Head Up Theory Question

[FishAndChips]

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why would you only want to win 42 dollars plus the antes if you have the second-best card out of a potential million.

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Hey, I would love to win a million with this hand too, but the structure of the game prevents this. You can only win what your expert opponent will put in the pot with an inferior hand.

The nature of this game, when 2 experts play heads up, is that you won't have any pots bigger than ~20 bets. Neither will raise 500,000 times or more without the nuts.

I think people are thrown off by the large denomination of hand values. 999,999 is a big number, but the number of raises you put in with your hand is not equivalent to the number of hands you initially beat. Remember to raise for value you must be the favorite when called. Your opponent will not make the mistake of raising if he's not a favorite.

The reason the number of raises is "so low" is because each value raise divides your opponents range of hands in half (or close to it). Thus , you can not value raise nearly as much as some people hope to.

[FishAndChips]

[ QUOTE ]
I polled a bunch of my friends about this question. I just wanted their intuition. Most of them don't play poker, and the ones that do play it for fun on an intuitive level. They may or may not be winning players but they have large leaks which are obvious to anyone with a hint of a mathematical understanding of the game.

Anyway, 23 out of 25 said to put in a million bets. They are almost all quite intelligent people. The two who got it close have never played poker in their life. (My wife said to put in 20 bets, and another friend said to put in something like some constant multiple of some log of a million). The math grad students of the bunch, and those who I think could if they wanted get a graduate degree in math, all understood immediately why their intuition was wrong after I explained it. The intuitive poker players of the bunch all had a very, very hard time accepting that their answer was extremely wrong. There is an overlap of two people in those groups, and those people had a difficult time understanding at first but could see why they were wrong after some effort.

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Very interesting. The answer is counter intuitive, I guess. There has got to be a way to exploit this and make some money with some proposition bets. Any ideas? [img]/images/graemlins/cool.gif[/img]

[marv]

[ QUOTE ]
I dont usually post on forums, but I've found this debate so interesting that I thought I'd throw my ideas into the hat.

I only have an answer for the first part of David's question, not the pot-limit part. Furthermore, I dont claim that this is definately the correct answer- I've just tried to put my thoughts down. I'd appreciate any comments about it. Also, I am still not sure where people such as reubenf are getting the figure 20 from. They may well be right, but I've not seen a post detailing the full logic behind their answer. Again, it would be great if they could explain it.

And so to my answer:

Both players have a strategy, S1 for the player (us), and S2 for the expert we are playing against. As a player, we want to devise a strategy such that it maximises our expected value from the hand. Maximising the EV of this hand is equivalent to maximising the size of your bank-balance once the hand has been completed i.e.

The player chooses strategy S1 to maximise:

BankBalance(Before Hand) + EV(S1,S2)

Where EV(S1,S2) is the expected value to player 1 given strategies S1 and S2.

Now, we know that the bank balance before the hand is played is infinity, therefore the player is choosing S1 to maximise:

Infinity + EV(S1,S2)


It is the case that Infinity + X (where X is any real number) equals Infinity. Therefore, whatever strategy player 1 plays (i.e. however many raises he makes) the expected value of his bankroll at the end of the hand is the same.

I therefore belive that it really doesnt matter how many raises you put in.

Feedback appreciated.

[/ QUOTE ]

I think the question DS really wants us find a solution to is

max_S1 min_S2 EV(S1,S2)

which will have a finite answer. (And I'm also pretty sure he doesn't know what it is!)

Marv

[eurythmech]

[ QUOTE ]
Why is there no folding if bluffs are done at a frequency minutely below optimal?

PairTheBoard

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I assume this is because after a while the pot odds make it so that bluffing someone is EXTREMELY hard. But still bluffing will always be part of optimum strategy. In the fixed limit example, the bluffing frequency will be very low after a few rounds of betting, but it will still be above 0. It quickly approaches a point where a minimal step below the optfreq is 0, or very close to 0.

Someone mathier can probably elaborate though.

[eurythmech]

[ QUOTE ]
I dont usually post on forums, but I've found this debate so interesting that I thought I'd throw my ideas into the hat.

I only have an answer for the first part of David's question, not the pot-limit part. Furthermore, I dont claim that this is definately the correct answer- I've just tried to put my thoughts down. I'd appreciate any comments about it. Also, I am still not sure where people such as reubenf are getting the figure 20 from. They may well be right, but I've not seen a post detailing the full logic behind their answer. Again, it would be great if they could explain it.

And so to my answer:

Both players have a strategy, S1 for the player (us), and S2 for the expert we are playing against. As a player, we want to devise a strategy such that it maximises our expected value from the hand. Maximising the EV of this hand is equivalent to maximising the size of your bank-balance once the hand has been completed i.e.

The player chooses strategy S1 to maximise:

BankBalance(Before Hand) + EV(S1,S2)

Where EV(S1,S2) is the expected value to player 1 given strategies S1 and S2.

Now, we know that the bank balance before the hand is played is infinity, therefore the player is choosing S1 to maximise:

Infinity + EV(S1,S2)


It is the case that Infinity + X (where X is any real number) equals Infinity. Therefore, whatever strategy player 1 plays (i.e. however many raises he makes) the expected value of his bankroll at the end of the hand is the same.

I therefore belive that it really doesnt matter how many raises you put in.

Feedback appreciated.

[/ QUOTE ]

The word infinity in these cases should not be understood as actual infinity, but rather a finite number so large that no matter how we look at it, it's size will be "big enough", that is. The number is so insanely huge that it really doesn't matter exactly HOW huge it is. All we need to know is that infinity is a good approximation.

[reubenf]

[ QUOTE ]
Also, I am still not sure where people such as reubenf are getting the figure 20 from. They may well be right, but I've not seen a post detailing the full logic behind their answer. Again, it would be great if they could explain it.

[/ QUOTE ]

Well initially I guessed it would be around 20 bets just because 2^20 > 1m and 2^19 < 1m. So if each bet approximately halves the ranges of hands, this is a good estimate. This post talks about that:

http://forumserver.twoplustwo.com/sh...;o=14&vc=1

Notice he concludes 9 raises because he's talking about how many raises YOU put in, not how many bets go in total. That's why he looks at powers of 4 instead of powers of 2.

pzhon posted a good looking strategy here: http://forumserver.twoplustwo.com/sh...;o=14&vc=1

[marv]

I've looked at this a bit more now and put optimal strategies for the 8-card game here http://web.onetel.com/~marv742/8cardpoker/

The summary is: for an 8-card deck, any optimal strategy for the first player *must* fold the 2nd highest card some of the time, but for a 100-card deck you can (there are optimal strategies which do) but you don't have to (the restriction that the first player may never fold when dealt the second highest card doesn't make the game any worse).

I think the OP may safely require that we never fold the 2nd nuts here and just ask how many raises we'd put in after an initial bet.

Marv

[eurythmech]

KK has maybe 80% equity against a random hand. The random hand is just as random after 110 bets as it was after 0 bets.
In this case you keep on betting for an infinity, even with 22.

[Stephen H]

[ QUOTE ]
To answer these questions more easily, just assume that any bluffs be done with a frequency minutely below optimum. Then there is no folding. Most game theory questions can be answered this way. The simplification is especially useful when analyzing the pot limit variety.


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I'll start by saying I'm a complete novice to game theory, so it seems a lot more likely that I'm misunderstanding what David is saying here as opposed to him being wrong.

I have the same questions as RaiseTheBoard here. As far as I can tell, if you're bluffing less than optimally, I can fold *more* often, as I know you have the good hand more often. Can someone explain this?

[David Sklansky]

If you play optimum game theory betting strategy, your opponent does equally well by folding or calling. Optimum calling strategy is to fold no more or less often than needed to keep it from being right for your opponent to always bluff.Thus to make calculations in multi round situations easier, assume one always at least calls, even though bluffs happen.