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JohnWilliams
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Dots on the unit circle

September 25th, 2007, 1:13 am

3 dots are randomly drawn on the unit circle with uniform probability (i.e. eual probability of being anywhere on the circle).What is the probability that they all lie in the same hemisphere?
 
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sevvost

Dots on the unit circle

September 25th, 2007, 2:26 am

QuoteOriginally posted by: JohnWilliams3 dots are randomly drawn on the unit circle with uniform probability (i.e. eual probability of being anywhere on the circle).What is the probability that they all lie in the same hemisphere?What does a hemishpere have to do with a circle?Perhaps you might be interested to take a look here: http://www.wilmott.com/messageview.cfm? ... adid=41363?
 
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iwanttobelieve
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Dots on the unit circle

September 28th, 2007, 6:54 pm

sevvost, a circle is a 2d-sphere...
 
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Advaita
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Dots on the unit circle

October 1st, 2007, 4:04 pm

Can't you always choose a semi-sphere?Construction:Draw a "line" from the center of the sphere to each of the three points. You have a "cone" made of three lines with the tip at the center of the sphere.It is easy to see that this cone will AlWaYs be "acute" if you look at it right. Then you can choose your semi-sphere. (Ignore the zero-content set of 3 points on the big circle.)So P = 1.
Last edited by Advaita on September 30th, 2007, 10:00 pm, edited 1 time in total.
 
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PaperCut
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Dots on the unit circle

October 1st, 2007, 9:20 pm

How, exactly, will this help me make money in Corn options this month?
 
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LegoLasVegas
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Dots on the unit circle

October 2nd, 2007, 11:33 am

0.75 is correct.
Last edited by LegoLasVegas on October 1st, 2007, 10:00 pm, edited 1 time in total.
 
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Traden4Alpha
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Dots on the unit circle

October 2nd, 2007, 12:29 pm

QuoteOriginally posted by: PaperCutHow, exactly, will this help me make money in Corn options this month?Perhaps you have some fancy-pants FFT trading system that resolves price action as the phases of the cycle. You believe that you may profit if the phases of three price series are gravitating to the same half of the 0-2*Pi phase circle but wonder what is the probability of that occurrence under the assumption that the phases are mutually independent. Using the answer, 3/4, gives you the null hypothesis for testing whether the observed phases really are spending a non-random amount of time being colocated in the same hemisphere.
 
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Advaita
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Dots on the unit circle

October 2nd, 2007, 6:03 pm

Sorry, for the weird reasoning, I read: "probability of 3 points on a sphere that can be contained in a hemi-sphere." Apologies.But here's a contructive proof for the actual problem I think you'll like:Take the 3 points --Look at the possible displacements. (Displacement = moving the point across the diagonal to its mirror image spot on the other side)As there are 3 mirror image spots to be filled (one for each point) -- there can be 2^3 = 8 displacements.So for instance: for (0,120,240) -> (0,120,240), (180,120,240), (0,300,240), ... ,(180,300,60)You will notice that there are exactly 2 displacements when the semi-circle does not fit.Thus for any configuration, 6 out of 8 displacements will fit on a semi-circle.So prob = 6/8 = 3/4EXCEPTIONs:When the points are EXACTLY on a semi-circle, this does not hold -- however we can safely ignore these points as they are zero-content (or something to that effect)
 
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Marsden
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Dots on the unit circle

October 3rd, 2007, 4:43 pm

I get 7/8.
 
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Marsden
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Dots on the unit circle

October 3rd, 2007, 4:49 pm

No; it's 3/4.
 
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PaperCut
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Dots on the unit circle

October 3rd, 2007, 10:49 pm

QuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: PaperCutHow, exactly, will this help me make money in Corn options this month?Perhaps you have some fancy-pants FFT trading system that resolves price action as the phases of the cycle. You believe that you may profit if the phases of three price series are gravitating to the same half of the 0-2*Pi phase circle but wonder what is the probability of that occurrence under the assumption that the phases are mutually independent. Using the answer, 3/4, gives you the null hypothesis for testing whether the observed phases really are spending a non-random amount of time being colocated in the same hemisphere.heh hehnice work
 
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amit7ul
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Dots on the unit circle

October 4th, 2007, 6:58 am

let the centre of unit radius circle be at (0,0). we can fix A, without loss of generality at (1,0).if B is at (cos t, sin t) with OB making angle t radians with +ve x-axis then required odds=1-2*integral(0,pi) t dt /(2*pi)^2=3/4 (same as 029's solution)this question was actually whats the probability[max(A,B,C)]<=90 degreesa more general problem and easily solvable problem could be probability[max(A,B,C)]<=X degrees
 
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tuchong
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Dots on the unit circle

October 11th, 2007, 2:01 pm

Probability of not forming a circle is:\int_0^2\pi \phi/so, the answer is 1-0.25 = 0.75
 
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Wilbur
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Dots on the unit circle

October 11th, 2007, 5:44 pm

Point 1 will be anywhere.Point 2 can be anywhere between directly across (180 from point 1) or directly on top of point 1 with equal prob.If point 2 is directly across, point 3 will be in a semicircle with prob. 1If point 2 is directly on top, point 3 will be in a semicircle with prob. 0.5(1 + 0.5)/2 = 0.75