This question is related to a previous question of mine:
Given any set A of sequences of natural numbers, every strategy (no matter for which player) is either winning (W), or losing (L), or neither(N) for A.
So depending on A, the set of all strategies available for any of both players can be, a priori, of any of the seven kinds: W (all winning), L (all losing), N (all neither), WL (some winning and the rest losing), WN (some winning and the rest neither), LN (some losing and the rest neither) and WLN (some winning, some losing, and the rest neither).
This makes a total of 49 situations (now taking into account both players). Of course, not all of them are possible because we have the following restrictions:
a. If one player has a winning (losing) strategy, the other one cannot have a winning (losing) strategy. b. If one player has only winning (losing) strategies, the other one only has losing (winning) strategies.
I don't know of any other restrictions (not derivable from them, for example it follows that if one player is WL for A, the other one can only be N).
This leaves us with the following possible situations:
(I'd draw a table here, but unfortunately I don't know how to edit it; I tried html without sucess)
- W for I and L for II
- L for I and W for II
- N for I and N for II
- N for I and WL for II
- N for I and WN for II
- N for I and LN for II
- N for I and WLN for II
- WL for I and N for II
- WN for I and N for II
- WN for I and LN for II
- LN for I and N for II
- LN for I and WN for II
- WLN for I and N for II
My question is, could one find examples (prove the existence of subsets of sequences of naturals) for each situation only assuming ZFC?
Some are obvious, like the empty set for 2 or the set of all sequences for 1, or like the set of all sequences with a 1 in the odd positions for 8, but others may be not.