Expected Entropy Based on Dirichlet Distribution The Dirichlet Distribution basically defines the probability that a sample came from a particular multinomial distribution if we assume that the prior probability of all multinomial distributions having generated the sample are equal. 
Each multinomial distribution has a corresponding categorical distribution, and the entropy of that categorical distribution is given by 
$$-\sum_x^{states}\Pr(x)\ln(\Pr(x))$$
Given a point $p=(p_1,p_2,p_3...p_n)$ randomly chosen according to a Dirichlet Distribution with parameters $k_1...k_n$, such that $\sum_ip_i=1$, the entropy of the corresponding categorical distribution is:
$$H(p)=-\sum_i^n p_i \ln(p_i)$$
What the expected value of $\text H(p)$?
In the special case where the Dirichlet Distribution is just defined by $k_1$ and $k_2$ and $p$ is 2-dimensional, the expected entropy $\text{H}(p)$ is given by the formula 
$$\frac{(k_1+k_2) H_{(k_1+k_2-1)}-k_1 H_{k_1}-k_2 H_{(k_2-1)}}{k_1+k_2}$$
Where $H_n$ is the $n$th harmonic number, however I haven't been able to calculate the answer for greater numbers of dimensions. 
 A: Here is a formal proof for general Dirichlet distributions $(\alpha_1, \dots, \alpha_m)$.
I use capital $P_i$ to indicate that we are working with random variables. By the linearity of expectations,
$$-E(\sum_i P_i \log P_i)=-\sum_i E(P_i \log P_i)$$
then $P_i \sim \textrm{Beta} (\alpha_i, A -\alpha_i)$  and working with the normalizing constant you can write
$$
-E_{\alpha_i, A-\alpha_i}(P_i \log P_i)= \frac{\alpha_i}{A}E_{\alpha_i+1, A-\alpha_i}(\log P_i) = \frac{\alpha_i}{A} [\psi_0 (A+1)-\psi_0(\alpha_i+1)]
$$
where the last step arises by a known result (see Wikipedia page on Beta distributions): if $X \sim \textrm{Beta}(\alpha, \beta)$ then $-E(\log X)= \psi_0(\alpha+\beta)-\psi_0(\alpha)$.
Summing over $i$ provides your general formula.
A: So as it turns out, the general closed-form solution is 
$$\text{Exp}(H(P))=H_A-\frac{1}{A}\sum _{i=1}^m \alpha _i H_{\alpha _i}$$
Where $m$ is the number of different states, $H(P)$ is the entropy of probability distribution $P$ where each state $s_i$ occors with probability $p_i$, the $\alpha _i$ are the distribution parameters of the dirichlet distribution $P$ is drawn from, and $A=\sum_{i=1}^m \alpha _i$. Each $\alpha_i = k_i+1$.
I believe the derivation is obvious. 
JK the derivation isn't obvious. I found a closed-form solution when $n=2$, when $n=3$, and when $n=4$, and then I fiddled with those until I came up with a general formula that looks elegant (see above), and then tested the formula against a monte-carlo estimation of the expected value. It works but I can't prove it. 
A: In a special case, there's a way to rewrite this.
In Entropy and Inference, Revisited by Nemenman et al, they note that:
$$ \mathbb{E}[\mathbb{H}[P]] = \psi(N_c\beta+1) - \psi(\beta+1) $$
where $N_c$ is the number of categories, $\psi$ is the digamma function, and $P\sim\text{Dir}(\beta)$ where $\beta:= \alpha_i$ parametrizes a Dirichlet with all parameters being equal (so not as general as your elegant formula).
Recalling that
$$ H_n = \gamma + \psi(n+1) $$
where $\gamma$ is the Euler-Mascheroni constant, we can reduce your formula as:
\begin{align*}
H_A - \frac{1}{A}\sum_i \alpha_i H_{\alpha_i} 
&= H_{N_c\beta} - \frac{1}{N_c\beta}\sum_i \beta H_\beta \\
&= \gamma + \psi(N_c\beta + 1) - \frac{1}{N_c}\sum_i H_\beta \\
&= \gamma + \psi(N_c\beta + 1) - (\gamma + \psi(\beta+1)) \\[2.1mm]
&=  \psi(N_c\beta + 1) - \psi(\beta+1) \\[2mm]
&= \mathbb{E}[\mathbb{H}[P]]
\end{align*}
