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Random Chords

Problem

A random chord is drawn by connecting two independent randomly selected points on the circumference of a circle.

If r random chords are drawn, find the probability that none of them intersect.


Solution

First we shall consider the number of ways in which r chords can be connected. Imagine a circle with n=2r points that are numbered sequentially 1 to n around the circumference of the circle.

The "first" chord is connected to 1 and (nminus1) other possible end points.
The start point of the "second" chord will be connected to the next free point: 2 unless first chord connected from 1 to 2. It will be able to connect to (nminus3) other end points.
In a similar way, the other chords will be able to connect to (nminus5), (nminus7), ..., 5, 3, 1 other end points.

Therefore, the number of ways of drawing r chords, D, is given by,
D = (nminus1)(nminus3)...5.3.1 = (2rminus1)(2rminus3)...5.3.1 = (2r)!/(r! 2r)

Next we shall consider the number of ways that no chords can intersect.

This diagram can be represented by the bracket string: 1(2(3)4(5(6)7)8), or more concisely as, ( () ( () ) ), with a left bracket signifying a start point and a right bracket signifying an end point.

For n = 2r points there will be r pairs of brackets, meaning there are C(2r,r) different bracket strings in total that can be formed. However, not all strings will represent non-intersecting chords; for example, () )( () does not.

By considering the number of non-intersecting bracket strings, C, for the first three cases of  r, we get:

r=1: C=1   ()
r=2: C=2   () (), ( () )
r=3: C=5   () () (), () ( () ), ( () )(), ( () () ), ( ( () ) )

It turns out the number of non-intersecting chords are the sequence of Catalan numbers: 1, 2, 5, 14, 42, ..., and C = C(2r,r)/(r+1). For completeness we shall derive this from first principles.

The trick to counting the number of brackets that correspond to non-intersecting chords is to subtract the number of "bad" bracket strings from the total number of bracket strings. We note that "bad" bracket strings occur when, working from left to right, the number of right brackets exceeds the number left brackets. For example, () )( ().

The first time we encounter a right bracket out of place, we shall invert every bracket thereafter: left becomes right and right becomes left. For example, the "bad" string, () )( (), becomes, () )) )(.

The result of this exercise is that we end up with (rminus1) left brackets and (r+1) right brackets, and as each one of these strings corresponds to a unique "bad" bracket string, we can determine that there are C(2r,r+1) "bad" bracket strings in total.

Hence there are C = C(2r,r)minusC(2r,r+1) non-intersecting chords.

C = C(2r,r)minusC(2r,r+1) = 
(2r)!

r! r!
minus 
(2r)!

(rminus1)!(r+1)!
  = 
(r+1)(2r)!

(r+1) r! r!
minus
r(2r)!

(r+1) r! r!
  = 
(2r)!

(r+1) r! r!
  
  = 
C(2r,r)

r+1
  
therefore P(no intersecting chords) = C/D = 
(2r)!/((r+1) r! r!)

(2r)!/(r! 2r)
  = 
(2r)!/((r+1)! r!)

(2r)!/(r! 2r)
  = 2r/(r+1)!
Problem ID: 195 (21 Dec 2004)     Difficulty: 4 Star

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