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 Same side story (Posted on 2017-03-16)
Suppose n (n>1) points are placed at random on a circumference of a circle.
If P(n) denotes the probability that all n points lie on the same side of some diameter - find P(2), P(3) and P(4).

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Possible Solution | Comment 1 of 5
Obviously P(2)=1 since the points cannot be more than 180 degrees apart.

For P(3) coordinatize.  After placing the first two points, rotate the circle so that one of the points is on the x axis and the other is between 0 and 180.  The second point is now equally likely to be at any angle from 0 to 180.
Each position of the second point gives the third point some probability of being on the same side of some diameter.
The function turns out to be linear.
If the second point is at 0, the probability is 1.
If the second point is at 180, the probability is 1/2.
The allows the probability to be found by integration (finding the area under a graph.)
P(3) = 3/4

For P(4) do the same thing.  For each position of the second point, consider every possibility of the third point from 0 to 360 and then the probability that the fourth point works.
Surprisingly it's linear again.
0 -> 1*3/4 = 3/4
45 -> 7/8*5/7 = 5/8
90 -> 3/4*2/3 = 1/2
135 -> 5/8*3/5 = 3/8
180 -> 1/2*1/2 = 1/4
P(4) = 1/2

This answer surprised me so I'm wondering if I did something wrong. My solution makes sense on paper but is hard to explain in words.

P(5) looks to be much harder but maybe I'll try later.

 Posted by Jer on 2017-03-16 09:17:28

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