Ants on a Triangle

There are three ants on a triangle, one at each corner. at a given moment in time, they all set off for a different corner at random. what is the probability that they don’t collide?

Solution

Consider the triangle ABC. We assume that the ants move towards different corners along the edges of the triangle.

Total no. of movements: 8 A->B, B->C, C->A A->B, B->A, C->A A->B, B->A, C->B A->B, B->C, C->B A->C, B->C, C->A A->C, B->A, C->A A->C, B->A, C->B A->C, B->C, C->B

Non-colliding movements: 2 A->B, B->C, C->A A->C, B->A, C->B

(i.e. the all ants move either in the clockwise or anti-clockwise direction at the same time)

P(not colliding) = 2/8 = 0.25

2026 Update: Symmetry, Probability, and Collision Problems

Ants-on-a-triangle problems (each ant chooses a direction randomly — what is the probability no two ants collide?) test probability over symmetric configurations. The key insight: ants collide if and only if they don’t all move in the same direction. For a triangle, P(no collision) = 2/2^3 = 1/4.

General formula: For n ants on an n-gon, each choosing clockwise or counterclockwise with probability 0.5, P(no collision) = 2 / 2^n.

def prob_no_collision(n_ants):
    """Probability that n ants on an n-gon don't collide."""
    # Only 2 configurations out of 2^n work: all clockwise or all counterclockwise
    return 2 / (2 ** n_ants)

for n in [3, 4, 5, 10]:
    print(f"{n} ants: P(no collision) = {prob_no_collision(n):.6f}")

The deeper insight — expected number of collisions: Use linearity of expectation. The expected number of colliding pairs is n * P(one pair collides) = n * (1 – P(they move apart)) — calculable without enumerating all configurations.

Modern application: Collision-avoidance probability calculations appear in distributed systems (two writes to the same key simultaneously), robotics (multi-robot path planning), and network packet collision analysis in CSMA/CD protocols. The symmetry argument is the same in all cases.