Points are chosen uniformly at random on a circle until their convex hull contains the center.

Neat puzzle.  I didn't find an elementary solution, or one that
generalizes nicely to random points on a sphere or higher dimension;
but the following, though not elementary (it uses calculus), is at
least simple:

To find the expected number of points (which is of course the same as
the number of sides), we determine for each n the probability that the
polygon has at least n points, and sum this from n=1 to infinity.  That
probability, in turn, is the probability that the first n-1 points do
*not* enclose the center in their convex hull.

For that to happen, the n-1 points must be contained in an arc of
length 2*pi*t for some t<1/2.  Let 2*pi*x be the shortest arc
containing the points.  We assume that n>2, else the probability is 1;
we also assume that the 1st and 2nd points are the left and right
endpoints of the arc -- since in fact any pair of points can play this
role we'll multiply at the end by (n-1)(n-2) to obtain the correct
probability.

Now the chance that the 1st and 2nd point form an arc of length between
x and x+dx is dx, and the chance that a further point falls in that arc
is x+c for 0<c<dx.  Thus the probability we're after is the integral
from 0 to 1/2 of (n-1)(n-2)x^(n-3) or (n-1)/2^(n-2).  Note that this
formula also holds in the excluded case n=2 (but not for n=1).

Therefore our sum is 1 + (1/1 + 2/2 + 3/4 + 4/8 + 5/16 + 6/32 + ...)
= 1 + (1+1/2+1/4+1/8+1/16+...)^2 = 1 + 2^2 = 5.

-- Noam D. Elkies <elkies@math.harvard.edu>

The analysis of Noam Elkies can be simplified in the following way.
Let c(n,t) be the probability that at least n points are needed that
are not contained in an arc of length 2*pi*t. So the first n-1 points
are contained in an arc of length 2*pi*t. Lets call this probability
p(n-1,t).

For t<=1/2 we have:
  p(n,t) = p(n,t|1) + p(n,t|2) + ... + p(n,t|n) = n*p(n,t|1)
where p(n,t|i) is the probability that the first n points are
contained in an arc of length 2*pi*t with the left endpoint is the
i-th point. This is the case as
  p(n,t|i,j) = 0    (for i != j)
which is the probability that the first n points are in an arc
beginning at the i-th and at the j-th point. This is not the case for
t>1/2. We see that
  p(n,t|1) = t^(n-1)   for 0<=t<=1 and n>=1.
Therefore we have
  p(n,t) = n t^(n-1) = d/dt t^n    for t<=1/2 and n>=1
  p(0,t) = 1.

The Expectation for the number of points that are not contained in an
arc of length 2*pi*t (for t<=1/2) is:

E(t) = c(1,t) + c(2,t) + c(3,t) + c(4,t) + ...
     = p(0,t) + p(1,t) + p(2,t) + p(3,t) + ...
     = 1 + (t^0 + 2t^1 + 3t^2 + 4t^3 + ...)
     = 1 + d/dt (t^1 + t^2 + t^3 + t^4 ...)
     = 1 + d/dt 1/(1-t)
     = 1 + 1/(1-t)^2

The special case t=1/2 gives E(1/2) = 5.

Question: What is E(t) for t>1/2?
  E.g. E(2/3) is the expected waiting time until the polygon contains
  the central circle of half the diameter.

Discrete Version:
-----------------
The random process generates points at the vertices of a regular
k-gon.  Let p(n,k,t) be the probability that the first n points
generated for the k-gon are contained in t vertices in a series.  If we
define p(n,k,t|i) in the same way we will have more complicated
formulas as p(n,k,t|i,j) is not zero as the probability that the i-th
and the j-th point are equal is not zero.  Therefore define p(n,k,t|i)
as the probability that the first n points generated for k-gon are
contained in set of vertices {i, i+1, ..., i+t-1} (cyclic numbering)
and vertex i is selected.
  p(n,k,t|i) = (t/k)^n - ((t-1)/k)^n   independent of i.
Then we have for 2t - 1 <= k and n>=1
  p(n,k,t) = k p(n,k,t|1) = k ((t/k)^n - ((t-1)/k)^n)
  p(0,k,t) = 1.

The expected waiting time for 2*t - 1 <= k is therefore:

E(k,t) = p(0,k,t) + p(1,k,t) + p(2,k,t) + p(3,k,t) + ...
       = 1 + k \Sigma_{n>=1} (t/k)^n - ((t-1)/k)^n
       = 1 + k \Sigma_{n>=0} (t/k)^n - ((t-1)/k)^n
       = 1 + k ( \Sigma_{n>=0} (t/k)^n - \Sigma_{n>=0} ((t-1)/k)^n )
       = 1 + k ( 1/(1 - t/k) - 1/(1 - (t-1)/k) )
       = 1 + k^2/((k-t)(k-t+1))

[Editor: the summation index was written k instead of n in the four
sums above]

Now evaluate the special cases:

case k odd:
  E(2m + 1, m + 1) = 5 + 1/(m(m+1))

case k even: (the border is allowed)
  E(2m, m) = 5 - 4/(m+1)

case k even: (the border is not allowed)
  As in this case 2t - 1 <= k is not satisfied, the p(n,k,t|i,j) have
  to taken into account. In this case p(n,2m,m+1|i,i+m) have to be
  subtracted to get p(n,2m,m+1).  The final result is
  E(2m, m+1) = 5 + 7/(2m-2) - 1/((2m-2)(2m-1))

The special case E(k, k-1) is covered by the coupon collector problem
and can be checked in this way. (E(k,k) is infinity by definition.)

  E(2,1) = 2*H(2) = 3
  E(3,2) = 3*H(3) = 5 1/2
  E(4,3) = 4*H(4) = 8 1/3

where H(n) is the harmonic series 1 + 1/2 + 1/3 + ... + 1/n.

Question: What is E(k,t) for 2t - 1 > k?

-- Torsten Sillke <Sillke@mathematik.uni-bielefeld.de>