Suppose for a given p, Aaron’s probability of winning on the infinite tree is x. Then looking at first two turns in the game, Aaron can win if at least one of the two sides of the tree has 3 edges marked A and BOTH subtrees Beren can choose between are winnable by Aaron (which we know has probability x, independently per subtree). So the equation x and p must satisfy is:

x = 2 * p^3 * x^2 - p^6 * x^4.

(The subtracted term is from the double counting of when Aaron can win on both sides of the tree).

We want to find the smallest positive p where this has a positive root x < 1. This can be done with the theory of discriminants, but also can be solved by noticing that as p increases, the graph of the quartic that is the right hand side of the above equation approaches the graph of f(x) = x from below, and the moment where it touches it will be tangent. So we can add the constraint that the derivative of the right hand side equals 1 at the point of equality:

1 = 2 * p^3 * (2x) - p^6 * 4x^3

These two equations are solved by x = 8/9, and p = (27/32)^(1/3). So the lowest nonzero probability Aaron can have of winning the game is the surprisingly high 8/9 chance!

Congrats to those of you who worked out this tricky math puzzle!