How to win some simple iteration games (Q5961723)

It was shown by \textit{D. A. Martin} and \textit{J. R. Steel} [J. Am. Math. Soc. 7, 1-73 (1994; Zbl 0808.03035)] that II has a winning strategy in the weak iteration game (of length \(\omega_1+1)\) on a countable \(M\) elementarily embeddable in some \(V_\alpha\). The result was used by \textit{J. R. Steel} [Ann. Pure Appl. Logic 65, 185-209 (1993; Zbl 0805.03043)] to obtain a comparison theorem for countable tame premice. It is an open problem whether II has a winning strategy in the (more general) full iteration game of length \(\nu\) on \(V\) for all \(\nu\). The authors introduce two new games: \({\mathcal G}\) (of length \(\omega_1+1)\), which is stronger than the weak iteration game, and \({\mathcal G}^+\) (of variable countable length), which is stronger than \({\mathcal G}\) but weaker than the full iteration game of length \(\omega_1\). In fact, \({\mathcal G}^+\) is played like the full iteration game except for the fact that player I must also play distinct natural numbers on the side. The game is over once I runs out of integers (provided neither player has lost by that time). It is shown that for a countable premouse \(M\prec V_\alpha \), a) II has a winning strategy in the game \({\mathcal G}\) on \(M\), and b) I has no winning strategy in the game \({\mathcal G}^+\) on \(M\). Each of these results yields a comparison theorem for non-tame mice. (According to the authors, \({\mathcal G}^+\) should ensure enough iterability to prove a comparison theorem for inner models with many strong cardinals overlapping Woodin cardinals, but would not suffice for hypotheses like a Woodin limit of Woodins). It is pointed out that by a result of Neeman, if there is a Woodin cardinal \(\lambda\) above a cardinal \(\kappa\) that is \(\lambda+1\)-strong, then for all countable \(M\), the game \({\mathcal G}^+\) on \(M\) is determined.