Expected number of non-empty regions Consider $d$ dimensional space cut by $n$ hyperplanes in general position, each one of which goes through the origin. The number of distinct regions created is known to be:
$$2\sum_{i=0}^{d-1} {n -1 \choose i}.$$
Let us assume the hyperplanes are created uniformly at random.  Now we sample $k$ points uniformly at random on the surface of the unit sphere centred at the origin.  How many distinct regions, on average will have at least one point in it?
 A: Here are the first two cases (both with $d=2$). I mention them and the (fairly simple) method in the hope that some more subtle analysis will yield more results. I don't know if one must derive the distribution of region sizes but, if so, that is non-trivial in itself even when $d=2.$
For $d=2$ and $n=1,$ the unit circle is split in two equal halves. The probability that all $k$ points are in the same half (as the first one) is $\frac{1}{2^{k-1}}.$ So the expected number is $$2-\frac{1}{2^{k-1}}.$$
For $d=2$ and $n=2,$ I get an expected number of $$ 4-\frac{8}{k+1}+\frac{4}{2^k(k+1)}=4\left(1-\frac{1}{2^{k+1}}\right)\left(1-\frac{2}{k+1}\right)-\frac{1}{2^{k-1}}  .$$
I'm not sure what the near factorization adds, but there it is. 
Analysis: Note: I did this first ,but was reassured when a simulation with $100000$ trials and $k=6$ gave close to the predicted values. The circle is split into four pieces of (relative) sizes $\alpha,\frac{1}{2}-\alpha,\alpha,\frac{1}{2}-\alpha$ with $\alpha$ uniformly distributed in $[0,\frac{1}{4}]$
So the probability that $k$ points all land in the same piece is
 $$p_1=\int_0^{1/4}\alpha^k+ (\frac{1}{2}-\alpha)^k+\alpha^k+(\frac{1}{2}-\alpha)^kd\alpha=\frac{4}{(k+1)2^k}.$$ 
There are six ways to pick two regions: four of combined size $\frac{1}{2}$ and one each of sizes $2\alpha$ and $1-2\alpha.$ Accordingly, the probability that all $k$ points land in exactly two regions is 
$$p_2=\int_0^{1/4}(2\alpha)^k+ 4(\frac{1}{2})^k+(1-2\alpha)^kd\alpha-3p_1=\frac{2}{k+1}+\frac{4}{2^k}-\frac{12}{(k+1)2^k}.$$
The subtraction is to compensate for the fact that an event that all the points actually land in one region  gets counted $3$ times by the integral as being in the union of two regions.
There are only two possible sizes for the combined length of $3$ regions leading to $$p_3=\int_0^{1/4}2(\frac{1}{2}+\alpha)^k+2(1-\alpha)^k d\alpha-2p_2-3p_1=\frac{4}{k+1}-\frac{8}{2^k}+\frac{8}{(k+1)2^k}$$ and $$p_4=1-p_1-p_2-p_3=1-\frac{6}{k+1}+\frac{4}{2^k}.$$ 
Then $p_1+2p_2+3p_3+4p_4$ gives the expected number.
For $k=6$ and $100,000$ trials this would predict frequencies of $$[893,32143,46429,20536]$$ (everything rounds up) and a simulation gave $$[924,32079,46445,20552].$$

For arbitrary $d,n$ the regions come in $R=\sum_{i=0}^{d-1} {n -1 \choose i}$ sizes with two of each size. A (highly?) nontrivial question is what the distribution function (over some $R-1$ simplex) is. For $d=2$, $R=n$ and for $d=3,$  $R=\frac{n^2-n+2}{2}.$ Perhaps for $n=3$ or $4$ iterated integrals as above could be pushed through. For $d=2$ and large $n$, one has a Poisson process.
A: Here is an answer for $d=2$ and general $n.$ Perhaps it simplifies, but I don't immediately see how. The approach might be more effective for  $d \gt 2$ than integrating over a high dimensional simplex. 
$$\frac{n}{2^k}\sum_0^k\binom{k}{\ell}\left(\frac{\ell}{n+\ell-1}+\frac{k-\ell}{n+k-\ell-1}\right).  $$  I don't immediately see how to simplify that. The reasoning seems valid and agrees with the answer for $n=2.$
One peculiarity of the problem is that we have regions whose sizes come from some distribution, but we have two of each size. Filtering out that feature makes the case $d=2$, which seems pretty challenging, at least by the approach I outlined in my first answer, much easier to solve. 
So, as before, $n$ hyperplanes in general position through the origin in $\mathbb{R}^d$ cut the surface of the unit sphere, each creating two hemispheres and collectively $2R$ regions. Each hemisphere cut into $R$ regions, one of each size. In this version choose a hemisphere (it matters not which), randomly choose $k$ points in that hemisphere, and ask: 

What is the expected number of regions with at least one point?

For $d=2$ this translates to: We have a unit interval and pick $n-1$ points splitting it into $n$ subintervals whose sizes come from a Poisson distribution. Then we pick $k$ sample points in the same interval and wonder how many subintervals are hit. Now forget the initial distribution and simply say that first we pick $n+k-1$ points and then randomly pick $n-1$ of them to be dividers and the remaining $k$ to be test points. Suddenly we have stars and bars (note that that reference switches the use of $n,k$) available to us.  If $x_i$ is the number of test points in region $i$ then we have an ordered sum of non-negative integers $\sum_1^nx_i=k.$ The number of such sums is $\binom{k+n-1}{n-1}.$ The number with all the $x_i \gt 0$ is $\binom{k-1}{n-1}$ and the number with exactly $j$ of the $x_i$ positive is $\binom{n}{j}\binom{k-1}{j-1}.$ So the expected number of non-empty regions is $$E(n,k)=\frac{\sum_1^n j\binom{n}{j}\binom{k-1}{j-1}}{\binom{k+n-1}{n-1}}=\frac{nk}{n+k-1}.$$  Comments:


*

*I'll leave out the ugly details of how I summed that with the assurance that calculations agree. 

*The final answer is so simple that I imagine there is an elegant derivation which escaped me. 

*Evidently $E(n,k)=E(k,n),$ the answer is unchanged by switching $k$ and $n$. In hindsight, that is clear from the stars and bars.
Here is the reasoning for the formula claimed at the top:
Go back to picking $k$ test points distributed over the entire sphere. Pick any one of the hyperplanes. Then the $k$ points are split into $\ell$ in one half and $k-\ell$ in the other. The chance of any particular split is $2^{-k}\binom{k}{\ell}.$ It follows that the expected number of regions is $$\frac{n}{2^k}\sum_0^k\binom{k}{\ell}\left(\frac{\ell}{n+\ell-1}+\frac{k-\ell}{n+k-\ell-1}\right).  $$
