Slaying a regrowing three-headed dragon

A dragon starts with $3$ heads and you must remove them all. Each attack does one of three things: cuts off $2$ heads, cuts off $1$ head (which always regrows immediately, so nothing changes), or misses and the dragon sprouts one extra head. Whenever the dragon has at least $2$ heads, cutting off $2$ and missing are equally likely. With $1$ head left, you cannot cut $2$. If the dragon ever reaches $5$ heads you lose. What is the probability you win?

$3/5$. Ignore the no-change attacks. From $h\ge2$ heads you move to $h-2$ or $h+1$ with probability $\tfrac12$ each, and from $1$ head you can only go to $2$, so $P_1=P_2$. Solving $P_2=\tfrac12+\tfrac12P_3$, $P_3=\tfrac12P_2+\tfrac12P_4$, $P_4=\tfrac12P_2$ gives $P_2=\tfrac45$ and $P_3=\tfrac35$.