[Paxinor]

@ DL: the positional advantage does not count in a nash equilibrium. there is basicly none, because the positional advantage is informational.

as you can see in the [0,1] games von neumann introduced in 1944 (those were as we know picked up by several pokerauthors) , position never matters.

there are some issues with impicit collusion though (in a 3+ player game) but let's just skip that as its kinda irellevant to the discussion right now.

@ question: basicly you will be not beatable if you play in a nash equilibrium which poker has (its been proven by nash himself) no matter what your opponents strategy is.

this strategy will be what we call in gametheory "mixed". meaning that you won't play a specific hand on a specific flop always similar. you might raise it 50% or might just call 50%. similar to roshambo where you dont always pick "rock"

it means that if a computer figures out a optimal strategy, he plays perfect poker. he will not maximize EV against every opponent but minimize his losses. he will be >= 0 against every player in the world.

but don't worry right now, its way to complicated to calculate the nash equilibrium even in limit HU poker

you might want to look in the "bot" thread in this forum there are further explanations...