[jogsxyz]

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For two-player zero-sum games, optimal strategies always exist. To see this, just specify strategies more robustly, including actions with each hand type in each possible sequence (betting action and cards that have come). Then make a matrix of these strategies. Obviously at least one cell must be the minimax, and therefore optimal.

jerrod

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That may be true for a simple two player game.
On the turn the values of the river matrices
are unknown. How is a minimax strategy calculated
when the cells contain unknown variables? Each
player is forced to make a decision on the turn
based on variables which are functions of the
river card. Hold'em is too complex.
Let's use lowball as an example.

Lowball. 2 player game. Simplify the game even
further. Each player is dealt and redealt cards
til they hold either a pat nine or a one card draw
to an eight, which they must play. Player X draws
one. Would he really defend with the same fixed
strategy against a player Y who draws one or plays
pat? X should call with a king against Y who
draws one and some smooth nine against Y who
stands pat. This two player game is really four
separate two player games. X pat, Y pat.
X pat, Y draws 1. X draws 1, Y pat.
X draws 1, Y draws 1.
There are separate optimal strategies for each
of the four games. Or is this considered one
compete overall optimal strategy where both X
and Y are allowed conditional mixed strategies
dependent on both draws?

In hold'em there are four streets and four bets
per street. The relative values of the two
player's hands can change with every street.
This is more complex than lowball. None of the
cell entries in the other games have variables.
How does one minimax strategies based on variables
which are a function of cards dealt on later
streets? The hold'em game requires recursion
for the solution. My simple lowball example is
really four separate games. How many more games
would hold'em be?

Also if there were an optimal strategy for every
zero-sum two player game and headsup hold'em is
a single zero-sum two player game, that would
mean the headsup hold'em game is solvable.
That just doesn't seem likely.