $x_k$ contains all other $n-1$ points. For one $k$, this probability is $\left(\frac\alpha{2\pi}\right)^{n-1}$, hence in total the desired probability is $$ n\cdot \left(\frac\alpha{2\pi}\right)^{n-1}.$$ In particular, $n=3$ and $\alpha=\frac\pi4$ lead to $\frac3{16}$, as you proposed.

Remark. The above method does not work for $\alpha>\pi$ as in that case there is a non-zero probability that we have a choice in picking the "first" point of an $\alpha$-arc containing all points.

  • 1
    $\begingroup$ Then how to compute probability for $\alpha>\pi$? Is there any simple solution for ir? $\endgroup$ – Seewoo Lee Jul 1 '16 at 6:40

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