Heads-up, one preflop bet only. Perfect plays.

[well]

Two players.
Both ante 1.
Both receive two cards from a full 52 card deck.

Player I either checks or bets a fixed bet b.
If player I checks, the board is dealt and the best hand takes the pot.
If player I bets, player II can either fold (give up the pot) or call.
In the latter case, the board will be dealt and the best hand takes the pot (2+2b).

Let A be the set of hands that player I will bet with.
Let B be the set of hands that player II will call with when bet into.

Let EV[A,B,b] be the value of the game for player I given the sets A and B and fixed bet b.

When player I raises all hands, there clearly is a set B* that minimizes EV.
Given that player II will call with such a set B*, there is a set A* that maximizes EV.
And so on...

Now with b=2 this method stabilizes quite rapidly, i.e. both players stop switching.
The EV will then be about 0.1515.
Here both player I and II play optimal strategies, right?

With b=3, b=4 and b=5 this is not the case.
The seqence of "improved" set starts alternating after some steps.
Let's call these sets A1, A2, B1 and B2.
If player II knows player I plays A1, the he will play B1.
But if player I knows player II plays B1, he will switch to A2.
As soon as player II realizes that, he will pick B2 to be his set.
When player I finds out, he will switch back to his original set A1.
And so on...

Now, my questions are:

- with a different startset, could a stable solution (always) be reached?
- could any information regarding optimal strategies be retrieved from the alternating sets?
- should I exclude hands from the sets that result in the least change of EV?
- my guess is that I should use mixed strategies, but then the algoritm fails to work. Is there an algoritm close to this one that will give proper results?
- how much of the optimal strategies will be mixed? Is it likely that both players take set they always play, and only one hand that they will play with a certain probability?


Thanks in advance,

Regards.

[danzasmack]

doesn't this game always favor player A?

[well]

It sure does.
If it wouldn't, player A could always just check and take the 0 EV.
Or only raise AA for slightly +EV.

[AaronBrown]

Yes, this always has an equilibrium. Let's start by simpifying the game. A and B each receive a number from a uniform distribution from 0 to 1. The betting structure is the same, except after a check or bet/call, the higher number takes the pot.

In your game each player can compute a probability of winning, but it has a more complex distribution than uniform 0,1. That complicates the formulae, but it doesn't change the basic idea.

Define P as (1+b)/(2+b), this is the percentage of time A will bet. These bets are split among her weakest and strongest hands. She will call if her number is less than (1 - P)b/(1 + 2b) or if it is greater than P + (1 - P)b/(1 + 2b).

B will call if his number is greater than P + (1 - P)b/(1 + 2b), the same as the strong hands A bets with.

A has an advantage of 1/2 + b/[2(2 + b)(1 + 2b)]. This is maximized when b = 1, which is the equivalent of pot limit. That's why pot limit is said to give the greatest advantage to aggressive play.

[well]

I am familiair with the [0,1]-games, did some research on it as well as you can find out by digging up the old posts about it.
The plan was to convert some of these games to hold'em hands, and see whether the [0,1]-results can in any way be applied to hold'em.
Also am I aware of the existence of equilibria, but it was more about the algoritm that worked for b=2, but failed for other values.
And things like the upper x part of the hands do no longer apply, as they depend on the set of hands your opponent plays.

So my question was about finding those equilibria rather than pondering about their existence.

I'll post when I found something new,

Regards.

[AaronBrown]

I apologize, I thought you were asking a simpler question.

There still is an equilibrium however. The difference between head's-up one-bet hold'em and zero/one games is the range of probabilities in hold'em go from something like 0.3 to 0.85 instead of 0 to 1.

A more subtle difference is the probabilities are not fixed, but depend on the hands your opponent will play. A pair of 7's has a small edge over suited Jack 10 if one of the sevens matches the suit. Suited Jack 10 has a small advantage over unsuited Ace 2. But unsuited Ace 2 beats 7's 70% of the time.

Despite the complications, there has to be a fixed point equilibrium. We can represent a strategy for the first player by the probability of checking for each of the 178 hold'em hands, the strategy for the second player is the probability of folding for the same set of hands. For every vector for the first player, there is a vector for the second player that maximizes expectation. Therefore, under the usual game theory assumptions, the first player knows the expected outcome for every vector, and can choose the maximum.

Of course, in real life, the second player does not know the first player's strategy and so the first player has to guess what he will do. With both players guessing, you can't predict the outcome.

[well]

I think the different equities of a hand against different sets of hands is not as subtle as you state it is.
The difference between JTs against different pairs of sevens should in my opinion not be considered. An optimal strategy does not distinguish this difference.
I agree that for each of your vector there are/is (a) best countervector(s).
But those vectors will - I think - never consist of numbers other than 0 and 1. Only don't cares could occur otherwise.
But the search is not for a perfect countervector as they are easily found. I called it "perfect" strategies, whereas I actually meant optimal ones.

O, and there are 169 hands to consider, not 178.

Looking forward for the next posts,

Regards.

[AaronBrown]

Sorry, of course you're right about the number of hands. I don't know where the 178 came from.

Non-transitivity in hand values can get you into a rock-paper-scissors situation. I agree it doesn't affect the existence of equilibrium, there is a 1/3, 1/3, 1/3 equilibrium in rock, paper scissors. But it does complicate the analysis.

I disagree that the optimal vector weights would be zero or one for all hands. That's true in the 0,1 game; if the hands were transitive, that would be true in heads-up one-bet hold'em as well (except for the trivial case of using an intermediate value to get an exact probability).

But suppose we had a six card deck with Win, Rock, Paper, Scissors and Lose. Win always wins, lose always loses and RPS play like the game against each other.

Clearly the first player bets with Win, as there is nothing to lose. He will also bet with enough of his RPS hands, the fraction depending on the bet size, in order to induce calls when he holds Win and his opponent holds RPS. But he'll have to randomize among RPS, he can't play one always and another never.

[well]

I think there's a slight misunderstanding here.
What I wrote was

[ QUOTE ]
I agree that for each of your vector there are/is (a) best countervector(s).
But those vectors will - I think - never consist of numbers other than 0 and 1.

[/ QUOTE ]

That means, a strategy that does best against a certain strategy.
I think the optimal strategy however, will be mixed. And that's why we can't use the approach I stated in the original post. All strategies found with that algoritm will be fixed.

Oh - and what was the sixth card?

Regards.

[AaronBrown]

I seem to have developed some kind of numeric dyslexia in this thread. Dysnumia? I can't keep my numbers straight. Five cards it is.

Now that (I think) I understand you, I think I agree.