Head Up Theory Question

[reubenf]

Uh, wanna play? Any stakes.

[Dissident]

I dont fold at all to the one outer. if unlimited money I would say 999,999 bets
"a dissident is here"

[catlover]

In order to attack this problem, we first need to attack the problem of how much confidence we need to have that our hand is good to make it correct to put in another raise.

The answer is certainly not that you raise if your hand is at least 50% likely to be good. If your hand is 50% likely to be good, you should just call. The reason is that if your hand is good, you will win 1 extra bet at most. But if your hand is no good, it will cost at least 1 bet and may cost 2 bets.

I can do this problem in the case where the pot is large enough that bluffing and folding are both pretty much out of the question. This won't quite answer David's original question, but at least it's a start.

The math isn't too bad, and it turns out the answer is that you should raise if the probability you are beat is less than or equal to (square root of 2) - 1. This is about 41%.

So in this simplified case, you should put in a raise if you are 59% confident that your hand is good.

Unfortunately, the question David actually asked is a lot harder. The problem is that bluffing and folding both come into play. I tend to think that these factors both make value raises a little less frequent. The reason is that when you raise, sometimes you will win 0 bets instead of 1 when your opponent folds a losing hand. But the math for finding the exact answer gets really messy.

[uphigh_downlow]

Decided to summarise and clean up the impromptou thining from my posts last night. Here goes an attempt.

MAIN PREMISES:
1)We can determine a value Xo, such that when in an unopened pot(just the antes), we open with the top Xo% of our hands, the opponent cannot exploit this.
2) Since we are playing against an expert player, we are trying to basically not give him any edge. Over a long run, only the player with a higher precision calulator will win [img]/images/graemlins/laugh.gif[/img]

Gameplay is still the same, but we look at a raise as an "Opening bet" which follows a call. So for example if you call, then you get a chance to open again.
This reduces the problem, to determining when you should open. I ignore bluffing, since the actual aim is to find out the number of max raises that we can call, and the possibility of these actions will likely not affect our answer by more than one or two bets.

Lets say after the antes are in your opponet has opened, which has now created a $p pot. You can call/fold. If you call, you will have the option of opening again. Also you have reason to believe that the opponent has a hand, that is randomly in the top X% of total possible hands.

Now when do you call? We'll only fold hands that fall in a certain bottom% of your opponents possible holdings for break-even play here. That is not too hard to determine. For example, if you think your oppnent has top 75% of hands, you can maybe call with top 80% of his range of hands, given the right pot-odds. This would be top 60% of all possible cards. In this case, we fold all hands in the bottom X/(p+1)% of his hand range.

Assumption is that once you put your oppnent on a range, you can assume that the cards outside of the range did not even exist in the deck.

If you decide to call, then you should open the pot(raise) again, with a holding that is in O(X,p)% of possible hands.
I define O(X,p) == X * f(p)
Notice that Xo is just a special case of O(X,p)

An example, if you think coming out betting is fine with 75% of top hands when the pot offers P:1, then if the pot still offers P:1, and you think the opponet has top 75% hands, then you should come out raising with top 75% of 75% (~56% of hands)

However the pot odds have changed, so you need to modify the range of raising hands slightly here, which I do with f(p). As p grows higher, f(p) gets closer to 1, and gradually becomes almost meaningless.

Now if the opponent comes back raising over the top, he must have a hand that is randomly spread out inside O((O(X,p),p+2)

This is X*f(p)*f(p+2).

This is a recurive relation and I came up with a few equations, but that might make this post too boring.

Well now you know which hands to fold with(although this range becomes very small very quick). ANd if you call, then you must reopen the pot, with a hand range as desribed by the recursive relation.

Anyway, if we can determine an Xo, as mentioned in the main premise, we can know when to raise again.

After having thought abbout it for a little but, I'm quite confused if it would actually be possible to determine such an Xo.

It seems that I would have to do this by hand, which is too cumbersome. Can anyone point out a technique to determine this?

Talk to me.

[PairTheBoard]

This is getting almost cruel watching peope trying to develop game theory from scratch. Will someone who knows how the well developed theory of games works please do this and put us out of our misery.

PairTheBoard

[M.B.E.]

What if the ante, instead of $1, were $10,000,000 (but bets and raises were still in increments of $1).

[catlover]

The problem is that even if you do know game theory, David's questions are not easy. Especially the ones about limit poker.

[felson]

David's question is slightly different, but I think this link is relevant, especially part 3.

http://tinyurl.com/4nelh

Regarding the original post in this thread, I think the halving approach is basically the right strategy. Except, you need to raise with less than the top half of your hands, because of the risk of getting reraised. So you need to be more selective than just halving, which means fewer than 20 raises.

[uphigh_downlow]

[ QUOTE ]
What if the ante, instead of $1, were $10,000,000 (but bets and raises were still in increments of $1).

[/ QUOTE ]

I purposefully made no refernce to any numbers.

It seems that there would be a lot more value betting, value calling and raising for value.

So Xo would be really high. And so, the reraising would go on for a really long time, each hand.

Can you clarify, where my strat breaks down, if antes were much higher than betting rounds.

I guess I will be reading about the [0,1] game tonight. mebbe that will give me a better perspective

[parappa]

Ok, I have read the thread and understand why something around 20 is much more likely to be right than the obvious 499,998 or so.

However, what I don't understand is whether the "correct" answer assumes that our opponent never overvalues his hand, or whether that doesn't matter. Most common-sense (incl me before I started thinking about this) would have you simply putting _at least_ 499,000 bets in. Some people would probably put in 999,998 bets, because there is an (although incorrect) common sense basis for doing so.

Are we to amend the original question to assume an opponent who won't overvalue his hand and plays correctly? If not, I think that the chances of our opponent overvaluing his hand and giving us too much action might just overwhelm the "correct" game-theoretical solution, because the basis for the halving method would not exist (i.e, you can't narrow down his range of holdings as quickly as you think you can if he's going to overvalue his hand, no?).

Like most of these Sklansky stumpers, I have gone from glancing at it and saying "silly question" to begging for someone to straighten my thinking out. [img]/images/graemlins/confused.gif[/img]