Let's play a game. You start with the ordinal $\alpha$ and I start with the empty sequence. Each turn, you decrease your ordinal, and I add a tree (where each node can have one of three labels), subject to the rule that no previous tree I choose can homeomorphically embed into the new one. Whoever moves last wins! What $\alpha$ should we choose to make the game fair, meaning whoever moves first will lose (so both players make the same number of moves)?

Note we can generalize this any number of labels $n$, or indeed, any well-quasi-ordered set instead of trees.

Well, note that the game can not go on forever, for two reasons. There is no infinite decreasing sequence of ordinals, so you must eventually hit $0$ and be out of moves. Likewise, due to Kruskal's tree theorem, I too can not go on forever. Indeed, we could have replaced trees with any other well-quasi-ordered set.

Now, what do we mean by fair? Well, since the tree picking game correspond to a game in the CGT sense, there is some ordinal $\alpha$ such that $T(3) - \alpha = 0$ (implying $T(3) = \alpha$, hence the title), where $T(3)$ is the tree picking game. $T(3)-\alpha$ is therefore the game described in the first paragraph. Since $T(3) - \alpha = 0$, whoever moves first in that game will lose (this is why we say whoever moves first will lose).

Arch. Math. Logic54(2014) 193–230), but I'm lost in a maze of twisty little papers all alike so I'm not sure this is really the right one. It's certainly relevant, though. $\endgroup$ – Gro-Tsen Oct 27 '17 at 22:27