TITLE: How many right angled triangles can a circle have?
QUESTION [5 upvotes]: Here's what I recall of the question from CNML Grade 11, 2010/2011 Contest #3, Question 7:

There are 2010 points on a circle,
  evenly spaced. Ford Prefect will*
  randomly choose three points on the
  circle. He will* connect these points
  to form a shape. What is the
  probability that the resulting shape
  will* form a right angled triangle?


I answered $\frac{1}{4} = 25\%$, but that's probably incorrect. (Right?)
When I got home, I thought it out in my head, and I got this:
$\frac{2010 * (2010/1005)}{2010 \choose 3}$
$\frac{2020050}{1351414120} = \frac{3015}{2017036} = 0.149476756984010201\%$
I'm probably wrong ...again. Can anyone tell how to get the right answer (if I'm not wrong :) )?
*in the past of the future of the perfect present present time double into ripple fluctuater byer doininger of the past future continuum...
EDIT: Realized my mistake in copying the question.

REPLY [2 votes]: Simplify it. There are 2010 points, but if you start with four points instead and pick 3 at random, you see the probability of getting a right triangle is 1 (draw it out, it helps). You can do it with six points too, and if you look at the probability for both figures you can create a formula to give the probability regardless of the number of points. 3/(n-1), where n is the number of points (I think it has to be even for this to work). With four points, 3/(4-1) = 1, six points 3/(6-1) = 3/5. If you have 2010 points then the probability would be 3/(2010-1) = 3/2009.