# Chords

• Dec 10th 2010, 09:51 AM
Veronica1999
Chords
Four distinct points A,B,C, and D, are to be selected from 1996 points evenly spaced around a circle. All quadruples are equally likely to be chosen. What is the probability that the chord AB intersects the chord CD?

My answer 1/5 but it's wrong.
The answer is 1/3. How could it be 1/3?

• Dec 10th 2010, 10:05 AM
TheEmptySet
Quote:

Originally Posted by Veronica1999
Four distinct points A,B,C, and D, are to be selected from 1996 points evenly spaced around a circle. All quadruples are equally likely to be chosen. What is the probability that the chord AB intersects the chord CD?

My answer 1/5 but it's wrong.
The answer is 1/3. How could it be 1/3?

Condiser the four points
$A,B,C,D$ The chord $AB$ intersects $CD$ if and only if they are arrainged around the circle as follow

$ACBD$ or $ADBC$

These simplest way way to get the probability is There are
$4!=24$ ways to arrange 4 letters
As show above there are two ways to have the chords intersect, but there are 4 different arrangements eg $ACBD \mapsto DACB \mapsto BDAC \mapsto CBDA$

You can do the same thing with the other to give 8. So the probability is

$\frac{8}{24}=\frac{1}{3}$
• Dec 10th 2010, 10:11 AM
Also sprach Zarathustra
It is 1/3.

Hint:
Compute:

(The total number of quads)/(The total number of intersecting quads) = ...