both gene tree probabilities ${\mathbb P}_\sigma( T)$ and clade probabilities ${\mathbb P}_\sigma(\mathcal C_g)$ arise from the multispecies coalescent on a species tree $\sigma=(\psi,\lambda)$.

Let $\sigma=(\psi,\lambda)$ be a binary species tree on $\mathcal X$, with edge lengths $\lambda_i>\epsilon \ge 0$. Under the multispecies coalescent model, suppose $\mathcal C_g\subset \mathcal
  X_g$ has clade probability $\mathbb P_{\sigma}(\mathcal
  C_g)\ge(1/3)\exp(-\epsilon)$. Then $\mathcal C$ is a clade on $\sigma$; that is, $\mathcal C\in \mathcal H(\sigma)$.

Furthermore, if $(1/3) \exp(-\epsilon)$ is replaced with any smaller number, this statement is no longer true for all such choices of species trees and non-trivial clades: For any $k<(1/3) \exp(-\epsilon)$, there exists a species tree $\sigma$on $\mathcal X$ and a taxon set $\mathcal C\subset \mathcal X$ with $1<|\mathcal C|<|\mathcal X|$ such that $\mathcal C$ is not a clade on $\sigma$, yet $\mathbb P_{\sigma}(\mathcal
  C_g)\ge k$.

If $\mathcal C$ is a trivial clade, there is nothing to show, so we may assume $1< |\mathcal C_g|<|\mathcal X|$. We prove the contrapositive: if $\mathcal C$ is not a clade on $\psi$, then $\mathbb P_{\sigma}(\mathcal C_g)<(1/3)\exp(-\epsilon)$.

Suppose $\mathcal C$ is not a clade on the species tree, so there exist $a,b \in \mathcal C$ and $c \in \mathcal
  X\smallsetminus\mathcal C$ such that $\psi$ does not 