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Strike a Chord (..Any Chord) (Posted on 2003-10-09) Difficulty: 4 of 5
What is the probability that a randomly drawn chord will be longer than the radius of the circle?

Prove it.

No Solution Yet Submitted by DJ    
Rating: 4.5263 (19 votes)

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Faith or Reason--no one will come to a final conclusion. | Comment 16 of 51 |
No one is going to come to a conclusion as to the "best" way of randomizing a chord.

In support of (√3)/2, the following comes from mathworld:
The latter interpretation is more satisfactory in the sense that the result remains the same for a rotated circle, a slightly smaller circle inscribed in the first, or for a circle of the same size but with its center slightly offset. Jaynes (1983) shows that the interpretation of "random" as a continuous uniform distribution over the radius is the only one possessing all these three invariances.

In addition to the cosmic ray argument presented in my previous post, I'd also give an example of a plane that is criss-crossed with random lines. A circle is then placed upon the plane and is intersected by a number of lines. The expected proportion that would form chords longer than the radius would be the (√3)/2. Those pre-existing random lines would be just like the random cosmic rays.

By the way, most random number generators are pseudo-random, actually being deterministic. But radiation, detected by a geiger counter, actually is random, so those cosmic rays do really hit randomly, having originated in quantum events.

As Aaron has pointed out, by taking two random points on the circumference, you are overcrowding chords that are at nearby endpoints, and overseparating them at far reaches.

While Federico Kereki claims that 2/3 is reached two ways, actually, choosing two points at random on the circumference is not really different from choosing a central angle--they are really the same thing, so of course the probability is the same.

I, however, do not understand DJ's objection to Bryan's answer:
"However, the analysis also depends on the assumption that any given range of angles gives an equal probability of random selection. That is to say, you need to assume the the chance of getting between 0 deg and 10 deg is the same as the chance of getting between 80deg and 90deg. How can you prove that this is true?"

Well, one can't prove that one's selection method is true. One just decides on a selection method. In this particular case someone has chosen a uniform distribution of angles (arclengths) from which to select a chord.
  Posted by Charlie on 2003-10-10 08:59:06
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