# Are there infinitely many $\alpha \times \beta$ Chomp boards where player 2 wins?

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.

• 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

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.
• Thanks for the answer! So $\beta$ exists and is unique given $\alpha$ -- that is a nice result, and the explicit answers are nice as well. Could you give some idea of where you learned these facts, even if you don't remember the exact reference? – 6005 May 25 '17 at 14:48
• I think Huddleston had something to do with those results. Google Huddleston three rowed ordinal chomp and it should (eventually) take you to the results. I've verified both of them personally, but not rigorously enough to be a formal proof. I've also tried the five row case, but the situation there is vastly more complicated. You can also consider three dimensional chomp, which has only one known result thus far, that $2 \times 2 \times \omega^3$ is a losing position. I'm currently working on figuring out if $3 \times 3 \times \omega$ loses, but it's a huge and incredibly complicated task. – Thomas Blok May 27 '17 at 5:28