Solution
= Solution
After the forced first step $A\to B$, let $h_i=P_i(T_A<T_F)$. Symmetry gives $h_C=h_D=x$, while
$$
h_B=(1+2x)/3,\quad x=(h_B+h_E)/2,\quad h_E=2x/3.
$$
Solving gives $h_B=2/3$. Thus the requested probability is $2/3$.
= Solution
After the forced first step $A\to B$, let $h_i=P_i(T_A<T_F)$. Symmetry gives $h_C=h_D=x$, while
$$
h_B=(1+2x)/3,\quad x=(h_B+h_E)/2,\quad h_E=2x/3.
$$
Solving gives $h_B=2/3$. Thus the requested probability is $2/3$.