# Probability that 3 vertices of a 2n+1 sided polygon chosen at random form vertices of an isosceles triangle

Consider a $$(2n+1)$$ sided regular polygon. Find the probability that three vertices chosen at random form the vertices of an isosceles triangle.

My Attempt

If I choose $$3$$ vertices containing two sets of $$r$$ consecutive sides then the triangle so formed by the $$3$$ chosen vertices(i.e. the $$2r$$ sides are all consecutive) is clearly isosceles. So number of ways to do so will be $$\sum_{r=1}^{n}r=\frac{n(n+1)}{2}$$So required probability$$=\frac{\sum_{r=1}^{n}r}{\binom{2n+1}{3}}$$.

Is it correct.

Next, we drop an axis of symmetry at the point. For each of the remaining $$2n$$ points, it suffices to pick a single vertex() from one side of the axis of symmetry(the other vertex will be determined my reflecting across the axis of symmetry).
The number of ways to choose 2 points at random frorm the remaininig points is $$\binom{2n}{2},$$ which gives an overall probability of $$\boxed{\frac{n}{\binom{2n}{2}}}.$$