Consider the following procedure for generating a random binary tree: Starting with a full binary tree (i.e., each node has either two or no children) we iterate over the leaves and (independently) for each leaf we add two children with probability $\theta(1+d)^{-\gamma}$, where $d$ is the depth of the leaf (the root node has depth $d=0$), $\theta\in (0,1)$, and $\gamma>0$. Repeat this process until none of the leaves have children added to them, at which point we return the tree.
Does this process always terminate (i.e., does it generate a finite binary tree with probability one)? Does it depend on the value of $\gamma$, or even $\theta$?
The expected number of nodes at level $d$ is as $$ 2\frac{\theta}{1^\gamma}\cdot2\frac{\theta}{2^\gamma} \cdots 2\frac{\theta}{d^\gamma} =\frac{(2\theta)^d}{(d!)^\gamma} = \Bigl( \frac{((2\theta)^{1/\gamma})^d}{d!}\Bigr)^\gamma$$ For any positive $\gamma$ the factorial in the denominator will eventually dominate the fraction, so the expected size of each level goes towards zero.
This expectation is also an upper bound for the probability that this particular layer contains at least one node, which is itself an upper bound for the probability that the tree is infinite.
So the tree is infinite with probability 0 and finite with probability 1.
Boundary cases: If $\gamma=0$, then the tree is finite with probability $\min(1, \frac{1-\theta}{\theta})$.
And if $\gamma<0$ and $\theta\ne 0$, then the recipe doesn't make sense at all, as $\theta(1+d)^{-\gamma}$ would not always be in $[0,1]$.