Searching for theorem about mixed strategies

[punter11235]

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I see. So there are multiple ways to break even against someone playing the optimal strategy.

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If optimal strategy is mixed there are infinitely many optimal counter-strategies which do equally well against it.

[alThor]

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There is a following theorem :
"In any 2 person game if one player is using mixed strategy A there exist pure strategy B which is optimal vs strategy A".
...This theorem is known as Yao Lemma in computer science but its stated for probabilistic algorithms. I need a source (or even better a short proof) with this theorem stated for general situation.
Can anybody help ?

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If this is a finite game (finite number of strategies) it is trivial: each strategy gives you a payoff, and one of them has to be best. There is nothing to prove here.

OTOH if it is not finite, you cannot prove anything without further assumptions on the payoff functions (e.g. continuous payoffs on compact strategy space; then, just use something like Weierstass's Theorem).

P.S. I'm not sure this has anything to do with Yao's lemma.

[mykey1961]

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No, Peter, the theorem is true for *every* strategy, including the optimal one, used by a non-responsive opponent. There *always* exists a pure strategy that achieves the maximum rate of return against him (though against the optimal strategy, it does only equally well as your own optimal strategy, not better.) That is one reason why optimal strategies are useful - they guarantee you a break-even performance even if you can't be bothered to study your opponent.

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I think some people confuse optimal with break-even.

If you are playing heads up holdem, and you're paying the big blind, an optimal strategy is not break-even.

[Shroomy]

mykey .. you stole my post. Optimal is not break even. It is just the strategy that if you told your opponent what your strategy was, the best he could do was to adjust to his optimal strategy, and if he did so, you would still not adjust your strategy. ( and your optimal strategy could still be -ev, just less -ev than other strategies.)

btw this is not the universally agreed upon definition for optimal strategy. Some would call it unexploitable strategy, and call the optimal strategy the best exploitive strategy for the actual strategy of the opponent.

So if the villain was playing optimal/unexploitable, your best exploitive strategy would also be your optimal/unexploitable strategy. (Nash's Equilibrium)

[Guruman]

a followup then so that I can be clear:

assume that player A is playing an unexploitable river bluffing strategy headsup against player B. Player B is unaware of this, but he ends up indifferent with his calls and folds because of player A's strategy. Player B doesnt raise without big hands.

Does this mean that all the following also occurs:

player A gets to valuebet with impunity unless player B starts raising lighter?
player A gets to watch for patterns in player B's river decisions and can eventually switch to a pure strategy?
there is no exploitative mixed strategy that will win more for player A than the unexploitable one even if B is playing in an exploitable fashion?

sorry, still wrapping my head around the exploitative ramifications of unexploitability. [img]/images/graemlins/smile.gif[/img]