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$N$ points chosen at random on the unit circle $x^2+y^2=1$. What is the probability that the center is enclosed by the convex polygon? Or rather the probability that the polygon formed by the $n$ points contains the origin? I of course tried the case for 3 points. This would mean both the second and third points have to be on opposite sides of the diameter formed by first point. I was trying to set up an integral over the arc length between the $n$ vertices but I got stuck.

Please help joriki especially

David Kleiman
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Randin
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  • Can we assume that the "convex polygon" refers to the convex hull of the $N$ points? Also: can you confirm that you are talking about the unit circle, not the unit disk? – A.G. Oct 15 '16 at 03:05
  • @A.G. He wrote $x^2 + y^2 = 1$ so it's the unit circle, not the disk. Also, since it is the circle, there's no need to take the convex hull. – David Kleiman Oct 15 '16 at 03:13
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    Here is the solution: https://books.google.ch/books?id=ywKyQz7_4-MC&pg=PA80&lpg=PA80&dq=random+points+on+a+circle+circumference&source=bl&ots=GOVGdB8gZQ&sig=K4wCaMt2RM2UaZ_uD3dWCGPZgL8&hl=de&sa=X&ved=0ahUKEwiUp_iR5tzPAhWB2hoKHUDAABAQ6AEIcTAJ#v=onepage&q=random%20points%20on%20a%20circle%20circumference&f=false – Christian Blatter Oct 15 '16 at 12:24
  • Yep x^2 +y^2=1 is the circle with center (0,0) not a " disk "! And yea it is convex hull – Randin Oct 15 '16 at 13:02

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