Suppose you're one of the nodes of this infinite tree, cut off for the purposes of the diagram, but you have no information whatsoever on which node you are. Region A is exactly like region B. And region B is exactly like region C.
So that you're in D must be at least as likely as that you're in B or C, but that you're in B is just as likely as that you're in A, and ditto for being in C. Hence that you're in D is at least twice as likely as that you're in A. But that you're in D obviously has the same probability as that you're in A.
Thus, P(A)=P(D)≥2P(A). Hence P(A)=0. But of course the whole tree is equal to three copies of A, plus the point 0. So if you can assign probabilities, then you're certain to be at 0. Which is absurd, especially since you can recenter the graph at another point and run the argument again.
Philosophically, this is a nice illustration of the severe limits to probabilistic reasoning. My eight-year-old son is looking at what I'm posting and says: "It's just probabilities and I'm not going to use probabilities on this tree. It's obvious." He wonders why I am posting such obvious things.
Mathematically, all that this displays is that of course there is a paradoxical decomposition of a regular tree, and hence that there is no finitely-additive symmetry-invariant probability measure on a regular tree.