[eastbay]

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What exactly do you mean when you say:

"...the house plays optimally in the game theory sense"?

John

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Here's a "practical" explanation:

The house is going to play the same strategy every time, that is, they will have set hands that they will call with, or at most, a set fraction of the time that they will call with any given hand.

Since we can watch the game, they know that we will eventually discover exactly how they play. They assume we are smart, and can find the best counter-strategy to however they play. So the best they can do is to find the strategy that minimizes the edge we can find by our best counter-strategy.

In game theory, this is called the "minimax" strategy, because it's the strategy that minimizes our maximum edge. It's the "optimal" strategy in the game theory sense.

Game theory tells us that such a strategy (that minimizes the edge we can get) in a 2-player, zero-sum game always exists.

In a 2-person, zero-sum game, the minimax strategy is the same as a "Nash equilibrium". A Nash equilibrium is a pair of strategies for which neither player can get any advantage by unilaterally changing his strategy. When playing the Nash equilibrium, if you do anything different, you do worse, and this holds true for both players. The Nash equilibrium is a deadlock.

At this deadlock point, the game may be fair, or it may favor one player or the other, depending on the rules of the game. The amount that one player benefits is called the "value" of the game, and when a game theory person talks about "solving" a game, they usually mean finding that "value" that exists when the two players are playing minimax aka Nash equilibrium strategies.

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