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Guest Message by DevFuse

Points on a circle

Best Answer bubbled, 06 May 2014 - 05:47 AM

Spoiler for My take

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16 replies to this topic

#11 bonanova

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Posted 06 May 2014 - 11:33 AM

Spoiler for My take

It's similar to DeGe's result, which is the probability that a particular point is a terminus of the arc.
A very simple words-only description also gives the result.
In this case it shoul explain why simulating a line segment works also when the ends are joined (circle.)
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#12 bubbled

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Posted 06 May 2014 - 09:47 PM

Spoiler for My take

It's similar to DeGe's result, which is the probability that a particular point is a terminus of the arc.
A very simple words-only description also gives the result.
In this case it shoul explain why simulating a line segment works also when the ends are joined (circle.)

I thought long and hard about a words/logic only answer. So I'd love to see one. I'd find it much more interesting than my prosaic approach.

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#13 bonanova

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Posted 07 May 2014 - 12:05 AM

Spoiler for Words

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#14 m00li

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Posted 07 May 2014 - 12:21 AM

Spoiler for Words

Spoiler for I am not so sure..

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#15 bubbled

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Posted 07 May 2014 - 05:43 AM

Spoiler for Words

Spoiler for Sticking with my answer

Edited by bubbled, 07 May 2014 - 05:48 AM.

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#16 plasmid

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Posted 07 May 2014 - 06:52 AM

I agree with m00li's skepticism of the answer in words as it's phrased, but the formula's results agreed with my simulation. It looks like I used essentially the same approach as bubbled, but having two different programs reach the same conclusion makes it unlikely that it's from a bug.
Spoiler for output from the program

Spoiler for code

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#17 bonanova

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Posted 07 May 2014 - 12:57 PM

Apologies to you both.     I tried to think late last night, and from memory. The better (euphemism for accurate) thought solution follows:

Place a set of n points pi on a circle.

The basic event of the puzzle is Ei = the event that a (edit: clockwise) semicircular arc beginning at pi contains no pj where i <> j.

The events for each point are disjoint, they all have obvious probabilities of (1/2)n-1 and they may be summed.to obtain p = n/2n-1.

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