For a simplified example, in a game with the only options all in or fold, and the only hands dealt being A7, A8, A9, AT, AJ, AQ, AK. If you pushed all in with AT or higher and your opponent could see your card, when you had AT, he would call with AJ AQ and AK
But what I’m trying to find is not the optimal for “face up strategy”, but instead the face down, assuming the opponent only knows how you play (what you push all in with and not the specific hand). If this was the case, and you pushed with AT and higher, your opponent would have to have AQ or AK to make sure that he had the advantage. If you pushed with A8 or higher your opponent would call with AJ, AQ, and AK.

One “optimal” strategy assumes your opponent KNOWS what you have but you don’t know your opponent, the other assumes your opponent KNOWS HOW you play, but not what you have.

I’ve been playing around with poker stove, trying to figure out some scenarios, but since I don’t know as much as I would like, it’s more of a trial and error, and I don’t get very far.
Basically, I have figured out that in the small blind if you shoved with any 2, you would lose over .8 big blinds per hand against an “optimal opponent”
If you shoved with any pair, any broadway, any ace,

If you could email me and let me know, I would greatly appreciate it. I’d be interested in an “all in or fold” strategy for cash games where everyone has 20 times the big blind, but if you could just show me an example of how to figure it out, that’d be great.

Thanks,
Mike

Since two-pair and three-of-a-kind each lose to a straight, your opponent has 8 cards out of 44 possible cards that will lead to a win against 12 of the wins that I wrote out above. I suppose I could make the hands more like this:

8♣ 8♦ 6♦ 3♦
K♦ Q♦ J♦ 10♦

That would be lead to a loss without using game theory bluffing: Out of 44 times, you’d win 12 (any A, 9 or ♦) times, and 12 times out of 44 (i.e. you get a high pair), you’d beat all but 8 of your opponent’s possible cards (any 6, 8 or 3). Overall that would be (12+12*(36/44))/44 times, which is a losing proposition with any fixed strategy, and yet within striking distance for an optimal bluffer. It would probably be a confusing example for a non poker-player, though.

Furthermore, you don’t consider an opponent who also optimally bluffs (overcalls), who in this case will win on average.

But, back to the discussion at hand… I was thinking that it would be an interesting project to create a game that demonstrates game theory without the complicated rules of poker. I’ll call it (for lack of a better name) “Matt’s Game Theory Game”. Here’s how it goes:

The game uses a deck of 101 cards, each containing a whole number between zero and 100. Each player is dealt 1 card face down. The player to the left of the dealer posts a blind bet. Play continues as in normal poker, with each player given the option to fold, call, or raise. Highest card wins.

The advantage to this game is that you can accurately compute the odds of winning, just based on the card you hold. For example, if you’re playing just one opponent (heads up), the percentage chance of winning (against a random card), is equal to the number on your card. If you hold a zero, you have no chance of winning a show-down, a 100 will always win, and a 50 wins half the time.

The probabilities of winning show-downs only stays easy against one opponent. Against two opponents, you expect to win half the time with a 67; against three, you win half the time with a 75 ( = 100 * N / (N + 1), where N is the number of opponents).

Anyways, I’ll see if I can find some time to put up an algorithm that can optimally play this game “heads up”. I’m still deep into learning Ruby, so that might be where I start.