Cool problem! (1 Viewer)

[Carrotsticks]

Normally, this would go into the 'Extracurricular Mathematics' section, but I figured this section would be more appropriate because (my method at least) requires knowledge of the Catalan Numbers (http://en.wikipedia.org/wiki/Catalan_number) unless you derive the result of it from scratch.

The question is as follows:

Consider a circle of any radius (does not matter). We select two randomly and independently selected points on the circumference and as a result, we have a random chord. If R random chords are drawn, show that the probability that NONE of them intersect is....

 

[HongKongPhooey]

After trying to bash it with double, triple, quadruple, etc integrals for a bit, I've found this solution. Maybe its the same as yours. For the case r chords, there are 2r points on the circle. Now we pick one of these to be one of the end points of a chord, then there are (2r-1) choices for the other end point. For the next chord, choose any of the points as an endpoint, then there are (2r-3) choices for the other end point. Continue this way to see that altogether given the 2r points, there will be (2r-1)(2r-3)...1 = (2r)!/((2^r)r!) ways of drawing the chords.

Now consider the points 1,2,...,2r. Suppose a chord was given by (i,j) for some 1 <= i < j <= 2r. Then any other chord must be (k,l) with: i < k < l < j; or 1 <= k < i and j < l <= 2r; or 1 <= k < l < i; or j < k < l <= 2r. It's clear then that there's a one to one correspondence between drawing chords so that none of them intersect and writing balanced strings of brackets like (())() - given the chord (), the cases correspond to, adding a pair of matching brackets, inside, surrounding the first, on the left, and on the right, respectively. This number of ways of drawing the r chords so that none intersect is then the r-th Catalan number, (2r)!/((r+1)!r!). Dividing this by (2r)!/((2^r)r!) gives the result.

 

[Carrotsticks]

The question actually was solved a while ago by a 2013'er. Will try to find later on.

 

[HongKongPhooey]

Oh okay, just saw no solution here so thought I'd have a go at it.