Triangle Sharing a Side with a Regular Polygon
Fix $n \geq 4$. You pick three distinct vertices uniformly at random from a regular $n$-gon and connect them to form a triangle.
What is the probability that at least one side of this triangle is also a side (edge) of the $n$-gon?
Compute the answer for $n = 10$.
Hints
- Since every subset of 3 vertices is equally likely, this reduces to counting favorable triangles over $\binom{n}{3}$.
- A triangle can share at most 2 sides with the polygon (exactly 2 means three consecutive vertices). Split the count into exactly 1 shared side and exactly 2 shared sides.
- For the 1-shared-side case, pick an edge ($n$ choices) then place the third vertex. How many vertices must you exclude so you don't accidentally create the 2-shared-side case?
Worked Solution
How to Think About It: Every set of 3 vertices out of $n$ is equally likely, so this is pure counting: favorable triangles over total triangles. The key observation is that a triangle can share 0, 1, or 2 sides with the polygon (never 3, since three consecutive edges of an $n$-gon with $n \geq 4$ don't close into a triangle). So count the triangles sharing exactly 1 side and exactly 2 sides separately, then add.
Quick Estimate: For $n = 10$, total triangles: $\binom{10}{3} = 120$. Triangles with 2 shared sides require 3 consecutive vertices -- there are 10 of those. Triangles with 1 shared side: pick one of 10 edges, then the third vertex can't be adjacent to that edge (or you'd get 3 consecutive), so you exclude 4 vertices, leaving $10 - 4 = 6$ choices, giving $10 \times 6 = 60$. Total favorable: $10 + 60 = 70$. Probability: $70/120 = 7/12 \approx 0.583$.
Approach: Count by cases -- triangles sharing exactly 2 sides and exactly 1 side with the $n$-gon -- then combine.
Formal Solution:
The total number of triangles is $\binom{n}{3}$.
*Case 1: Exactly 2 shared sides.* This requires three consecutive vertices of the polygon. There are exactly $n$ such triples (one for each starting vertex, cycling around).
*Case 2: Exactly 1 shared side.* Pick the shared edge: there are $n$ edges. The third vertex must not be adjacent to either endpoint of that edge (otherwise we'd get 3 consecutive vertices, which is Case 1). That rules out the 2 endpoints themselves plus the 2 vertices adjacent to the endpoints on the outside, so 4 vertices are excluded. This leaves $n - 4$ choices for the third vertex. So there are $n(n - 4)$ such triangles.
Adding both cases:
$$\text{Favorable} = n + n(n-4) = n(n-3)$$
The probability is:
$$P = \frac{n(n-3)}{\binom{n}{3}} = \frac{n(n-3)}{\frac{n(n-1)(n-2)}{6}} = \frac{6(n-3)}{(n-1)(n-2)}$$
Answer: For general $n$, the probability is $\dfrac{6(n-3)}{(n-1)(n-2)}$. For $n = 10$:
$$P = \frac{6 \cdot 7}{9 \cdot 8} = \frac{42}{72} = \frac{7}{12}$$
Intuition
This is a clean example of case-based counting on a symmetric structure. The polygon's cyclic symmetry means every edge and every vertex play the same role, so you can count configurations for one edge and multiply without worrying about overcounting asymmetries. The only subtlety is avoiding double-counting: when you count triangles with exactly 1 shared side, you must exclude the cases that would give you 2 shared sides (three consecutive vertices). That exclusion of 4 vertices is the step most people trip over -- they forget to remove the neighbors of the chosen edge's endpoints.
In practice, problems like this appear in combinatorial probability interviews to test whether you can organize a case split cleanly and handle the overlap between cases. The general formula $6(n-3)/((n-1)(n-2))$ also has a nice sanity check: as $n \to \infty$, the probability goes to $6/n$, which makes sense because each of the 3 sides of your triangle has roughly a $2/n$ chance of landing on a polygon edge, and for large $n$ these events are nearly independent.