How many right angled triangles can a circle have? 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.
 A: Hint: You can choose $3$ points in ${2010}\choose {3}$ ways. To form a right angled triangle, we need to choose a pair of diametrically opposite points (how many ways to do this?) and then any third point will form a right angled triangle with the chosen pair. 
A: 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.
A: Let the points be $P_1$, $P_2$, and $P_3$.  Letting $P_1$ be arbitrary, we win if $P_2$ is the point diametrically opposite to $P_1$, probability $1/2009$.  Otherwise, with probability $2008/2009$, we still win if $P_3$ is diametrically opposite either $P_1$ or $P_2$, which it is with probability $2/2008$.  So the probability of a win is $1/2009 + (2008/2009)(2/2008) = 3/2009$.
You can also get this as follows:  The number of possible wins is the number of diameters times the number of remaining points, or 1005*2008. The number of possible triples is $2010\choose 3$. Dividing the first by the second gives $3/2009$.
Even easier:  The probability that a given pair of points lie on a diameter is $1/2009$.  With three points, you have three pairs, hence $3/2009$.  Here it's OK to add probabilities because it's not possible to have overlap; if one of the pairs lies on a diameter, no other pair can lie on a diameter.
BTW, I do not agree with the comment thread that says we must consider the possibility that the points are not distinct.  Sometimes there can be an ambiguity, but in common language "choose three points from 2010 points" means "without replacement".  For example, I am not incorrect in saying that $n\choose r$ is the number of ways of choosing $r$ objects out of a set of $n$.  Although I admit that if I were grading the test and this issue were pointed out to me, I might accept the "with replacement" answer as well.
The only other alternative answer I might accept is 1, i.e., that the points are certain to lie on a right triangle.  After all, it is Ford Prefect, so I have no way of knowing whether the Infinite Improbability Drive has been activated.
