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HU cash, you always have AA. HU cash, you always have AA.

09-10-2009 , 02:52 AM
Yeah I'm going to keep a little quiet about global strategy for the most part from now on especially as other people's money is on me, and I've solved a few very significant situations. I will have a LOT to say about this game by the end of it, and also a decent amount on why this is a worthwhile exercise for our normal grind.

btw, as for that specific hand, I definitely should be betting more on flop/turn.
HU cash, you always have AA. Quote
09-10-2009 , 05:39 AM
Quote:
Originally Posted by TNixon
Where does this come from?
I got that in Mathematics of Poker, chapter 19 :


- They show that optimal bet-sizing is the one that satisfies geometrical growth of the pot, i.e. betting a same fraction x of the pot on every street, getting all-in by the river.
The size of the pot (when bets are called) then grows by 2x+1 on each street. So for the final pot to be 200, that gives you 2*(2x+1)^3 = 200.

- To find out the EV :

Player A = clairvoyant player (atc), Player B = bluffcatching guy (AA)
When player A plays GTO, player B is indifferent between the strategies :
(1) c/f
(2) c/c c/f
(3) c/c c/c c/f
(4) c/c c/c c/c

On the river, when player A bets x*pot, player B gets (x+1):x odds. Since he's indifferent between calling and folding that must be exactly the valuebet:bluff ratio in player A's betting range. So if A has y value hands, he'll have x/(x+1)*y bluffs, for a total of (2x+1)/(x+1)*y betting hands.

Similarly on the turn by the indifference between (2) and (3), the ratio (hands A bets turn & river with) : (hands that bluff turn and give up on river) is again (x+1):x. And A bets the turn with a total of ((2x+1)/(x+1))^2 *y hands.

Same on the flop by indiff. (1)-(2). So player A bets flop with ((2x+1)/(x+1))^3 *y hands.

Now to calculate the EV, since player B is indifferent between (1)-(4), we can actually assume he always chooses (1), c/f everytime on the flop. In that case player A just wins the pot with every hand that bets on the flop.
So player A's EV is (money in the pot) * ((2x+1)/(x+1))^3 *y.
(Except of course if ((2x+1)/(x+1))^3 *y > 1, in which case player A doesn't have enough bluffs, and just bets flop with his entire range and B c/f every time).
HU cash, you always have AA. Quote
09-10-2009 , 06:24 AM
aaaaaaaah the math nerd in me wants to correct errors I see in your application and add what a few correct statements imply but I still have to keep my math shut so that TNixon doesn't play too damn smart and make me have to think really hard. At any rate, just one time I want someone to post part of Chapter 19 of mathematics of poker and have it be followed by the michael jackson popcorn gif.
HU cash, you always have AA. Quote
09-10-2009 , 06:40 AM
Quote:
Originally Posted by mersenneary
aaaaaaaah the math nerd in me wants to correct errors I see in your application and add what a few correct statements imply but I still have to keep my math shut so that TNixon doesn't play too damn smart and make me have to think really hard. At any rate, just one time I want someone to post part of Chapter 19 of mathematics of poker and have it be followed by the michael jackson popcorn gif.
Would you mind telling me what they are by PM ? I promise I won't try to sell the info to TNixon
HU cash, you always have AA. Quote
09-10-2009 , 06:52 AM
Money sent.

Crush him mers
HU cash, you always have AA. Quote
09-10-2009 , 07:15 AM
Quote:
Originally Posted by checktheriver
Would you mind telling me what they are by PM ? I promise I won't try to sell the info to TNixon
TNixon wouldn't buy

I think you would want to post more about what I have to say afterward and I don't want to have to put a gag order on you or anything. I'll talk about this post specifically when I can, it's a very good one but it's missing a couple of things (imo, I could very easily be thinking about it wrong. there have already been like a million paradigm shifts in my analysis. The "when A plays GTO B is indifferent" general concept is great).

btw, both bettors have been shipped and both bets are BOOKED against TNixon. I will do my best to make them both winners.

Last edited by mersenneary; 09-10-2009 at 07:25 AM.
HU cash, you always have AA. Quote
09-10-2009 , 07:52 AM
Sorry any cliffs notes on how and when this is happening? Or I guess hands will be posted? (replayer?)
HU cash, you always have AA. Quote
09-10-2009 , 08:03 AM
Quote:
Originally Posted by RoSeeker
Sorry any cliffs notes on how and when this is happening? Or I guess hands will be posted? (replayer?)
TNixon wrote a program that works through AIM. Sejje operates it. 50 hands complete, I'm at +5.2 bbs. 300 hands will be played total for the purposes of this bet. Sounds like we'll be able to get another 50-100 in tomorrow. Wildly different opinions about optimal strategy and which side is the sucker side itt. fwiw, TNixon and I have come to the conclusion that it's much closer than both of us thought when we first eagerly got the bet down.
HU cash, you always have AA. Quote
09-10-2009 , 08:11 AM
Okay cool; will you be posting all the major hands ITT after the bet? Seems very interesting.
HU cash, you always have AA. Quote
09-10-2009 , 08:29 AM
At first, I thought this:

Quote:
Statement 1: It's +EV for AA-man given that AA-man is more intelligent and faster at picking up patterns in the oppenent's game and changing gears accordingly. It's -EV for AA-man given the other way round. Same intelligence, therefore, means neutral EV for both.

This is because:

Let's assume that AA-man is non-thinker and ATC-man is basic thinker. Then EV is + for ATC-man neutral since given this strategy:

When ATC-man is ahead of AA (~1/8th time or whatever) OTF he bets xBB such that he will be winning over loadsa hands. Since AA-man is non-thinker, he stacks all flops.

Now, let AA-man similar intelligence to ATC-man. AA will recognize ATC's strategy and adjust to will break even.
Etc etc...
So whoever is more intelligent will adjust faster and hence win.
But, then:

Given this information, then AA-man will not play an ATC-man more intelligent than him, and ATC-man will not play an AA-man anyone more intelligent than him.

Therefore, in theory, then only 2 ppl who would sign up for tis 'challenge' are 2 people who think they are just as intelligent as one another.

Hence, it is neutral EV for both AA-man and ATC-man, by Statement 1.

QED
HU cash, you always have AA. Quote
09-10-2009 , 08:49 AM
No.
HU cash, you always have AA. Quote
09-10-2009 , 08:58 AM
By your theory, no fish would ever sit vs a good player in poker...
HU cash, you always have AA. Quote
09-10-2009 , 11:44 AM
HEADS UP

If you are more intelligent than villain, heads up, you are ALWAYS favourite, imo. I don't see how this cannot be the case
HU cash, you always have AA. Quote
09-10-2009 , 01:27 PM
Quote:
Originally Posted by checktheriver
I got that in Mathematics of Poker, chapter 19 :


- They show that optimal bet-sizing is the one that satisfies geometrical growth of the pot, i.e. betting a same fraction x of the pot on every street, getting all-in by the river.
The size of the pot (when bets are called) then grows by 2x+1 on each street. So for the final pot to be 200, that gives you 2*(2x+1)^3 = 200.

- To find out the EV :

Player A = clairvoyant player (atc), Player B = bluffcatching guy (AA)
When player A plays GTO, player B is indifferent between the strategies :
(1) c/f
(2) c/c c/f
(3) c/c c/c c/f
(4) c/c c/c c/c

On the river, when player A bets x*pot, player B gets (x+1):x odds. Since he's indifferent between calling and folding that must be exactly the valuebet:bluff ratio in player A's betting range. So if A has y value hands, he'll have x/(x+1)*y bluffs, for a total of (2x+1)/(x+1)*y betting hands.

Similarly on the turn by the indifference between (2) and (3), the ratio (hands A bets turn & river with) : (hands that bluff turn and give up on river) is again (x+1):x. And A bets the turn with a total of ((2x+1)/(x+1))^2 *y hands.

Same on the flop by indiff. (1)-(2). So player A bets flop with ((2x+1)/(x+1))^3 *y hands.

Now to calculate the EV, since player B is indifferent between (1)-(4), we can actually assume he always chooses (1), c/f everytime on the flop. In that case player A just wins the pot with every hand that bets on the flop.
So player A's EV is (money in the pot) * ((2x+1)/(x+1))^3 *y.
(Except of course if ((2x+1)/(x+1))^3 *y > 1, in which case player A doesn't have enough bluffs, and just bets flop with his entire range and B c/f every time).
Ouch.

Now I wish I wouldn't have asked. Cause I had already figured out a lot of that.

Suffice it to say that pots are going to get a lot bigger, mers. :/ I was hoping to surprise you with it though, and be able to bluff more often the first few times.

god my metagame sucks. :/
HU cash, you always have AA. Quote
09-10-2009 , 03:07 PM
Quote:
Originally Posted by jewbinson
If you are more intelligent than villain, heads up, you are ALWAYS favourite, imo. I don't see how this cannot be the case
Because Stephen Hawking would be the favorite against Phil Ivey, amirite?
HU cash, you always have AA. Quote
09-10-2009 , 03:48 PM
Quote:
Originally Posted by bluemage55
Because Stephen Hawking would be the favorite against Phil Ivey, amirite?
Actually, given that Hawking does a few days/weeks study, I'm confident he could beat Ivey in a head's-up match. And by that I mean at least 200 games.
Bit unrealistic though

And what I meant to originally say was " if you are more intelligent at poker". So yeah
HU cash, you always have AA. Quote
09-10-2009 , 03:55 PM
Quote:
Originally Posted by checktheriver
I got that in Mathematics of Poker, chapter 19 :
- They show that optimal bet-sizing is the one that satisfies geometrical growth of the pot, i.e. betting a same fraction x of the pot on every street, getting all-in by the river.
This tilted me in MoP because I don't believe they provided a sufficient argument to demonstrate this. My complaint is that equity is hot/cold on the river but not on earlier streets. If we were to play a number game where each person gets a number between 1 and 100, and then we do three rounds of betting, this would make perfect sense to me and I'm pretty sure that given an hour I could prove that any other betting strategy performs worse in position. However, we're playing a number game where there's a third number generated in each round of betting that has a rewiring function on the other two numbers.
HU cash, you always have AA. Quote
09-10-2009 , 04:00 PM
Quote:
Originally Posted by jewbinson
Actually, given that Hawking does a few days/weeks study, I'm confident he could beat Ivey in a head's-up match. And by that I mean at least 200 games.
Bit unrealistic though

And what I meant to originally say was " if you are more intelligent at poker". So yeah
LOL at anyone beating phil ivey at this lifetime
HU cash, you always have AA. Quote
09-10-2009 , 04:03 PM
Quote:
Originally Posted by jewbinson
HEADS UP

If you are more intelligent than villain, heads up, you are ALWAYS favourite, imo. I don't see how this cannot be the case
But I thought the point of this particular game (AA vs ATC) and others like it (where interesting conditions or constraints are imposed on HU NLHE) is to figure out if the altered structure gives a potentially insurmountable edge to one side or the other. So poker intelligence comes into play if the difference between the two opponents is large enough to negate such an edge. Otherwise, the player on the side that benefits from said edge, if he's not sufficiently less intelligent than his opponent, will ultimately prevail.
HU cash, you always have AA. Quote
09-10-2009 , 04:07 PM
Quote:
Originally Posted by Des_Astar
LOL at anyone beating phil ivey at this lifetime
I'd work for 2 days, solve the game, then spend 3 days studying the charts for each board type, then sit there and destroy him. Anyone who thinks this is anything but a computing problem is mistaken.
HU cash, you always have AA. Quote
09-10-2009 , 04:13 PM
Quote:
Originally Posted by skates
I'd work for 2 days, solve the game, then spend 3 days studying the charts for each board type, then sit there and destroy him. Anyone who thinks this is anything but a computing problem is mistaken.
I am talking about standard HU game not this special version than any bot could crush
HU cash, you always have AA. Quote
09-10-2009 , 04:21 PM
Quote:
Originally Posted by skates
I'd work for 2 days, solve the game, then spend 3 days studying the charts for each board type, then sit there and destroy him. Anyone who thinks this is anything but a computing problem is mistaken.
Agree. The calculations are complicated, though, especially if AA chooses to play poker. I'm quite sure one side does have an unexploitable static mixed strategy that leads to it being +EV (and more +EV adjusting to opponent's responses), it's just difficult with all of the different boards to figure out and play correctly.

Yesterday I went from realizing something for ATC (namely overbets = good, I know TNixon knows this now) and thinking oh **** I'm screwed to realizing something else and liking AA's side again.

The bet is a little unfair with concern to the initial question because of the hand limit. TNixon and I have both figured out that AA gains more and more the shorter the match is. If I just keep it close over the next 200 hands I'll be the favorite.
HU cash, you always have AA. Quote
09-10-2009 , 06:11 PM
Quote:
Originally Posted by lagdonk
But I thought the point of this particular game (AA vs ATC) and others like it (where interesting conditions or constraints are imposed on HU NLHE) is to figure out if the altered structure gives a potentially insurmountable edge to one side or the other. So poker intelligence comes into play if the difference between the two opponents is large enough to negate such an edge. Otherwise, the player on the side that benefits from said edge, if he's not sufficiently less intelligent than his opponent, will ultimately prevail.
Well then, I don't think anyone has an edge due to the altered structured game, if this answers your query. This is because, imo, of the following:

Let's say you have a strategy which you think will give you an advantage over villain. You win the first 20 hands and this is the point villain knows, almost for certain, your strategy.

However, since play is ongoing, let's say that it takes 60 hands for villain to adjust his strategy to beating yours.

Now, given that you have the same poker intelligence as villain, it takes about 60 hands to adjust to villain's strategy. Etc.


There is no argument against it.
If you were to say:
"you could plan ahead so that if villain adjusts to our strategy by doing a, then we do b. If he does x, we do y. If then, he adjusts to our 2nd strategy, b or y, then we have 2 new options for his response to both b and y (not necessarily 2, this is an example). Then we can plan so far ahead that surely we are GARUNTEED to win."

But this goes against the idea that we are as intelligent as villain.
Any head's-up match involves adjusting to strategy and if you and your opponent are = intelligence, you only lose rake.


What I'm basically trying to say is that unless the problem involves every hand fixed e.g. AA vs 72o with both players knowing each other's cards, which involves very little strategy, then it is impossible to have a winning strategy.
Even if it were 72o vs ATC, given same intelligence, villain and hero would lose only to rake imo.

Last edited by jewbinson; 09-10-2009 at 06:20 PM.
HU cash, you always have AA. Quote
09-10-2009 , 06:20 PM
lol. You're not using math. Your argument works if the game is KK v. ATC, that game does not have the same value as AA v. ATC, therefore your argument is invalid.
HU cash, you always have AA. Quote
09-10-2009 , 06:28 PM
Great topic guys. I think the edge definitely goes to ATC in unlimited texas hold thems. Im sure you crazy math guys can probably solve this pretty easily but isn't much of the edge in bluffing? I mean you can't just raise with hands that beat aces, you have to mix in semi bluffs and bluffs as well. I think if it is limit hold em, aces would have an edge no?
HU cash, you always have AA. Quote

      
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