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Wynton
08-01-2005, 03:12 PM
Here's a modest little question to keep the math people out there occupied.

Assume that players can be categorized in three ways: (1) loose or tight preflop; (2) passive or aggressive preflop; (3) passive or aggressive post-flop.

Now consider all the possible starting hand combinations.

At a table with six opponents, how many possible situations can arise?

And for extra credit, answer the same question at a full, 10 person table.

(By the way, I haven't a clue myself.)

AaronBrown
08-01-2005, 04:11 PM
There are eight types of players. With n opponents, there are 8^n possible combinations of types, if position matters. That's 262,144 with six opponents and 134,217,728 with nine.

Wynton
08-01-2005, 04:13 PM
Ok, that's a start. But I was also asking you to take into account all the possible starting hand combinations for the player types too.

AaronBrown
08-01-2005, 05:09 PM
Well, there are 169 starting hands. It's true that not all combinations are possible, three players cannot hold AA for example. But it's still pretty close to true that there are 8*169 = 1,352 possible situations per opponent. Raise to the power of the number of opponents. It will be a big number.