For a fixed p, let f(p) be the probability that the tree has an infinite path of sum zero, and g(p) the probability that the tree has an infinite path of sum at most 1. This month’s problem is asking to find the p such that g(p) = 1/2.
Using the decomposition of a binary tree into its root and its two subtrees, we see
f(p) = p·(2f(p) - f(p)2)
which simplifies to
f(p) = 2 - 1/p.
Further, we see
g(p) = p·(2g(p) - g(p)2) + (1-p)·(2f(p) - f(p)2).
Plugging in g(p) = 1/2 we end up with a cubic equation in p:
3p3 - 10p2 + 12p - 4 = 0
with approximate solution 0.5306035754…
Congrats to this month’s solvers!