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MagnetoHydroDynamics comments on Probability is in the Mind - Less Wrong

60 Post author: Eliezer_Yudkowsky 12 March 2008 04:08AM

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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:

  1. 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.
  2. 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.
  3. 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.


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