Re: Game Theory question

> The better for players 1 and 2 is to play the same strategy, getting an
> average pay-off (-1/2).

Yes, that's exactly what I got as well. Actually, that's the same game
as the game between only 1 player and player 3. In that case the optimal
(mixed) strategy for player 1 is the entire domain of mixed strategies
lambda*S1 + (1-lambda)*S2 (lambda \in [0,1]) because the pay-off
will be -1/2 anyway.

Okay, it seems this problem is a bit ill-defined, I left out one more bit,
which is that player 1 knows exactly the outcomes of all the games
ever played, and while player 2 doesn't, they could take advantage of
this (ie., cooperate) to maximise their pay-offs...


.

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