I like this question a lot. It provides an interesting way of talking about
some of the ideas connected with the maximality principle and the
modal logic of forcing.
Let me make several observations.
First, Alice can clearly win, in one move, with any forceably
necessary statement $\sigma$, which is a statement for which
$\newcommand\possible{\Diamond}\newcommand\necessary{\Box}\possible\necessary\sigma$
holds in the modal logic of forcing, meaning that one can force so
as to make $\sigma$ remain true in all further forcing extensions.
She should simply force to make $\necessary\sigma$ true, and then
Bob cannot prevent $\sigma$ in any further extension, including the
limit model. Under the maximality
principle, all such
statements are already true.
Let me point out that there are some subtle issues about
formalization in the question. For example, the strategies here
would be proper class sized objects, and so one must stipulate
whether one is working in ZFC with only definable classes or
whether one has GBC or KM or whatever and whether global choice
holds. Another difficulty concerns the determinacy of the game,
since even open determinacy for class
games is
not a theorem of GBC.
Meanwhile, here is something positive to say. I shall consider only the direct-limit version of the game.
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 holds, then CH will be such a statement
$\sigma$, even though this is a switch, because the collapse
forcing will collapse $\kappa$ and the CH will hold in $V[G]$, as a
residue of the GCH in $V$. Thus, the GCH implies that Alice can win
the CH game.
A dual analysis is:
Theorem. If the class of cardinals $\kappa$ of countable
cofinality for which the collapse forcing
$\text{Coll}(\omega,\kappa)$ forces $\sigma$ is stationary, then
Alice can defeat any strategy of Bob in the $\sigma$ game.
Proof. For any strategy for Bob, there is a club of cardinals
$\theta$ such that $V_\theta$ is closed under the strategy, in the
sense that if Alice plays a poset in $V_\theta$ then Bob's strategy
will reply with a strategy in $V_\theta$. So by the stationary
assumption of the theorem, there is a $\kappa$ of countable
cofinality that is closed under the strategy. Alice can now play so
as to collapse more and more of $\kappa$, and Bob will always reply
with a poset below $\kappa$. So the limit forcing will again be the
collapse of $\kappa$, which forces $\sigma$. So Alice can defeat
this strategy.
$\Box$
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.
Theorem. From suitable consistency assumptions, it is
consistent with GBC that the CH game is not determined with respect
to class strategies.
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.
$\Box$
I guess the argument isn't just about CH, but rather any statement $\sigma$ such that there is a stationary/co-stationary class of $\kappa$ of countable cofinality such that $\sigma$ holds after the collapse of $\kappa$. In such a case, neither player can have a winning strategy.
The ideas appear to culminate in answer to question 1.
Theorem. The following are equivalent for any statement $\sigma$.
- Alice has a winning strategy in the $\sigma$ game.
- 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 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.
$\Box$
Best Answer
I discussed this with Arvind Singh a while ago and I think we can show the non trivial inequality $p_{opt}\leqslant\frac{3}{8}$ with simple arguments. The proof relies on the symmetry of the problem and the intuition is that one can not find a strategy wich is good simultaneously for a configuration and its inverse.
It will be simpler to work with the sets of indices such that the coin is on $1$: $$A=\{1\leqslant i\leqslant n : A_i=1\},\quad B=\{1\leqslant i\leqslant n : B_i=1\}.$$ We want to bound $$G=\mathbb P (f_a(B)\in A, f_b(A) \in B).$$ Introducing the function $$g(A,B)=\frac{1}{4}\left(1_{f_a(B)\in A, f_b(A) \in B}+1_{f_a(B^c)\in A^c, f_b(A^c) \in B^c}+1_{f_a(B)\in A^c, f_b(A^c) \in B}+1_{f_a(B^c)\in A, f_b(A) \in B^c}\right),$$ we get by symmetry (since for example $A^c,B$ has the same law than $A,B$): $$G=\mathbb E [g(A,B)].$$ But there are some incompatibilities in $g$: the first term and the third term can not be both equal to $1$ since one contains $f_a(B)\in A$ and the other $f_a(B)\in A^c$. The same thing applies for the second and the fourth. Thus $g(A,B)\in\{0;\frac{1}{4};\frac{1}{2}\}$ almost surely.
On the event $E_1=\{f_b(A)\in B, f_b(A^c)\in B\}$, only the first and the third term can be non vanishing and since they are incompatible $g(A,B)$ is at most $1/4$ (in fact it is equal to $1/4$). Besides, by first conditionning on $A$ we see that $E_1$ is of probability at least $1/4$ (the probability that $B$ contains (one or) two elements).
The same applies to $E_2=\{f_b(A)\in B^c, f_b(A^c)\in B^c\}$. If we consider the event $$E=\{f_b(A)\in B, f_b(A^c)\in B\}\cup\{f_b(A)\in B^c, f_b(A^c)\in B^c\}.$$ we have built an event such that $g1_E\leqslant \frac{1}{4}$ and $\mathbb P(E)\geqslant \frac{1}{2}$ (the union is disjoint). Thus since $g\leqslant \frac{1}{2}$, $$G=\mathbb E[g(A,B)]\leqslant \mathbb E[g1_E]+\mathbb E[g1_{E^c}]\leqslant \frac{1}{4}\mathbb P(E)+\frac{1}{2}(1-\mathbb P(E))\leqslant \frac{1}{4}\frac{1}{2}+\frac{1}{2}(1-\frac{1}{2})=\frac{3}{8}.$$