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 * (2*x*) - *p*^6 * 4*x*^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!**