TITLE: An elementary question in bond percolation
QUESTION [9 upvotes]: Consider a locally finite, connected graph and "bond (edge) percolation" on this graph. Each edge is open with probability $p.$ There is a parameter $\alpha$,  $0<\alpha<1.$
The question: Is it possible to divide the set of nodes into "sources" and "receivers" so that the following two conditions are fulfilled?
(i) For any "receiver", the probability that it ends up in a connected component with a "source" is larger or equal than $\alpha$;
(ii) For any "source", the probability that it ends up in a connected component with another "source" is smaller than $\alpha$.
I can prove, by elementary methods, that this is true for linear graphs, finite or infinite, complete graphs, and  star graphs (graphs, in which $n-1$ nodes have degree $1$ and one node has degree $n-1$).
Also, there are many other situations in which existence is trivial. In a finite graph, when $p$ is close to zero, the solution is that every node is a source, and the set of receivers is empty. If the graph is $\mathbb{Z}^2,$ then, if percolation function $\theta(p)>\sqrt{\alpha},$ then making one node a source and all other nodes receivers is a solution. (The case of an infinite graph might be very different. Interestingly, it is possible to have $p>\frac{1}{2}$ and a solution with an infinite number of sources on $\mathbb{Z}^2$.)
I wonder if this is true for any finite (and locally finite) graph?

REPLY [2 votes]: To indicate the non-triviality of this problem, consider a symmetric star graph with 3 rays (X,Y, and Z), each having 2 vertices (thus the graph has in total 7 vertices, including the center $C$, let us denote them $X_1,X_2,Y_1,Y_2,Z_1,Z_2$,and $C$). For this graph, the answer is "yes" for all possible values of $\alpha$ and $p$.
Star graph with 3 rays
One of the following SEVEN possible configurations will work (I indicate which vertices are sources):
(i)  $\{C\}$
(ii) $\{X_2,Y_2,Z_2\}$
(iii) $\{X_2,Y_2,Z_2,C\}$
(iv) $\{X_1,X_2,Y_2,Z_2\}$
(v) $\{X_1,X_2,Y_1,Y_2,Z_2\}$
(vi) $\{X_1,X_2,Y_1,Y_2,Z_1,Z_2\}$
(vii) $\{X_1,X_2,Y_1,Y_2,Z_1,Z_2,C\}$
It turns out that this is the minimal number of configurations needed to cover all possible values of the parameters. Especially difficult cases, requiring non-symmetric (!) configurations are $p=0.8, \alpha=0.95$ and $p=0.74, \alpha=0.9$.
The plot below indicates areas of possible pairs $(p,\alpha)$ covered by the seven configurations above.
Phase diagram for the pair $(p,\alpha)$