Bertrand, creatively

[bonanova]

What is the average length of a random chord constructed on a unit circle? I'm looking for creative definitions of random that produce the greatest and the smallest answers.
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[MNOWAX]

A unit. why? It's halfway between 0 and 2 units.

Go ahead. disprove that.
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MNOWAX

[The Potter]

Funny... I was thinking sqrt(2) because it is halfway.
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[Zag]

Creative definition of random:

1. Choose any point on the circle.
2. Choose any point on the intersection of the circle and the line that passes through the first point and the center of the circle, and isn't coincident with the first point.

My average chord is pi!

edit: I meant 2, of course. pi would be the arc.

Do I win for the random function that produces the maximum?

[The Potter]

chord length as a function of theta is 2sin(theta/2).

If we integrate over the range of 0 to 2π

-4cos(theta/2) |(2π and 0) = 8

For an average of 8/2π
=1.2732
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[Zag]

You've made an assumption about what "a random chord" means: that you choose two points on the circle where every point on the circle has an equal chance of being selected.

I realize that my definition of "a random chord" above was silly, but I do think it would be reasonable to define it as:

Choose two points within a circle such that every point in the circle has equal probability of being selected, and draw the line which connects them. Where that line intersects the circle defines the chord. I don't know what the average length will be, but I can trivially show that it is longer than the one you've chosen.

The random function you chose is the same as saying, "Choose two points within a circle such that every point in the circle has equal probability of being selected, draw the two radii which pass through these two points. The two points where the radii intersect the circle define the chord." Note that this chord is always smaller than the one that my random function makes, given the same two points within the circle selected.

[The Potter]

As far as I know he was seeking a cord on the unit circle. Lets pretend there are just 10 points equally spaced around the circle. The first point will be at 0 (we will define it that way). The second point is equally likely to fall on any of 0 to 9. There is no reason for me believe any number is more likely then the next.

Then the average cord length can be found by summing the lengths and dividing by the number of points. With 10 discrete points the average cord length will be 1.2628.

0
0.618033988749895
1.17557050458495
1.61803398874990
1.90211303259031
2
1.90211303259031
1.61803398874990
1.17557050458495
0.618033988749895
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[Thok]

The minimum is 0. This uses the xkcd random function: randomly choose a point, and then whenever somebody asks for a chord, pick the chord connecting that point to itself.

[Zag]

The Potter, you've missed my point entirely. I agree that your integral solution was correct, for the problem that you were solving. However, I don't agree that you're solving the right problem. The definition of 'a random chord on a unit circle' is not as clear-cut as you seem to think.

I could say, "choose a random integer from 0 to 10" and you might assume that I meant to choose such that each of the 11 integers has the same chance of being selected. But you satisfy the requirements by rolling two dice and subtracting 2 from the result, making a result of 5 far more likely than 0 or 10. Similarly, there are many ways to choose your random chord, and the different selection processes will yield different results. Some of these selection processes (like my first one and like Thok's) are ridiculous if you want to maintain the term "random" but some are perfectly reasonable.

[The Potter]

The chord is constructed on the unit circle...
So both the points we care about have to fall on the unit circle. How far apart they are is only a function of the angle they form with the center point.
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[The Potter]

Ah, I see what you are saying now. But I don't feel that is a very good way of (I guess that is part of the point I missed) cord construction for random numbers. A picture will clear things up
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[bonanova]

[LordKinbote]

[L'lanmal]

Consider any method of choosing a random chord on a circle, where every chord has a non-zero probability of being selected. Observe that this has a (not necessarily well-behaved) density function over the space of chords in a circle.

Now consider some method of choosing a random chord on a circle (every chord has a non-zero probability of being selected) whose pdf is equal to the original scaled by the chord length of each outcome and renormalized. Alternatively, scale by (2 - chord length).

Iterate.

It seems like you can generate methods (where every chord has a non-zero probability of being selected) which have an average chord length arbitrarily close to 2. Alternatively, 0.

I'm guessing you'd want one that is easily described.

[L'lanmal]

Obviously individual chords can have a 0 probability of being selected. I meant that distribution function is non-zero on that chord.

[bonanova]

[Nsof]

removed a non creative idea
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[Jack_Ian]

Choose two chords A & B.
Toss a coin to select the appropriate chord.
Select initial chords depending upon whether you want a large or small final average.

[Zag]

Ah ha! I've just realized that I can win this game easily, at least for longest average chord length.

Step 1. Choose any process for randomizing which chord you choose such that there is a non-zero chance of choosing a diameter.
Step 2. Draw your circle around a black hole.

Your average chord length is infinity!

[Nsof]

choose a point outside the circle.
there are two tangent lines from that point to the circle.
the "random" chord is the one connecting the two points where the tangent lines intersect the circle.

depending on how you "randomly" choose that point, you can get different average results.
e.g.: if unit circle is centered on origin and point is defined as (n,0) where n, an integer, is uniformaly taken from [1,1000] the average result would be close to two.
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[groza528]

[Amb]

What happens if you draw a circle on a sphere? Then either side of the line, is "inside" the circle depending on your perspective. Could you think shrink the circle to small, and increase the sphere to infinite? Then you could get interesting large and small chords?

[jadesmar]