Let $\alpha$ and $\beta$ be nonzero ordinals. Infinite chomp (called ordinal chomp by Wikipedia) on an $\alpha \times \beta$ board is played as follows. We consider the set $\alpha \times \beta$, partially ordered under the natural ordering: $(x,y) \le (z,t)$ iff $x \le z$ and $y \le t$. On your turn, you choose an uneaten square of chocolate $(x,y) \in \alpha \times \beta$ and "eat" all squares $(x',y') \ge (x,y)$. The square $(0,0)$ is poisoned, so the person who eats it loses.
This game can only progress a finite number of moves, so one player must have a winning strategy. For finite ordinals $\alpha$ and $\beta$, not both $1$, it is well-known that player 1 wins. On the other hand, player 2 wins for the $1 \times 1$ and $2 \times \omega$ boards.
Question: Are there infinitely many pairs $\alpha, \beta$ where player 2 wins on an $\alpha \times \beta$ board?
Bonus: Classify as many boards as possible where player 2 wins.
Additional observations:
- If $\alpha$ and $\beta$ are both successor ordinals, the well-known strategy-stealing argument applies: If player 2 had a winning strategy, player 1 could eat the largest chocolate and then copy player 2's supposed strategy. So player 1 must win.
References:
My question was motivated by this answer, in which a winning strategy is presented for player 2 in $2 \times \omega$ chomp, and for player 1 in $\alpha \times \omega$ chomp which works for any ordinal $\alpha > 2$.
This article by Ian Stewart covers finite Chomp and the $2 \times \omega$ case.
$\alpha \times \beta$ chomp is a special case of a poset game, namely the poset game on the poset $(\alpha \times \beta) \setminus \{(0,0)\}$.
This Wikipedia reference could be helpful:
p. 482 in: Games of No Chance (R. J. Nowakowski, ed.), Cambridge University Press, 1998
A google books link is here, but I cannot find electronic access to the book's contents.
Yes, for every nonzero ordinal $\alpha$ there is an ordinal $\beta$ such that $\alpha \times \beta$ is a second player win (I can't find the source sorry). So far the only pairs that are known are $1 \times 1$ (trivial), $2 \times \omega$, $3 \times \omega^\omega$, and $4 \times \omega^2$.
The situation with five or more rows gets incredibly complicated, and it seems likely that the $\beta$ for 5 rows is a lot bigger than $\omega^\omega$, although bounds could be found by constructively applying the proof of existence.
It seems (from a very rough first guess) that even numbers will have a much smaller $\beta$ than odd numbers.