Number of isosceles triangles in a regular heptagon > If any three vertices are chosen from $7$ sided regular polygon (a heptagon), then find the probability that the chosen vertices form an isosceles triangle. **My attempt:** The total number of ways $\displaystyle n(S) = \binom{7}{3}$. The number of favorable cases $\displaystyle n(A) = 7$ (in which two sides are common with a heptagon). How can I calculate other isosceles triangles?

There are seven vertices you can consider as the top of three possible isosceles triangles. For instance, if the vertices in circular order are called $X_1, X_2, \ldots, X_7$, the triangles $\Delta X_1X_4X_7$, $\Delta X_2X_4X_6$ and $\Delta X_3X_4X_5$ are all isosceles triangles. As such, the probability that a chosen triangle is an isosceles one equals:

$$\frac{{7 \choose 1} {3 \choose 1}}{7 \choose 3} = \frac{21}{35} = \frac{3}{5}$$