1
$\begingroup$

Suppose you have $4$ randomly chosen points on the circle, then the probability that the center is contained is their convex hull is $1/2$. This is not hard to see, in fact the following holds:

If you choose $n$ point uniformly on a circle, then the probability that the center is contained in the convex hull is $1-n2^{1-n}$.

Proof: Suppose we clockwisely order the points on the circle $p_0,p_1,\ldots, p_n=p_0$, and denote by $L_i$ $0\leq i\leq n-1$ the corresponding arcs. Then the probability that the center is not contained in the convex hull is equal to the event at least one of the $L_i$ is at least half the perimeter. Then event $A_i=\{L_i$ is at least $\pi\}$, has probability equal to $2^{-n+1}$. And since $A_i\cap A_j=\emptyset$, we conclude. $\blacksquare$

Now, plugging $n=4$ we see that the probability is, indeed, $1/2$.

Question: In what other ways can we prove this result? I feel that there should be a more elegant argument to show this.

The previous result is not a coincidence, namely it generalizes for an $n-$sphere.

Theorem: If you choose $N$ point uniformly on $S^n$, then the probability that the center is contained in the convex hull is $2^{-N+1}\sum_{k=0}^{n-1}{N-1 \choose k}.$

Quote from Jen $``$ This problem is discussed in J. G. Wendel; A Problem in Geometric Probability, Mathematica Scandinavica 11 (1962) 109-111. Wendel showed, that the probability of $N$ random points lying on the surface of the unit sphere in dimension $n$ all lie on one hemisphere is

$2^{-N+1}\sum_{k=0}^{n-1} {{N-1}\choose k}$

I've found this here.$"$

Therefore, for $N=2n$ we see that $2^{-N+1}\sum_{k=0}^{n-1} {{N-1}\choose k}=1/2.$

Question: Can the proof for $S^2$ generalize in this case? In what other ways(easy?) can we prove this result ?

$\endgroup$
  • $\begingroup$ I think your first proof is elegant $\endgroup$ – Henry Jan 25 '18 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.