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Theorem. Suppose there is a closed unbounded class of cardinals $\kappa$ such that the statement $\sigma$ is forced by the collapse forcing of $\kappa$ to $\omega$. Then Alice has a winning strategy in the $\sigma$ game.

Proof. Let $C$ be the class club of such $\kappa$, and let Alice simply play always to collapse the next element of $C$ above the size of the previous forcing played by Bob. It follows that the limit forcing $\mathbb{P}$ will collapse all the cardinals up to an element $\kappa\in C$, and since $\kappa$ will have cofinality $\omega$, it will also collapse $\kappa$ itself. Since the forcing will also have size $\kappa$, in the direct limit case, it follows that the forcing is isomorphic to $\text{Coll}(\omega,\kappa)$, and so $\sigma$ holds in the model $V[G]$, so Alice has won. $\Box$

For example, if the GCH fails on a closed unbounded class of cardinals with countable cofinality (this contradicts SCH), then Alice can win with $\neg$CH. And if it fails on a stationary class of such cardinals, then Bob cannot win the $\neg$CH game.

Proof. Using the Foreman-Woodin theorem that it is relatively consistent that GCH fails everywhere, we can perform additional forcing by first adding a generic class of cardinals, and then collapsing cardinals. The result will be a model of GBC where the class of cardinals $\kappa$ of cofinality $\omega$ at which the GCH holds is both stationary and co-stationary. By the theorem above, considered from either Alice's or Bob's perspective, either player can defeat any strategy of the other player. So the game is not determined. $\Box$

Proof. Statement 2 implies statement 1 by the proof of the first theorem above. Conversely, if statement 2 fails, then there is a stationary class of such $\kappa$ where $\sigma$ fails in the collapse extension. In this case, Bob can defeat any strategy for Alice, by Bob's analogue of the second theorem. $\Box$

Theorem. Suppose there is a $\omega$-closed unbounded class of cardinals $\kappa$ such that the statement $\sigma$ is forced by the collapse forcing of $\kappa$ to $\omega$. Then Alice has a winning strategy in the $\sigma$ game.

Proof. Let $C$ be the class $\omega$-club of such $\kappa$, and let Alice simply play always to collapse the next element of $C$ above the size of the previous forcing played by Bob. It follows that the limit forcing $\mathbb{P}$ will collapse all the cardinals up to an element $\kappa\in C$, and since $\kappa$ will have cofinality $\omega$, it will also collapse $\kappa$ itself. Since the forcing will also have size $\kappa$, in the direct limit case, it follows that the forcing is isomorphic to $\text{Coll}(\omega,\kappa)$, and so $\sigma$ holds in the model $V[G]$, so Alice has won. $\Box$

For example, if the GCH fails on a $\omega$-closed unbounded class of cardinals (this contradicts SCH), then Alice can win with $\neg$CH. And if it fails on a stationary class of such cardinals with countable cofinality, then Bob cannot win the $\neg$CH game.

Proof. Using the Foreman-Woodin theorem that it is relatively consistent that GCH fails everywhere, we can perform additional forcing by first adding a generic class of cardinals, and then forcing certain instances of GCH by collapsing cardinals. The result will be a model of GBC where the class of cardinals $\kappa$ of cofinality $\omega$ at which the GCH holds is both stationary and co-stationary. By the theorem above, considered from either Alice's or Bob's perspective, either player can defeat any strategy of the other player. So the game is not determined. $\Box$

Proof. Statement 2 implies statement 1 by the first theorem above. Conversely, if statement 2 fails, then there is a stationary class of such $\kappa$ where $\sigma$ fails in the collapse extension. In this case, Bob can defeat any strategy for Alice, by Bob's analogue of the second theorem. $\Box$

1. Alice has a winning strategy in the $\sigma$ game.
2. The class of cardinals $\kappa$ of countable cofinality, such that the collapse forcing of $\kappa$ to $\omega$ forces $\sigma$, contains a class club.

Proof. Statement 2 implies statement 1 by the first theorem above. Conversely, if statement 2 fails, then there is a stationary class of such $\kappa$ where $\sigma$ fails in the collapse extension. In this case, Bob can defeat any strategy for Alice, by Bob's analogue of the second theorem. $\Box$

1. Alice has a winning strategy in the $\sigma$ game.
2. The class of cardinals $\kappa$ of countable cofinality, such that the collapse forcing of $\kappa$ to $\omega$ forces $\sigma$, contains a class $\omega$-club.

Proof. Statement 2 implies statement 1 by the proof of the first theorem above. Conversely, if statement 2 fails, then there is a stationary class of such $\kappa$ where $\sigma$ fails in the collapse extension. In this case, Bob can defeat any strategy for Alice, by Bob's analogue of the second theorem. $\Box$

The ideas appear to culminate in answer to question 1.

Theorem. The following are equivalent for any statement $\sigma$.

1. Alice has a winning strategy in the $\sigma$ game.
2. The class of cardinals $\kappa$ of countable cofinality, such that the collapse forcing of $\kappa$ to $\omega$ forces $\sigma$, contains a class club.

Proof. Statement 2 implies statement 1 by the first theorem above. Conversely, if statement 2 fails, then there is a stationary class of such $\kappa$ where $\sigma$ fails in the collapse extension. In this case, Bob can defeat any strategy for Alice, by Bob's analogue of the second theorem. $\Box$