Microstates, Distribution of Particles, and the Probability of an Empty Compartment If I have a closed system composed of $N$ particles and $p$ compartments, the total number of microstates available to that system is
$$
p^N
$$
Now say I want to find the probability that any one of the $p$ compartments is empty. My first instinct is to say that the probability of one specific compartment being empty is $p^{-N}$, and thus the probability of any of the $p$ compartments being empty is $ \frac{p}{p^N} = p^{-(N-1)}$. This seemed like a nice answer because it was certainly a very small number.
But, then when I think about it, claiming that the probability of one specific compartment being empty is $p^{-N}$ is equivalent to stating that there is only one microstate that corresponds to the $i^{th}$ compartment being empty - but of course there many microstates that can occur for only one compartment being empty; namely, there are $(p-1)^N$ microstates according to my initial statement.
So then for one specific compartment being empty, the probability is
$$
\frac{(p-1)^N}{p^N}
$$
and so the probability of any of the compartments being empty is
$$
\frac{p(p-1)^N}{p^N}
$$ 
But for some reason this number seems too big. I realize that for any sufficiently large $p$ or $N$ the number will be small, but for whatever reason it doesn't feel small enough for me to be confident about this answer, even though I'm pretty sure the logic is sound.
I'm sure at this point I'm overthinking it, so any input would be appreciated. Thanks!
 A: The probability for one compartment being empty is actually the probability that at least one compartment is empty. Let's call it $\Pi^{(1)}$,
$$
\Pi^{(1)} = \frac{(p-1)^N}{p^N}
$$ 
When you write the probability for any of the compartments to be empty as 
$$
\Pi = \sum_{j=1}^{p}{\Pi^{(1)}} = p\frac{(p-1)^N}{p^N}
$$ 
you are overcounting.
Longer argument:
Denote $\pi_{i_1i_2..i_k} = \pi^{(k)}$, $1\le k\le p$,  the probability that compartments $i_1$, $i_2$, …, $i_k$ are empty, but all others are non-empty. The probability that compartment $j$ is empty reads
$$
\Pi_j = \pi_j + \sum_{i_1}'{\pi_{ji_1}} + \sum_{i_1i_2}'{\pi_{ji_1i_2}} +…+ \sum_{i_1i_2i_3…i_{p-2}}'{\pi_{ji_1i_2i_3…i_{p-2}}} = \\
= \pi^{(1)} + (p-1)\pi^{(2)} + \left(\begin{array}{c} p-1\\2 \end{array} \right)\pi^{(3)} + … + \left(\begin{array}{c} p-1\\p-2 \end{array} \right)\pi^{(p-1)} = \sum_{k=0}^{p-2}{\left(\begin{array}{c} p-1\\k \end{array} \right)\pi^{(k+1)}}
$$
where $\sum'$ means that summation variables jump over $j$. Obviously, we must also have 
$$
\Pi_j \equiv \Pi^{(1)} = \frac{(p-1)^N}{p^N} = \sum_{k=0}^{p-2}{\left(\begin{array}{c} p-1\\k \end{array} \right)\pi^{(k+1)}}
$$
If the probability that one arbitrary compartment is empty is now calculated as
$$
\Pi = \sum_j \Pi_j = \sum_{k=0}^{p-2}{p\left(\begin{array}{c} p-1\\k \end{array} \right)\pi^{(k+1)}} \equiv \sum_{k=1}^{p-1}{p\left(\begin{array}{c} p-1\\k-1 \end{array} \right)\pi^{(k)}}
$$
it can be easily checked that terms coming from multiple empty boxes are counted multiple times. 
If you indeed want the probability that at least one compartment is empty, then it is
$$
\bar\Pi = \sum_{k=1}^{p-1}{\left(\begin{array}{c} p \\k \end{array}\right)\pi^{(k)}}
$$
Notice that $p\left(\begin{array}{c} p-1\\k-1 \end{array} \right) = k\left(\begin{array}{c} p\\k \end{array} \right)$ and so your initial attempt overestimates $\bar\Pi$ by
$$
\Pi - \bar\Pi = \sum_{k=1}^{p-1}{(k-1)\left(\begin{array}{c} p \\k \end{array}\right)\pi^{(k)}}
$$
Short argument:
Given the above, also note that if $\pi^{(0)}$ is the probability that no compartment is empty, then 
$$
\pi^{(0)} + \bar\Pi = \sum_{k=0}^{p-1}{\left(\begin{array}{c} p \\k \end{array}\right)\pi^{(k)}} = 1
$$
meaning that 
$$
\bar\Pi = 1 -\pi^{(0)}
$$ 
Side note:
If you need the probability that any one compartment is empty while the other $(p-1)$ are non-empty, then you'd have
$$
\tilde\Pi = p\pi^{(1)}
$$
but of course $\pi^{(1)} \neq \Pi^{(1)}$. 
You'll have to check with somebody who knows combinatorics better, but I think the probabilities $\pi^{(k)}$ that exactly $k$ labeled boxes are empty are given by the number of ways to split $N$ objects into the remaining $(p-k)$ non-empty compartments, which is $(p-k)! \mathcal S_N^{(p-k)}$, with $\mathcal S_N^{(p)}$ the Stirling number of the 2nd kind,
$$
S_N^{(p)} = \frac{1}{p!}\sum_{j=0}^p{(-1)^{p-j}\left(\begin{array}{c} p \\j \end{array}\right)j^N}
$$ 
In this case we have
$$
\pi^{(k)} = \frac{(p-k)!\mathcal S_N^{(p-k)}}{p^N}\\
\bar\Pi = 1 -\pi^{(0)} = 1 - \frac{p!\mathcal S_N^{(p)}}{p^N}
$$
In the limit of large $N$ it is known that $\lim_{N \rightarrow \infty}\frac{\mathcal S_N^{(p)}}{p^N} = 1/p!$ and so
$$
\lim_{N \rightarrow \infty} \bar\Pi \rightarrow 0
$$
Note added in proof:
The reason I wasn't 100% sure of the combinatorial result at first is because it implies a certain identity involving the Stirling numbers of the 2nd kind which I could not corroborate at the time. Namely, from
$$
\sum_{k=0}^{p-1}{\left(\begin{array}{c} p \\k \end{array}\right)\pi^{(k)}} = 1
$$
it follows that 
$$
\sum_{k=0}^{p}{\left(\begin{array}{c} p \\k \end{array}\right)k!{\mathcal S}_N^{(k)}} = p^N
$$
where after substituting $\pi^{(k)} = \frac{(p-k)!\mathcal S_N^{(p-k)}}{p^N}$ and multiplying both sides by $p^N$, I used $(p-k) \rightarrow k$ and extended summation over $k=0$ on account of ${\mathcal S}_N^{(0)} = \delta_{N,0}$. It turns out that the identity is correct: it appears for instance in eq.(29) of this paper on "Close Encounters with the Stirling Numbers of the Second Kind".
Also, it may be useful to note that the connection between Stirling numbers of the 2nd kind and the number of non-empty partitions was first noticed in RC Kao, LH Zetterberg, The American Mathematical Monthly, vol. 64, no. 2, 1957, pp. 96-100.
A: First of all, if there are $p^N$ possible states, this means that the particles and the compartments are distinguishable. For indistinguishable particles (and distinguisable compartments, which is probably reasonable  physically speaking) the number of possible states, or configurations, has to be divided. For instance the configuration $(n_1,n_2,\dots,n_p)$ of indistinguishable particles corresponds to 
$$\frac{N!}{n_1!n_2!\dots n_p!}$$
configurations of distinguishable particles.
Note that we have of course
$$\sum_{{\substack{n_1,n_2,\dots,n_p\\n_1+n_2+\cdots+n_p=N}}}
\frac{N!}{n_1!n_2!\dots n_p!}=p^N.$$
Let us start with the answer to the question "how many configurations have no compartment empty ?" 
With indistinguishable particles, we place $N$ particles in $p$ compartments, with at least one particle in each. 
To count this number, we use the bullet representation of
the integer $N$ like this
$$\bullet\bullet\bullet\bullet\cdots\bullet$$
there are $N$ bullets. We now place the compartment separations, denoted as $|$. One possible configuration with $N=10$ into $p=5$ compartments is
$$\bullet|\bullet\bullet\bullet|\bullet\bullet\bullet|\bullet|\bullet\bullet$$
which is a reprensentation of $(1,3,3,1,2)$. We have to place $p-1$ divisors at any $N-1$ positions. Therefore the number of possible divisions is $\binom{N-1}{p-1}$ (basically, we have counted the number of partition of the integer $N$ into sums of non-zero integers when order matters). There are therefore
$$C_0(N,p)=\binom{N-1}{p-1}\quad\text{(indistinguishable particles)}$$
configurations with no empty compartments.
In the case of distinguishable particles, this actually more difficult, as @ndrv pointed out. We have $p^N$ possibilities, but we have to substract 
configurations with one empty compartment. We can pick up any of the $p$
compartment as empty and substract $p\times (p-1)^N$, but then we have substracted the configurations with two empty compartments two times. So must add $\binom{p}{2}(p-2)^N$. But by doing this we have also added twice the configurations with three empty compartments, so must add them back. This goes on until we have added or substracted the configuration with $p-1$ empty compartments, since there are no further configurations.
 As a conclusion, we have
$$C_0(N,p)=\sum_{j=0}^{p-1}(-1)^j\binom pj(p-j)^N
\quad\text{(distinguishable particles)}$$
This number is related to the Stirling number of second kind by $C_0(N,p)=p!\genfrac\{\}{0pt}{}Np$. 
We can answer to the question "How many configurations have exactly one empty compartment ?" The answer is, in both cases
$$pC_0(N,p-1)$$
because there remains only the choice of the empty compartment among the $p$
possible compartments.
We can even answer to the question "How many configurations have exactly $k$ empty compartments ?" From the same reasoning, we find
$$\binom pkC_0(N,p-k)$$
If we now want to allow one or more empty compartments, we can sum over $k$:
$$C(N,p)=\sum_{k=1}^{p-1}\binom pkC_0(N,p-k).$$
but we can also remark that 
$$C(N,p)=p^N-C_0(N,p).$$
