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 Eight Points (Posted on 2016-08-24)
Eight points are selected at random from the interior of a unit circle.

What is the probability that these eight points constitute the eight vertices of a convex octagon?

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Just for curiosity, random points in a square Comment 3 of 3 |
If the random points were anywhere in a square rather than a circle the probability would be lower (about 1 in 137 rather than the 1 in 114).

76 10000 .0076 done
52 10000 .0052 done
45 10000 .0045 done
79 10000 .0079 done
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65 10000 .0065 done
73 10000 .0073 done
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65 10000 .0065 done
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74 10000 .0074 done
75 10000 .0075 done
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76 10000 .0076 done
89 10000 .0089 done
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89 10000 .0089 done
79 10000 .0079 done
61 10000 .0061 done
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79 10000 .0079 done
80 10000 .008 done
78 10000 .0078 done
91 10000 .0091 done
71 10000 .0071 done
77 10000 .0077 done
67 10000 .0067 done
86 10000 .0086 done
52 10000 .0052 done
71 10000 .0071 done
62 10000 .0062 done
85 10000 .0085 done
73 10000 .0073 done
81 10000 .0081 done
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77 10000 .0077 done
66 10000 .0066 done
71 10000 .0071 done
78 10000 .0078 done
62 10000 .0062 done
77 10000 .0077 done

That's 3658 hits in 500,000 trials for about 1 in 137 or 0.73%.

Results from commenting out the requirement that points be within the unit circle:

' Do
xp = 2 * Rnd(1) - 1
yp = 2 * Rnd(1) - 1
' Loop Until xp * xp + yp * yp < 1

 Posted by Charlie on 2016-08-25 09:34:10
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