(1-q_{B})^{N - M_{N,2}} \\
        \nonumber
        & {} + \frac{1}{2} q_{A}^{M_{N,2}} (1-q_{A})^{N - M_{N,2}} q_{B}^{M_{N,1}} (1-q_{B})^{N - M_{N,1}} \\
        \label{eq:efpf_two_bern}
        &= \efpf(N, M_{N,1}, M_{N,2}).\end{aligned}$$ Here, $\efpf$ is some function of the number of indices $N$ and the feature sizes $(M_{N,1}, M_{N,2})$ that we note is symmetric in $(M_{N,1}, M_{N,2})$; i.e., $\efpf(N,M_{N,1},M_{N,2}) = \efpf(N,M_{N,2},M_{N,1})$.

When the feature allocation probability admits the representation $$\label{eq:efpf}
    \mbp(\labrandfa_{N} = \labdetfa_{N}) = \efpf(N,|\blockfa_{1}|, \ldots, |\blockfa_{K}|)$$ for every ordered feature allocation $
    \labdetfa_{N} = (A_{1},\ldots,A_{K})
$ and some function $\efpf$ that is symmetric in all arguments after the first, we call $\efpf$ the *exchangeable feature probability function* (EFPF). We take care to note that the exchangeable partition probability function (EPPF), which always exists for partitions, is not a special case of the EFPF. Indeed, the EPPF assigns zero probability to any multiset in which an index occurs in more than one feature of the multiset; e.g., $\{\{1\},\{2\}\}$ is a valid partition and a valid feature allocation of $[2]$, but $\{\{1\},\{1\}\}$ is a valid feature allocation but not a valid partition of $[2]$. Thus, the EPPF must examine the feature indices of a feature allocation to judge their exclusivity and thereby assign a probability. By contrast, the indices in 