A triangle is formed by picking $3$ points at random from the vertices of a regular $2n+1$-gon. Find the probability for it to be an equilateral triangle.

Question

ganeshie8 · Answer

for first few values of $n$ im getting  below probabilities :

n = 1, probability = 1
n = 2, probability = 0
n = 3, probability = 0

thomas5267 · Answer

My guess would be for 3n-gon, $P=\dfrac{n}{\binom{n}{3}}$ and P=0 for all other regular polygon.

ganeshie8 · Answer

Thats right! could you share how you arrived at that

ganeshie8 · Answer

pretty sure you meant $P=\dfrac{n}{\binom{\color{red}{3}n}{3}}$

thomas5267 · Answer

I am absolutely confused now lol.  I thought the answer would be $\dfrac{3n}{\binom{3n}{3}}$.

For all regular polygon of not having 3n vertices, rotation of 120° would not constitute a valid symmetry.  So that restricts us to 3n-gon.  For 3n-gon, there are $\binom{3n}{3}$ choices of 3 vertices.  Now I have to check my numerator!

thomas5267 · Answer

Ah yes, $\dfrac{n}{\binom{3n}{3}}$ is indeed the correct answer as there are n equilateral triangle in 3n-gon.

ganeshie8 · Answer

Looks nice, there are just some repetitions to be accounted for..