Comment author:[deleted]
21 July 2013 03:08:34PM
*
2 points
[-]

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.

## Comments (188)

Old*2 points [-]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:

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