21 replies to this topic

[bonanova]

What is the length of a random chord drawn through a unit circle? There are at least four answers.

[Prof. Templeton]

Should we randomly chose the definition of random? I know our friend laid out three ways to draw a random chord while leaving out, arguably, the most intuitive way. I thought I remember reading that there were more ways than that even. Math is just a hobby for me so I am not immersed in it.

[brifri238]

I am thinking the average length of a random chord is pi/4 (0.7854) as a first thought.

[curr3nt]

Spoiler for I'm getting an average of...

pi / 2 * radius

based upon a range of distances from the center of the circle

Edited by curr3nt, 15 November 2012 - 04:25 PM.

[curr3nt]

Spoiler for interesting...

This took me to find Bertrand's paradox. Looks like my answer covers method 2. WIll try to calculate method 1 and 3. What's the fourth?

[Prof. Templeton]

Spoiler for interesting...

This took me to find Bertrand's paradox. Looks like my answer covers method 2. WIll try to calculate method 1 and 3. What's the fourth?

He left out from his three what I thought was the most intuitive way to make a chord

Spoiler for Namely

Two points are located randomly and independently in the interior of the circle. An extended straight-line connection of the two random points determines a random chord.

[CaptainEd]

He left out from his three what I thought was the most intuitive way to make a chord

Spoiler for Namely

Two points are located randomly and independently in the interior of the circle. An extended straight-line connection of the two random points determines a random chord.

Spoiler for Professor, I'm uncomfortable with the weight

My problem with the method of randomly choosing two points is that it seems to me to result in an estimate of the length squared.

Here's my argument. Choose two points in the interior, B and C. Extend to form the chord intersecting the circle at A and D. Now we have a length: the distance from A to D.

Notice, no matter what two points we could have chosen between A and D, we would arrive at the very same chord. So our overall average is going to be weighted by the probability of getting two points on chord AZ.

But, if we had chosen two points L and M resulting in a much shorter chord KN, it would have been weighted by the (smaller) probability of getting two points on chord KN.

So our overall average would incorporate the length of the chord twice, once for the length of the chord, and once for the probability of picking out that particular chord.

I would think that a "randomly chosen chord" would uniformly choose between chords, with each different chord having the same weight. Unfortunately, I can imagine (before looking at Bertrand's paradox) at least two ways to choose a chord randomly, and I still see differential weighting.

Both my suggested methods involving designating one point P on the circumference of the circle. After all, we know we want something that touches the circumference. It doesn't matter where, because it's a circle. Let's just start with one point. Now:
(method 1) uniformly choose an angle between 0 and 180, measured from a tangent to P. Compute the length of the chord, average over all angles.
(method 2) uniformly choose a point Q from the circumference of the circle. Compute the length of the chord, average over all points Q.
I assume they give different results. Sorry...:-(

[Prof. Templeton]

Spoiler for Professor, I'm uncomfortable with the weight

My problem with the method of randomly choosing two points is that it seems to me to result in an estimate of the length squared.

Here's my argument. Choose two points in the interior, B and C. Extend to form the chord intersecting the circle at A and D. Now we have a length: the distance from A to D.

Notice, no matter what two points we could have chosen between A and D, we would arrive at the very same chord. So our overall average is going to be weighted by the probability of getting two points on chord AZ.

But, if we had chosen two points L and M resulting in a much shorter chord KN, it would have been weighted by the (smaller) probability of getting two points on chord KN.

So our overall average would incorporate the length of the chord twice, once for the length of the chord, and once for the probability of picking out that particular chord.

I would think that a "randomly chosen chord" would uniformly choose between chords, with each different chord having the same weight. Unfortunately, I can imagine (before looking at Bertrand's paradox) at least two ways to choose a chord randomly, and I still see differential weighting.

Both my suggested methods involving designating one point P on the circumference of the circle. After all, we know we want something that touches the circumference. It doesn't matter where, because it's a circle. Let's just start with one point. Now:
(method 1) uniformly choose an angle between 0 and 180, measured from a tangent to P. Compute the length of the chord, average over all angles.
(method 2) uniformly choose a point Q from the circumference of the circle. Compute the length of the chord, average over all points Q.
I assume they give different results. Sorry...:-(

I was just pointing out that Bertrand left out that method for choosing a chord. However,

Spoiler for There are more ways to skin a cat

Pick a random point inside the circle and spin a random angle relative to the radius and pivoting on the point creating a chord.

[CaptainEd]

Indeed, more than one way.

Bonanova, what is the goal of this puzzle?
a) show us that the notion of "randomly chosen" depends on how you define it?
b) have us show lots of ingenuity in defining truly different ways to define it?
c) elicit from us the really-really-best way to define it?
d) all of the above
e) none of the above
f) other

[Caliban27]

I was just pointing out that Bertrand left out that method for choosing a chord. However,

Spoiler for There are more ways to skin a cat

Pick a random point inside the circle and spin a random angle relative to the radius and pivoting on the point creating a chord.

I agree with CaptainEd. Jaynes' solution to the Bertrand paradox shows that random radius (method 2) is the only robust theoretical means of choosing random chord.