Probability that $n$ random points on a circle, divided into $m$ fixed and equal sized slices are contained in less than $m/2$ adjacent slices. What is the probability that $n$ random points on a circle, divided into m fixed and equal sized slices, are contained in less than $m/2$ adjacent slices?
This is related to finding the probability that $n$ random points on a circle all fall in a semicircle, but it is not yet clear to me how to extend the solution to this case.
 A: Adapting this elegant answer from the linked question:
Let $k = \lceil \frac m2 \rceil -1  $ .  We are interested in the event that the occupied slices are contained in some cluster (set of adjacent slides) of size $k$. 
Let $X_{i,t}$ ($t,i=1,2 \cdots m$) be the indicator value that takes value $1$ if the cluster of $t$ slices starting at position $i$ (say, clockwise) covers all the occupied slices,  $0$ otherwise.
Then notice that $$Z=\sum_i X_{i,k} -  \sum_i X_{i,k-1} \tag{1}$$ (i.e., number of covering $k-$clusters minus number of covering $k-1$-clusters) is $1$ if our event of interest happens, $0$ otherwise.
Then we want $$P(Z=1)=E(Z)=\sum_i  E[X_{i,k}]-E[X_{i,k-1}]=m(P[X_{1,k}=1]-P[X_{1,k-1}=1]) \tag{2}$$
But $$P[X_{1,t}]=\left(\frac{t}{m}\right)^n \tag{3}$$
Then
$$P(Z=1) = m \left(\left(\frac{k}{m}\right)^n -\left(\frac{k-1}{m}\right)^n\right)=\frac{k^n-(k-1)^n}{m^{n-1}}$$
A: I'll adapt the last paragraph of my answer to the related points in a semicircle question. I suggest you read that first, or read it if you feel you're getting lost in this answer here.
You did not say whether $m$ is even or odd, so I suppose it may be both. If you want to use less than $\frac m2$ adjacent slices, you may use up to
$$k:=\biggl\lfloor\frac{m-1}2\biggr\rfloor=\biggl\lceil\frac m2\biggr\rceil-1$$
slices, which is the largest integer smaller than $\frac m2$. Suppose your points are called $A_1$ through $A_n$. Now what are the chances that all your $A_i$ will lie in the $k$ slices starting with the one that contains $A_1$ and continuing in a counter-clockwise fashion? $A_1$ definitely lies in that range (unless $m\le2$ and $k=0$), and every subsequent point will lie in that range with a probability of $\frac km$. So the total probability of having all the points in these slices would be $\left(\frac km\right)^{n-1}$.
But this depends on picking the right point as the starting point. In the continuous case of points on a semicircle, we could argue that this would be the case only for a single starting point, so we could simply add probabilities. This is no longer the case here, as we might have multiple points lying in the first segment. Just adding probabilities would count these multiple times, so we have to compensate for that. And here is a formula that does so:
$$p=\sum_{i=1}^n \frac ni\binom{n-1}{i-1} \left(\frac 1m\right)^{i-1}\left(\frac{k-1}m\right)^{n-i}$$
The counter $i$ indicates how many points are in the first slot.
The numerator $n$ stands for the $n$ ways how you can pick one of the points as the first. Then there will be $i-1$ more points in that same slot, each with a probability of $\frac 1m$, and $n-i$ points in the subsequent $k-1$ slots, each with a probability of $\frac{k-1}m$. But you don't need the second, third and so on point to be in the same slot; any choice of $i-1$ points from the $n-1$ ones that follow the first may end up in the first slot. That's what the binomial coefficient $\binom{n-1}{i-1}$ expresses. And lastly, this counts every such scenario $i$ times, once for every choice of first point among those in the first slot, so we divide by $i$ to account for it exactly once.
I did some brute force computations for small $m$ and $n$, checking all $m^n$ possible distributions the naive way, and comparing them against the result of that formula. Looks correct. Note that the formula depends on the fact that the choice of start segment is unambiguous. If you were to allow $k=\tfrac23 m$ or some such, this assumption would no longer hold. The choice of $k$ is relevant here. Also note that the formula does not hold for $m<3$, as $\frac{k-1}m$ would be negative in these cases, but should be zero instead. You can simply take $p=0$ if $k<3$.
According to my computer algebra system, the formula can be simplified to
$$
p=\sum_{i=1}^n \binom{n}{i}
\left(\frac 1m\right)^{i-1}\left(\frac{k-1}m\right)^{n-i}=
m\left(\left(\frac km\right)^n-\left(\frac{k-1}m\right)^n\right)
$$
I'd be very interested to hear an intuitively understandable direct derivation of that formula from the problem statement.
