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What is probability that $3$ random point in $2n$ reguler polygon contains center of gravity

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I contain the diagram to make up my explanation. Left one is contain the center of gravity,while right one doesn't.

$$\binom{2n}{n}$$ is the whole case and I found that when $n=1$,Probablity $P_{1} =$ Not defined.And $n=2$,$P_{2}=1$. I also get $P_{3}=\frac{7}{10}$ and $P_{4}=\frac{4}{7}$, but I can't certain about this.

Also I found $$\lim_{n\rightarrow\infty}P_{n} =\frac{1}{4}$$

Plz help me to find $P_n$

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  • $\begingroup$ made me think about this : youtube.com/watch?v=OkmNXy7er84 probably not that related but good video though $\endgroup$ – P. Quinton Sep 4 '18 at 14:28
  • $\begingroup$ @P.Quinton Yes I think about this from that vidio~!!!!. But I can't do my own to do this...... I tried about to get all cases using symmetry, It's value is very awkward.... $\endgroup$ – user366725 Sep 4 '18 at 14:30
  • $\begingroup$ What if the center of gravity lies on the border of the triangle? Does that count as contained? $\endgroup$ – Mike Earnest Sep 4 '18 at 14:41
  • $\begingroup$ @MikeEarnest Yes, that's why I wrote that $P_2=1$ $\endgroup$ – user366725 Sep 4 '18 at 14:42
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In order for the three points to not contain the center in their convex hull, two of the points must be at most $2\pi\cdot\frac{(n-1)}{2n}$ radians anti-clockwise from the third point. There are $2n$ possibilities for the location of the third point, and for each of these are $\binom{n-1}2$ ways to choose two points from the $n-1$ points in that arc. Therefore, $$ 1-P_n = \frac{2n\binom{n-1}{2}}{\binom{2n}3}=\frac{3(n-1)(n-2)}{(2n-1)(2n-2)}=\frac{3(n-2)}{2(2n-1)}, $$ $$ P_n = \frac{n+4}{4n-2}. $$

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