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[In any given head-to-head showdown, pocket aces have the best odds of not losing, at 85.5% and 3-2 offsuit have the worst odds at 35.3%.]

If the first player to act does not have a top 10% hand, he should fold them, and the next player's hand should be the best out of nine possible hands, or in the top 11.1%. If people continue to fold in this manner, we eventually arrive at the second to last person, whose hole cards should be in the top 50% of all possible poker hands (there are 169 possible hands).

Now, this is all supposing that every player is in fact playing against every other player. In other words, it is an ideal "game".

On the other hand, suppose that 2 of the players are husband and wife, and they only care whether their combined efforts are at least break even.

In this case, what is the correct percentage of hands that they should both play from first position? At first, I thought you should be able to double the percentage of hands to 20% to come up with the best result. But then I thought that they should be able to play even more hands than this because .2*.2=.04, or: two 20% probability events taken together equal one 4% event. In order to find the most optimal percentage in this spot, don't you do sqrt(.1)=.312, meaning that they should both play 31.2% of their hands?

Likewise, if either of them is in the second to last position (aka "small blind") in this table folding situation, shouldn't they both play sqrt(.5)=.707, or 70.7% of their hands?

I've been thinking about getting into some games for quite awhile now, but I've just been trying to understand the math before I get started. I've come to realize that it would be insane to try to make a living at it all on my own, given my low bankroll.

I think that I have a good handle on "ideal", non-cooperative game percentages, but now I'm just wondering how the dynamics change when 2+ people are on the same "team", as it were!

(If anyone is interested in getting in on this, let me know!)