simple stuf, beyond me.

[Oink]

Not trying to be a smartass or anything (well maybe a little), but

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Most (all?) zero-sum games have only one optimal solution (proof exists in game theory text books).

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Whether the game is zero-sum or not doesnt affect whether there exist optimal strategies for the players and/or whether there exist Nash-equilibria.

1) If a player's strategy set is finite, as it is in limit hold em, there will always exist at least one optimal strategy for a player given his opponents choice of strategies.

2) If the strategy set is not finite, as it is in NL holdem, there doesn't necessarily exist optimal strategies. However, if the strategy set is compact and convex and the payoff function is continuous, optimal strategies exist.

If by "optimal solution" you mean Nash-equilibria , Nash has shown that in any finite game (like limit hold em) there exist equilibria with mixed strategies.

Oh, and if you are really interested in game theory: Any n-player NON zero-sum game can be reduced to a n+1 player zero-sum game, where the set of equilibria in the non zero-sum game is equivalent to the set of equilibria in the zero-sum game. So the fact that the game is zero-sum really hasn't any implications.


Sorry for the Hijack, just responding to the question mark in Heisenberg's reply.

[Heisenb3rg]

Thanks for clearing that up.. Wasn't sure if it was at least one or exactly one. Think my statement is technically correct though , even though redundent :P

NL's strategy set isn't finite?
Aren't you limited by the size of your stack?

[Oink]

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NL's strategy set isn't finite?
Aren't you limited by the size of your stack?

[/ QUOTE ]

Yeah but if you can choose any number between 0 and your stacksize, your strategy set is a continuum and thus infinite.

Come to think of it, most poker sites doesnt allow you to choose any number but restrict the number of decimals. So yeah, the strategy set in NL is finite as well.

[Guy McSucker]

[ QUOTE ]
Thanks for clearing that up.. Wasn't sure if it was at least one or exactly one. Think my statement is technically correct though , even though redundent :P


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No, because your statement says "only one". There's at least one. There may be more.

Nit successfully picked.

Guy.