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  #1  
Old 03-19-2007, 09:42 PM
Bruce D Bruce D is offline
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Default Question on game theory

Let us say that I am straight forward, but weak player. Let us assume that I will only pfr AA and AK.

Let us say that on a dry board or draw heavy board, I always raise a bettor in front with AA.

Let us say that if I had AK, I would never raise.

I would therefore be turning my cards face up for a villian who is paying attention to my actions.

There are 16 combos for AK, and 6 for AA. Would it therefore be optimal to play 5 of my AK hands like I would AA? I merely took (16+6)/2 and I get 11. I took this as I should play 11 of my AK normally and screwplay 5 of them?

Is this a valid idea? I understand that we have to mix it up, but I am looking for optimization in a vacuum.
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  #2  
Old 03-20-2007, 02:32 AM
Pontus Pontus is offline
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Default Re: Question on game theory

I've though about this for a while now, trying to recall all the text in "Game theory and bluffing" or what the chapter was named.

I have one question:
Why would you add them together and divide them? I'm a little behind here and have maybe forgot something.

Though:
(A very loose one, just speculations though):

To get AA before the flop is 4/52*(3/51)= about 6%?
To get AK before the flop I get it to: 8/52*(4/51)=8%?

That means that I would get AK 1/3 more then AA.

If the flop comes out 2, 7, Q with no draw and he has the queen, the pot is $10. If you would be raising with both hands, he would win more then half the times. (33% more?)

Let's say that there are no more cards to come, and the only available bet to make is $10. No more, no less and no raises. If you make that bet, he would get 2-1 odds. That means that he would figure to win more then he looses the times you have AA. (if you bluff All AK.)

If you play this hand 100 times exactly and all odds are working exactly. (you get your AA 6 times and AK 8 times).

If the odds are 2-1, you should be bluffing about 2-1 too, right? Which means that you should be bluffing once every three times, which comes down to 2 times in total (on random of course), then he would get the odds 2-1 for a call that he figures to win 1 in 3 (2-1 against). That would not be profitable but it would not be unprofitable. Though, the times he folds, you gain if you had AK.

Warning! This may be wrong
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  #3  
Old 03-20-2007, 06:56 AM
senjitsu senjitsu is offline
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Default Re: Question on game theory

Might want to check that math.. your odds of picking up aces are 4*3/52*51 = 12/2652 = 221 to 1 (around .45 percent)

Youll pick up AK about 8*4/51*52 or 32/2652 or around 1.2 percent of the time (2 2/3 times as often). There are 6 ways to make AA and 16 to make AK, if it helps to think about it like that.

[ QUOTE ]
I've though about this for a while now, trying to recall all the text in "Game theory and bluffing" or what the chapter was named.

I have one question:
Why would you add them together and divide them? I'm a little behind here and have maybe forgot something.

Though:
(A very loose one, just speculations though):

To get AA before the flop is 4/52*(3/51)= about 6%?
To get AK before the flop I get it to: 8/52*(4/51)=8%?

That means that I would get AK 1/3 more then AA.

If the flop comes out 2, 7, Q with no draw and he has the queen, the pot is $10. If you would be raising with both hands, he would win more then half the times. (33% more?)

Let's say that there are no more cards to come, and the only available bet to make is $10. No more, no less and no raises. If you make that bet, he would get 2-1 odds. That means that he would figure to win more then he looses the times you have AA. (if you bluff All AK.)

If you play this hand 100 times exactly and all odds are working exactly. (you get your AA 6 times and AK 8 times).

If the odds are 2-1, you should be bluffing about 2-1 too, right? Which means that you should be bluffing once every three times, which comes down to 2 times in total (on random of course), then he would get the odds 2-1 for a call that he figures to win 1 in 3 (2-1 against). That would not be profitable but it would not be unprofitable. Though, the times he folds, you gain if you had AK.

Warning! This may be wrong

[/ QUOTE ]
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  #4  
Old 03-20-2007, 07:09 AM
senjitsu senjitsu is offline
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Default Re: Question on game theory

It depends on your goal... If your goal was only deception, that is to say if your opponent had to guess your hand (AA or AK) after the action on the flop, and you got points according to wether his guess was right or wrong, it wouldn't really matter how often you raised with one hand or the other, so long as you raised them both with the same frequency. For example, if you raise every time with AA, then the optimal game strategy would also to be to raise all the time with AK.

Raising the AK 5/16 times, so that your raising villian exactly half the time (on average), adds some deception, but hed still eventually figure out that around half (6/11) of your raises are AA but _all_ of your calls are AK... which is to say youre still giving away a lot of information, though more with your calls than raises.

Of course, the real problem is the first condition of the question -- that youll only pfr AA or AK.... once you accept that, its very difficult to add any deception to your post flop play.


[ QUOTE ]
Let us say that I am straight forward, but weak player. Let us assume that I will only pfr AA and AK.

Let us say that on a dry board or draw heavy board, I always raise a bettor in front with AA.

Let us say that if I had AK, I would never raise.

I would therefore be turning my cards face up for a villian who is paying attention to my actions.

There are 16 combos for AK, and 6 for AA. Would it therefore be optimal to play 5 of my AK hands like I would AA? I merely took (16+6)/2 and I get 11. I took this as I should play 11 of my AK normally and screwplay 5 of them?

Is this a valid idea? I understand that we have to mix it up, but I am looking for optimization in a vacuum.

[/ QUOTE ]
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  #5  
Old 03-20-2007, 08:56 AM
mvdgaag mvdgaag is offline
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Default Re: Question on game theory

These calculations are mostly made when you want to optimise your EV, but you'll have to know more about the current odds you lay.

If you always bluff he will always call and if you never bluff he will never call. So in both cases you are far from optimal. Now what you want to do is make the EV the same for when he calls or folds, so it doesn't matter what he does.

Let's take another example first (very hypothetical)
-magic 72o which always loses
-magic AA which always wins

Now if you bet $4 in a $2 pot only with AA and fold 72o he's going to fold all the time and you win nothing

If you bet all of them you are going to have 72o 12 times and AA 6 times and he's always going to call, which makes your EV 12*-$4 + 6*$6 = -$12.

How often will you have to bet 72o to optimise your play? This is where the odds against your bluffing are the same as the odds you lay, which is 2-3 ($4 to win $6). So for the 6 aces you'll need four bluffing 72o's.

So EV if he calls = 4*-$4 + 6*$6 = $20.
And if he fols = 10*$2 = $20.

Whatever he does, with this distribution you'll have an EV of the current potsize ($20 / (4+6)) for ten out of 18 deals.

--

In your example there is no pot yet, so if he folds your EV = 0. To balance AK and AA so you'll have an EV of 0 when he calls had no point. But..

AA has an equity of 80% against a possible pocker pair caller and AK 45% (this is a pocket pair, but you could check your equity against his range and do the same). If you were raising all your AA's or none of your AK's you'd be giving information and he'll call if you didn't raise giving you an equity of only 45%. The solution here is to sometimes not raise AA (If you sometimes raise AK, the times you don't raise still signal your hand being AK). Suppose if you do not raise you want an equity equal to when you do raise. Here's the trick:

You want your (optimal playing) opponent to have the same expectation when you raise or when you don't so he cannot adjust his play to your strategy. To do this you need the same equity when you raise as when you don't. So you raise half your aces and half your AK's for an equity of 54% and you check half your aces and half your AK's for an equity of 54%. This will be the same equity as always raise or always call and is the optimum. If you want to play al your AA and AK hands than the only point is not to give information so he can adjust his strategy to yours.

In practise players will not play an optimal game against you and either fold too often or call too often. If they call too often you should bluff less and if they fold too often you should bluff more. Same goes for optimal strategies, if they fold too often to a raise add more of the possible low equity hands to the raising group and vice versa.

GL
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  #6  
Old 03-20-2007, 09:01 AM
Bruce D Bruce D is offline
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Default Re: Question on game theory


The purpose would be to force my opponent into making errors on laters streets. That is, to fold a winning hand.

I guess it gets complicated in that we have to analyse the post size.

I think I should add a condition that my opponent originally knows that I will only raise with AA, or AK. In that respect he can bet to the river and have complete information as to where he is in the hand.

He would bet any board that didn't have an A or a K on it. At some point I would have to start calling down his river bets with A high and that is a losing proposition.

I don't want to see the river with AK. When he comes out betting with either a bluff, or a semi-bluff I want to gain value from it. Throwing in a raise using game theory while holding AK will a) reduce the value of his bluffs and b) reduce the value of his legitimate pair hands.

Basically I am looking for the correct frequency to raise AK such that he will not always know the correct play if I end up raising the flop.
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  #7  
Old 03-20-2007, 09:09 AM
mvdgaag mvdgaag is offline
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Default Re: Question on game theory

If we assume he calls preflop and will give up with an A or K on the flop (which is very hypothetical and just for the sake of this threat, no way it is this simple in real poker). You want to maximise your EV in the other situations:

you will flop an A or K about 1/3 of the time.
The other 2/3 of the time you want him to have to give up as well.

So you'll continue with AK a little less often as you do with AA when you didn't hit the flop. You were dealt 6x AA and 16x AK, so you could for example continue even if you didn't hit with all the suited AK's and AsKc. Now if you continue you'll have AA 6 times and AK 5 times, so even if your opponent knows your strategy he will still be in bad shape.

Again: In practise your opponents do not know and you'll have to continue less often most of the time, because they will call too often in general.
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