$\mathcal C_g$ will be a clade on a gene tree can be made as close to $(1/3)\exp(-\epsilon)$ as we wish by taking the branch above this subtree to be long.

Setting $\epsilon=0$ yields Corollary \[cor:third\].

\[cor:third\] Let $\sigma$ be a binary species tree on taxa $\mathcal X$, with positive edge lengths. Under the multispecies coalescent model, suppose $\mathcal C\subset \mathcal
  X$ is such that $\mathbb P_{\sigma}(\mathcal C_g)\ge1/3.$ Then $\mathcal C$ is a clade on $\sigma$.

Furthermore, this statement is no longer true for $1< |\mathcal C|<
  |\mathcal X|$ if $1/3$ is replaced with any smaller number.

If the species tree is not binary, a slightly weaker result holds.

Suppose the species tree $\sigma$ is not necessarily binary, and $\mathcal C\subset \mathcal X$ is such that $\mathbb
  P_{\sigma}(\mathcal C_g)>1/3$. Then $\mathcal C$ is a clade on $\sigma$.

Furthermore, this statement is no longer true for $1<|\mathcal
 C|<|\mathcal X|$ if $1/3$ is replaced with any smaller number.

To show $\mathcal C$ is a clade, we suppose $c\in \mathcal X \smallsetminus \mathcal C$ and demonstrate that $c\notin {\operatorname{desc}_\mathcal X}({\operatorname{MRCA}}(\mathcal C))$. Choose $a,b\in \mathcal C$ such that ${\operatorname{MRCA}}(\mathcal C)={\operatorname{MRCA}}(\{a,b\})$. Note that $\mathbb P_{\sigma}( ((A,B),C) )\ge \mathbb
P_{\sigma}(\mathcal C_g)>1/3$ implies that the rooted triple $((a,b),c)$ 