Say you have a circle. On this circle you draw the inscribed equilateral triangle.
Simple, right?
Okay. For a random chord in this circle, what is the probability that the chord is longer than the side in the triangle?
So, to choose a random chord, there are three obvious methods:
Pick a point on the circle perimeter, and draw the triangle with that point as an edge. Now when you pick a second point on the circle perimeter as the other endpoint of your chord, you can plainly see that in 1⁄3 of the cases, the resulting chord will be longer than the triangles’ side.
Pick a random radius (line from center to perimeter). Rotate the triangle so one of the sides bisect this radius. Now you pick a point on the radius to be the midpoint of your chord. Apparently now, the probability of the chord being longer than the side is 1⁄2.
Pick a random point inside the circle to be the midpoint of your chord (chords are unique by midpoint). If the midpoint of a chord falls inside the circle inscribed by the triangle, it is longer than the side of the triangle. The inscribed circle has an area 1⁄4 of the circumscribing circle, and that is our probability.
WHAT NOW?!
The solution is to choose the distribution of chords that lets us be maximally indifferent/ignorant. I.e. the one that is both scale, translation and rotation invariant (i.e. invariant under Affine transformations). The second solution has those properties.
Say you have a circle. On this circle you draw the inscribed equilateral triangle.
Simple, right?
Okay. For a random chord in this circle, what is the probability that the chord is longer than the side in the triangle?
So, to choose a random chord, there are three obvious methods:
Pick a point on the circle perimeter, and draw the triangle with that point as an edge. Now when you pick a second point on the circle perimeter as the other endpoint of your chord, you can plainly see that in 1⁄3 of the cases, the resulting chord will be longer than the triangles’ side.
Pick a random radius (line from center to perimeter). Rotate the triangle so one of the sides bisect this radius. Now you pick a point on the radius to be the midpoint of your chord. Apparently now, the probability of the chord being longer than the side is 1⁄2.
Pick a random point inside the circle to be the midpoint of your chord (chords are unique by midpoint). If the midpoint of a chord falls inside the circle inscribed by the triangle, it is longer than the side of the triangle. The inscribed circle has an area 1⁄4 of the circumscribing circle, and that is our probability.
WHAT NOW?!
The solution is to choose the distribution of chords that lets us be maximally indifferent/ignorant. I.e. the one that is both scale, translation and rotation invariant (i.e. invariant under Affine transformations). The second solution has those properties.
Wikipedia article)