NL HU all-in/fold game theory Q

[eastbay]

NL, HU hold'em, you're on the SB (every hand) and have only two options: push in or fold. Every hand starts with a specific stack size that both you and the BB have.

The "house" is on the BB, and the house plays optimally in the game theory sense.

The blinds are 200/400. What conditions do you require of the stack sizes to play this game against the house profitably?

eastbay

[eastbay]

[ QUOTE ]
NL, HU hold'em, you're on the SB (every hand) and have only two options: push in or fold. Every hand starts with a specific stack size that both you and the BB have.

The "house" is on the BB, and the house plays optimally in the game theory sense.

The blinds are 200/400. What conditions do you require of the stack sizes to play this game against the house profitably?

eastbay

[/ QUOTE ]

This was a hard problem. Here's my answer:

For deep money, this game is unfair to the SB. Below a critical stack size, the advantage shifts to the SB. This critical stack size is about 7.8 BB. The edge for the SB then increases to a maximum at about 4 BB, and then sharply decreases again.

#BB +/-chips for SB
12 -33.0
11 -25.9
10 -18.1
9 -10.1
8 -1.6
7 +6.7
6 +15.3
5 +22.5
4 +26.4
3 +20.6
2.5 +11.8
2 +3.8

At the breakeven point, the optimal strategies are:

sb push hands: 22+,A2+,K2+,Q4o+,Q2s+,J7o+,J2s+,T7o+,T4s+,97o+,95s +,87o,84s+,76o,74s+,64s+,53s+,43s (0.631976)

bb call hands: 22+,A2+,K2+,Q7o+,Q3s+,J9o+,J7s+,T9o,T8s+,98s (0.457014)

There are no mixed stratgies at this point, interestingly enough.

At the point of maximum edge for the small blind, the optimal strategies are:

sb push hands: 22+,A2+,K2+,Q2+,J2+,T5o+,T2s+,96o+,93s+,86o+,84s+, 76o,74s+,64s+,54s (0.737557)

bb call hands: 22+,A2+,K2+,Q2+,J2+,T5o+,T2s+,96o+,93s+,86o+,84s+, 76o,74s+,64s+,53s+ (0.740573)

Again, there are no mixed strategies, and moreover, within one hand they are playing the same strategy! I suspect this may point to a deeper principle of some kind, but I'm not sure what it might be.

So, my answer to the original question is: "I would insist that my stack be less than 3120 chips, and if I get to choose my stack, I choose 1600 chips."

eastbay

[HonestIago15]

For those of us who aren't at all sure how to even begin to solve a problem like this, could you explain your thought process?

[bigjohnn]

What exactly do you mean when you say:

"...the house plays optimally in the game theory sense"?

John

[fnord_too]

[ QUOTE ]
What exactly do you mean when you say:

"...the house plays optimally in the game theory sense"?

John

[/ QUOTE ]

He means that the house plays optimally against an optimal strategy by player. With a given stack size, player will have a specific range that he can profitably push. Given that range, house will properly call when it is profitable to do so.

[eastbay]

[ QUOTE ]
What exactly do you mean when you say:

"...the house plays optimally in the game theory sense"?

John

[/ QUOTE ]

Here's a "practical" explanation:

The house is going to play the same strategy every time, that is, they will have set hands that they will call with, or at most, a set fraction of the time that they will call with any given hand.

Since we can watch the game, they know that we will eventually discover exactly how they play. They assume we are smart, and can find the best counter-strategy to however they play. So the best they can do is to find the strategy that minimizes the edge we can find by our best counter-strategy.

In game theory, this is called the "minimax" strategy, because it's the strategy that minimizes our maximum edge. It's the "optimal" strategy in the game theory sense.

Game theory tells us that such a strategy (that minimizes the edge we can get) in a 2-player, zero-sum game always exists.

In a 2-person, zero-sum game, the minimax strategy is the same as a "Nash equilibrium". A Nash equilibrium is a pair of strategies for which neither player can get any advantage by unilaterally changing his strategy. When playing the Nash equilibrium, if you do anything different, you do worse, and this holds true for both players. The Nash equilibrium is a deadlock.

At this deadlock point, the game may be fair, or it may favor one player or the other, depending on the rules of the game. The amount that one player benefits is called the "value" of the game, and when a game theory person talks about "solving" a game, they usually mean finding that "value" that exists when the two players are playing minimax aka Nash equilibrium strategies.

eastbay

[eastbay]

[ QUOTE ]
For those of us who aren't at all sure how to even begin to solve a problem like this, could you explain your thought process?

[/ QUOTE ]

Yeah, but it won't be too enlightening. My thought process was "use a computer to solve the linear program defined by a statement of the minimax principle."

It may be possible to intuit an answer, at least approximately, but I don't know how. Otherwise, this is a fairly involved computational problem.

I wonder if Sklansky, who likes to keep computation at bay, has one of his hand-wavy arguments for this one.

eastbay

[bigjohnn]

Sorry, I dont think my post explained what I was getting at. I understand the concepts of Nash Equilibria etc (but I'm sure others will benefit from your explanation). I just felt that saying that the house played "optimally" was a little vague, especially for those who maybe arent as well versed in game-theoretic concepts. I thought maybe saying that the house was a "rational player" would have been a better way of putting it. I understood what you meant, but thought that others might benefit a different description.

John

[alThor]

[ QUOTE ]
At the point of maximum edge for the small blind, the optimal strategies are:

sb push hands: 22+,A2+,K2+,Q2+,J2+,T5o+,T2s+,96o+,93s+,86o+,84s+, 76o,74s+,64s+,54s (0.737557)

bb call hands: 22+,A2+,K2+,Q2+,J2+,T5o+,T2s+,96o+,93s+,86o+,84s+, 76o,74s+,64s+,53s+ (0.740573)

Again, there are no mixed strategies, and moreover, within one hand they are playing the same strategy! I suspect this may point to a deeper principle of some kind, but I'm not sure what it might be.

[/ QUOTE ]

I thought this might be some kind of interesting mathematical result, but now I can't see how it could be true that SB and BB play almost the same hands. I'll do some quick computations by rounding the numbers, and ignoring ties.

We're talking about when the stacks are around 4BB (I'll assume that includes the posted chips). Your strategies have the players playing roughly 3/4 of their hands.

Let's look at the BB's "worst" hand. Whatever it is, he should be indifferent between folding and calling the all-in when his pot odds leave him indifferent. This means he should have a 3/8 chance (37.5%) of winning across SB's range of hands, with that worst hand.

Look at SB's worst hand (whatever it is). He should be indifferent between pushing all-in and folding. Since BB folds 1/4 of the time, one can verify that his worst hand must have around a 11/24 chance (45.8%) of winning (based on elementary EV calculations).

Those two probabilities are so far apart, I have a hard time believing that both players would use the same range of hands in equilibrium. Which of us made a mistake somewhere?

alThor

[eastbay]

[ QUOTE ]
[ QUOTE ]
At the point of maximum edge for the small blind, the optimal strategies are:

sb push hands: 22+,A2+,K2+,Q2+,J2+,T5o+,T2s+,96o+,93s+,86o+,84s+, 76o,74s+,64s+,54s (0.737557)

bb call hands: 22+,A2+,K2+,Q2+,J2+,T5o+,T2s+,96o+,93s+,86o+,84s+, 76o,74s+,64s+,53s+ (0.740573)

Again, there are no mixed strategies, and moreover, within one hand they are playing the same strategy! I suspect this may point to a deeper principle of some kind, but I'm not sure what it might be.

[/ QUOTE ]

I thought this might be some kind of interesting mathematical result, but now I can't see how it could be true that SB and BB play almost the same hands. I'll do some quick computations by rounding the numbers, and ignoring ties.

We're talking about when the stacks are around 4BB (I'll assume that includes the posted chips). Your strategies have the players playing roughly 3/4 of their hands.

Let's look at the BB's "worst" hand. Whatever it is, he should be indifferent between folding and calling the all-in when his pot odds leave him indifferent. This means he should have a 3/8 chance (37.5%) of winning across SB's range of hands, with that worst hand.

Look at SB's worst hand (whatever it is). He should be indifferent between pushing all-in and folding. Since BB folds 1/4 of the time, one can verify that his worst hand must have around a 11/24 chance (45.8%) of winning (based on elementary EV calculations).

Those two probabilities are so far apart, I have a hard time believing that both players would use the same range of hands in equilibrium. Which of us made a mistake somewhere?

alThor

[/ QUOTE ]

Using blinds 1,2 and stacks 8:

For the BB:

EV_fold = EV_call
-2 = 8*pwin +(1-pwin)(-8)
pwin = 3/8

For the SB:

EV_fold = EV_push
-1 = 3/4[ 8*pwin +(1-pwin)(-8)] +1/4(2)
-1 = 3/4(16*pwin - 8) +1/2
-1 = 12*pwin -6 +1/2
pwin = 9/24 = 3/8

Did you maybe set the LHS for SB to zero rather than -1?

Moreover, 76o which I eyeballed as likely to be BB's weakest hand against SB's range, has an equity of .376.

eastbay