Consider a rooted tree of height $h$, such that all the leaves are at last layer. We perform the following random process: each edge is deleted with probability $0.5$, and otherwise it is retained. We are interested in the probability that after the process ends, there remains a path from the root to one of the leaves. In particular, we are interested in whether the probability goes to zero as $h$ goes to infinity or not.


It is known that if the tree is a complete $d$-ary tree for $d > 2$ then this probability is greater than $0$ regardless of $h$. However, I am interested in the case where the graph is irregular, and in particular may have internal vertices with a large number of children and may have internal vertices with only one or two children.


Suppose that we know that the *average* degree in each layer is large: Say, there is some large constant $d$ such that in every layer, the average number of children of vertices in that layer is at least $d$. Suppose that we also know that the maximal degree of a vertex is bounded, i.e., the number of children that a vertex can have is at most some constant $D$ which is independent of $h$ (and may be significantly larger than $d$, e.g., $D= 2^d$). Can we prove that, the probability that there is a path from the root to a leaf does not go to $0$ when $h$ goes to infinity (assuming $d$ is sufficiently large)? If not, can we say something about the rate of convergence (e.g., the probability is at least $\frac{1}{\log h})$?.



Percolation on finite irregular trees
You still need more information on the structure of the tree. The 1-3-tree (Example 1.2 in Lyons & Peres: [Probability on trees and networks](http://mypage.iu.edu/~rdlyons/prbtree/prbtree.html)) shows that the probability can go to 0, even if the maximum degree is bounded. 


Here is a quick proof sketch: It is not hard to see that the subtree hanging off of any other than the rightmost vertex in every layer only contains finitely many rays. Simply observe that there are $\theta(3^n)$ vertices at level $n$ of the rightmost subtree and only $2^n$ vertices at this level in total. Thus eventually all vertices with 3 descendents will be contained in the rightmost subtree. 


So the probability for percolation in any subtree except the rightmost one is $0$. If percolation happens, then all edges of the rightmost branch must be open, which of course happens with probability $0$.


Analogously one can construct a $1$-$n$-tree with an average number of $n-1$ successors and see that the probability for percolation is $0$.


