mathschallenge.net

Problem

A 5 by 5 square lattice is formed by drilling holes in a piece of wood. Three pegs are placed in this lattice at random.

Find the probability that three randomly chosen points of a 5 by 5 lattice will form a triangle.

Solution

There are ²⁵C₃ = 2300 ways of picking three points from twenty-five. However, some of the sets will be collinear, and this can happen in a number of ways.

Along each vertical, horizontal, and main diagonal:
12⁵C₃ = 1210 = 120

Along the other diagonals:
4(⁴C₃+³C₃)=4(4+1)= 20

Then, we have twelve more:

2300 (120+20+12) = 2148, hence the probability of three random points forming a triangle will be 2148/2300 = 537/575 0.934.

What about a 6 by 6 lattice?
Although there is no general formula, investigate other sized lattices to find an algorithm.