In common-payoff games the denominator is not zero, in general. For example, suppose that SA=SB={a,b}, uA(a,a)=uA(b,b)=1, uA(a,b)=uA(b,a)=0, uB≡eA, α=β=δa. Then aB/A(α,β)=1, as expected: current payoff is 1, if B played b it would be 0.
You’re right. Per Jonah Moss’s comment, I happened to be thinking of games where playoff is constant across players and outcomes, which is a very narrow kind of common-payoff (and constant-sum) game.
I don’t think in this case aB/A should be defined to be 1. It seems perfectly justified to leave it undefined, since in such a game B can be equally well conceptualized as maximally aligned or as maximally anti-aligned. It is true that if, out of some set of objects you consider the subset of those that have aB/A=1, then it’s natural to include the undefined cases too. But, if out of some set of objects you consider the subset of those that have aB/A=0, then it’s also natural to include the undefined cases. This is similar to how (0,0)∈R2 is simultaneously in the closure of {xy=1} and in the closure of {xy=−1}, so 00 can be considered to be either 1 or −1 (or any other number) depending on context.
In common-payoff games the denominator is not zero, in general. For example, suppose that SA=SB={a,b}, uA(a,a)=uA(b,b)=1, uA(a,b)=uA(b,a)=0, uB≡eA, α=β=δa. Then aB/A(α,β)=1, as expected: current payoff is 1, if B played b it would be 0.
You’re right. Per Jonah Moss’s comment, I happened to be thinking of games where playoff is constant across players and outcomes, which is a very narrow kind of common-payoff (and constant-sum) game.
I don’t think in this case aB/A should be defined to be 1. It seems perfectly justified to leave it undefined, since in such a game B can be equally well conceptualized as maximally aligned or as maximally anti-aligned. It is true that if, out of some set of objects you consider the subset of those that have aB/A=1, then it’s natural to include the undefined cases too. But, if out of some set of objects you consider the subset of those that have aB/A=0, then it’s also natural to include the undefined cases. This is similar to how (0,0)∈R2 is simultaneously in the closure of {xy=1} and in the closure of {xy=−1}, so 00 can be considered to be either 1 or −1 (or any other number) depending on context.