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Andreas Blass
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Since you were nice enough to ask for any bounds on $\gamma$ better than countable: $\gamma$ is a recursive ordinal, because the game tree, starting from any finite position, is recursive.

Correction: As pointed out by Joel David Hamkins in the comments (see also my subsequent comment), the recursiveness of the game tree implies only that, for every position $p$ from which White has a forced win, there is a recursive ordinal $\alpha$ such that White wins in $\alpha$. A uniform bound $\gamma$ that works for all such $p$ simultaneously would thus be the first non-recursive ordinal $\omega_1^{CK}$.

Since you were nice enough to ask for any bounds on $\gamma$ better than countable: $\gamma$ is a recursive ordinal, because the game tree, starting from any finite position, is recursive.

Since you were nice enough to ask for any bounds on $\gamma$ better than countable: $\gamma$ is a recursive ordinal, because the game tree, starting from any finite position, is recursive.

Correction: As pointed out by Joel David Hamkins in the comments (see also my subsequent comment), the recursiveness of the game tree implies only that, for every position $p$ from which White has a forced win, there is a recursive ordinal $\alpha$ such that White wins in $\alpha$. A uniform bound $\gamma$ that works for all such $p$ simultaneously would thus be the first non-recursive ordinal $\omega_1^{CK}$.

Source Link
Andreas Blass
  • 73.1k
  • 8
  • 191
  • 290

Since you were nice enough to ask for any bounds on $\gamma$ better than countable: $\gamma$ is a recursive ordinal, because the game tree, starting from any finite position, is recursive.