In a number of blog posts I found the claim that the game described below cannot be lost, which is to say, every possible strategy is a winning strategy. In each case, a sketch proof is given that involves ordinal arithmetic, and the claim is made that Peano arithmetic is not strong enough to prove the result.
The problem is that I must have misunderstood something rather basic, because by playing around with pen and paper I was quickly able to find strategies for which the number of "heads" grows indefinitely, which seems to mean the claimed result is incorrect. My question is whether my misunderstanding is in the mechanics of the Hydra game, or in the nature of the claim being made about it.
Here are the rules of the game, as I understand them. A hydra is a rooted tree. Its leaves are called "heads". The player (Hercules) must kill the hydra (i.e. reduce it to only its root node) by chopping off its heads. To chop of a head, we first remove it from the tree. If this creates a new leaf node, it becomes a new head. We then find the grandparent node of the original head that we removed, and look at all of the subtrees whose root is the grandparent node. Each of these is duplicated $n$ times, where $n$ is any arbitrary natural number, such that the new subtrees each have the grandparent node as their root. (If the original head didn't have a grandparent because its parent was the root node, no new subtrees are generated.)
Here is a figure to illustrate this process, taken from one of the posts linked above.
The claim is that every strategy is a winning strategy, which as I understand it means that no matter which head is removed at each time step, and no matter what value of $n$ is chosen at each time step, the hydra will always be reduced to its root node in a finite number of moves.
Now, let us write $x$ for a head and put nodes in parentheses if they share a parent. Then $((xxx))$ is a hydra, whose root node has one child, which in turn has three children, which are heads. Removing one head yields $((xx))$. The head's grandparent was the root node, so duplicating the subtree (with $n=2$) yields $$ ((xx)(xx)). $$ Removing one of these heads, again with $n=2$, yields $((xx)(x))$, and duplicating again with $n=2$ gives $$ ((xx)(xx)(x)(x)). $$ But now the previous step is a subtree. By chopping off the leftmost head we obtain $((xx)(xx)(x)(x)(x)(x))$ and then $((xx)(xx)(x)(x)(x)(x)(x)(x)(x)(x))$, and by induction after the $k^\text{th}$ move we will always have the $(xx)(xx)$ subtree, followed by $2^k$ copies of $(x)$.
This then appears to be a losing strategy, contradicting what I understood to be the claim that no such strategy exists. Moreover, there appears to be no winning strategy for this tree as long as we always choose $n>1$, because the number of subtrees will always increase. What have I misunderstood?
