midpoint. This means that all the surface points have a finite probability to be connected to the midpoint. Hence at $p_{c2}=1$, a finite fraction of the surface point is connected to the midpoint (in this trivial case, this fraction is 1). The number of surface points is $B=2^{L}$, and one can alternatively ask what is the chance for finding a connection from the midpoint to just one of the surface points. This is just $P\cdot
B=p^{L}2^{L}=(2p)^{L}$, which becomes finite for $p=1/2$ and defines the first percolation threshold $p_{c1}=1/2$. Thus the binary tree possesses two separate thresholds: the lower one $p_{c1}$ defined as the first appearance of a connection from the midpoint to the surface and the second one, $p_{c2}$, defined as the first appearance of midpoint to surface connections which reaches a finite fraction of the surface points.

We apply the same method to obtain the thresholds for the EBT case: two possible routes from a surface point to the midpoint are drawn in Fig. \[fig:latt\](b). The difference with the binary tree is that a route for the EBT can jump between the various tree branches. Thus there are now many possible routes and the chance that one of them 