Here is an amusing concrete non-determined game, under the assumption that the dependent choice principle fails.
Assume DC fails. This means that there is a set $X$ and a binary relation $R$ on $X$, such for every $x\in X$ there is $y$ with $x\mathrel{R} y$, but there is no sequence $\langle x_n\mid n\in\omega\rangle$ with $x_n\mathrel{R} x_{n+1}$ for all $n$. In other words, the tree of finite sequences that accord with $R$ has no leaves — every node can be extended one more step — but there is no way to iterate these steps and the tree has no infinite branch.
Consider the game on $X$ where player I plays a point $a$ from $X$ and player II responds with $b$, and player II wins just in case $a\mathrel{R} b$; otherwise player I wins. So the game is over very quickly, after just one move for each player. (If you insist that games should have infinitely many plays, then you can continue with infinitely many irrelevant moves.)
Player I obviously cannot have a winning strategy, since whatever point $a$ is played, there is a way for II to defeat it by playing some $b$ with $a\mathrel{R} b$.
But player II also cannot have a winning strategy, since any such strategy would provide a choice function on the $R$-successors of the points in $X$, and any such function could be iterated to thread $R$, whereas there is no such infinite thread.
Since the game is an open game (in fact clopen), this argument shows that one cannot prove open determinacy without DC. The usual proof of open determinacy uses DC, when showing that if a position does not have an ordinal rank, then the closed player can avoid losing by maintaining rank. But in order to do so, that player needs DC to find the infinite thread of rank-maintaining moves.
Update. As noted in wojowu's comment below, this example can be adapted to prove the following theorem.
Theorem. ZF proves that there is a non-determined set. Specifically, in ZF we can prove that $\text{AD}_{P(\mathbb{R})}$ fails.
Proof. First, let $F$ be any family of non-empty sets, and consider the game where player I selects some $A\in F$ and player II selects some $a\in A$, and the first player to violate either requirement loses. Clearly, there can be no winning strategy for player I, since if such a strategy directed player I to play a particular $A$ in $F$, then it would be defeated by the strategy for player II to always play some particular element $a\in A$. Conversely, observe that if player II has a winning strategy for the game, then we would have a choice function for $F$. So any violation of AC leads immediately to a non-determined game. In particular, if choice fails for families of nonempty sets of reals, then we get a violation of $\text{AD}_{P(\mathbb{R})}$. On the other hand, if choice holds for such families, then the reals are well-orderable, and we may construct a non-determined set for binary play. So in either case, $\text{AD}_{P(\mathbb{R})}$ fails. QED
Regarding the remarks below about constructing an "explicit" game, one can easily do this as follows.
Corollary. In ZF we can write down an explicit definition of a non-determined set.
Proof. Consider the game where player I chooses either (1) a well-ordering of the reals, and then the rest of the payoff set is as would be constructed for a non-determined set by the usual argument given such a well-ordering; or (2) a family $F$ of nonempty sets of reals having no choice function, with player II then picking an element of the family and player I playing an element of that set, as in the theorem (but with the players reversed). This gives altogether one big payoff set, explicitly defined.
Note that player I cannot have a winning strategy, since he cannot have a strategy for the resulting play in case (1), as the game is undetermined, and he cannot have a strategy that wins by placing him in case (2), since any such strategy would have to provide a choice function for $F$.
Conversely, player II cannot have a winning strategy, since player I can play either a family with no choice function, in which case it is hopeless for II to play with particular element of the family, or else AC holds for all such families in which case player I can play a well-ordering and a non-determined game, for which player II can have no responding strategy. QED
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
As a supplement to Joel's answer, you may want to look at this nice paper of Bollobas, Leader, and Walters concerning continuous games. As a starting point they discuss the classical Lion and Man game introduced by Rado. In this game there is a lion chasing a man inside the unit disk. Both have identical maximum speeds. The lion wins if he catches the man, and the man wins if he is never caught by the lion. If the lion chooses to always run directly toward the man, then he will get arbitrarily close to the man, but never catch him. On the other hand, if the lion instead moves at top speed so that he is always on the radial vector from the centre to the man, it was 'clear' that this was a winning strategy. Proof: without loss of generality, the man stays on the boundary of the disk. However, in 1952, Besicovitch exhibited an ingenious winning strategy for man! Thus, staying on the boundary is with loss of generality for man. Nonetheless, one can ask the perplexing question if lion also has a winning strategy? In this particular game, it turns out that the answer is no. But by changing the metric space, Bollobas, Leader, and Walters prove that there are games in a similar vein where both lion and man have winning strategies!