= Solution
If the urn contains $g$ green balls, it contains $g+2$ red balls: both update rules preserve the difference $R-G=2$. Until absorption at $g=0$, the green count is therefore a birth-death chain with
$$
p_g=\mathbb P(g\to g-1)=\frac{g}{2g+2},
\qquad
q_g=\mathbb P(g\to g+1)=\frac{g+2}{2g+2}.
$$
The <function>
$$
h(g)=\frac1{g+1}
$$
is harmonic, since
$$
p_gh(g-1)+q_gh(g+1)
=\frac1{2(g+1)}+\frac1{2(g+1)}=h(g).
$$
Let $\tau_0$ and $\tau_N$ be the hitting times of $0$ and $N$. The <optional sampling theorem for a supermartingale>, applied to the bounded stopped martingale $h(G_{n\wedge\tau_0\wedge\tau_N})$, gives
$$
h(m)=\mathbb P_m(\tau_0<\tau_N)
+\frac1{N+1}\mathbb P_m(\tau_N<\tau_0).
$$
Solving,
$$
\mathbb P_m(\tau_0<\tau_N)
=\frac{N-m}{N(m+1)}.
$$
Letting $N\to\infty$, the events on the left increase to eventual termination. The <harmonic hitting probability for the balanced-difference urn> is therefore
$$
\boxed{\mathbb P(\text{the process terminates})=\frac1{m+1}}.
$$
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