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 06-28-2012, 03:42 PM #1 Administrator     Join Date: Aug 2002 Posts: 9,908 Ante Only NLH Game Theory Question Everyone folds to you in the cutoff. There is x dollars in the pot and if you don't put in at least one dollar the button wins it automatically. You and the button both have \$100 stacks. How big does x have to be such that you should fold none of your hands?
 06-28-2012, 05:17 PM #2 newbie   Join Date: Jun 2012 Posts: 19 Re: Ante Only NLH Game Theory Question For further information, are you referring to an open-fold, or do folds to a re-raise by button count?
 06-28-2012, 05:19 PM #3 stranger   Join Date: May 2012 Posts: 4 Re: Ante Only NLH Game Theory Question The size of the antes (and hence, the pot) surely come in to play as well.
06-28-2012, 05:30 PM   #4
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Re: Ante Only NLH Game Theory Question

Quote:
 Originally Posted by hshy307 The size of the antes (and hence, the pot) surely come in to play as well.
The pot, which contains only the antes, is the "x" specified in the OP.

 06-28-2012, 06:54 PM #5 newbie   Join Date: Jun 2012 Posts: 48 Re: Ante Only NLH Game Theory Question \$2? giving you 2 to 1 odds.
06-28-2012, 09:12 PM   #6
newbie

Join Date: Jan 2011
Location: Mars
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Re: Ante Only NLH Game Theory Question

Quote:
 Originally Posted by David Sklansky Everyone folds to you in the cutoff. There is x dollars in the pot and if you don't put in at least one dollar the button wins it automatically. You and the button both have \$100 stacks. How big does x have to be such that you should fold none of your hands?
Speaking strictly on a pot-odds level? If so I'd think x could be any dollar amount to justify a call. Even if every ante is a dime, you're still getting nearly 1:1 on your dollar; your hand would have to be terrible to give up the pot without protest.

Or am I mistaken?

06-29-2012, 12:05 AM   #7
journeyman

Join Date: May 2012
Posts: 399
Re: Ante Only NLH Game Theory Question

Quote:
 Originally Posted by MGMDonk Speaking strictly on a pot-odds level? If so I'd think x could be any dollar amount to justify a call. Even if every ante is a dime, you're still getting nearly 1:1 on your dollar; your hand would have to be terrible to give up the pot without protest. Or am I mistaken?
In a game with typical blinds you're getting 1.5 to 1 on your money and still often open fold the cut-off, don't you?

 06-29-2012, 02:06 AM #8 newbie     Join Date: Jan 2011 Location: Mars Posts: 29 Re: Ante Only NLH Game Theory Question That's a fair point, but I'm not sure it helps in this scenario. In a game with typical blinds you'd obviously fold weaker hands from the CO for the same reason you'd fold them anywhere else: You've got a minimum of 3 (possibly 4 if you're in a game that allows straddling) people behind you left to act whom you also have absolutely no hand information on. In this case, the table is already isolated to you and the button. No matter what the button has there doesn't seem to be much incentive for it to raise here because (as I see it) one of three scenarios happens. (Since we're trying to find the worst possible hand to bet a minimum of a dollar here, I'm going to assume that I'm working with some kind of middling, non-connecting, non-suited cards. Think J-6 or Q-7.) Anyway, as best I can see it, either: 1. The Button is just as weak as you are, if not weaker. Your bet might actually take down the pot outright and spare you any further thought for the hand. 2. The Button has some kind of middle of the road hand or a hand like AQs. There's no incentive to risk more money with those hands here, so Button will probably call and you can see if you've got anything post flop. 3. Button has some kind of crazy good hand. There's still little incentive to raise though, because why drive out the only player left to pay off his big hand? The worst case scenario is that the Button raises anyway and you're out 1% of your stack. (Technically it's even less since you have \$100 after your ante is already in play.) That seems worth the risk to me, especially when you can get what essentially equates to a free round of hands worth of antes. There's something else to consider too. In my initial question to Sklansky's premise, I assume the antes are .1% of your stack; that's preposterously low. Even if you raise it to \$0.25 per ante, you're now getting 2.25:1 to make a dollar bet. The more I look at this, the more it seems logical to get involved with any starting hand, possibly excluding the three or four bottom rung hands. Last edited by MGMDonk; 06-29-2012 at 02:08 AM. Reason: Math clarification
06-29-2012, 02:59 AM   #9
journeyman

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Quote:
 Originally Posted by Bolivian Ram For further information, are you referring to an open-fold, or do folds to a re-raise by button count?
Have to assume open folding as he's stated at least a dollar and obviously at 100 stacks you have to have a raise fold range. Seems unanswerable as youd need to solve for 100 BBs with post flop play.

 06-29-2012, 03:01 AM #10 journeyman   Join Date: Aug 2007 Location: Sydney, Australia Posts: 306 Are these \$100 stacks after the ante ?
 06-29-2012, 01:30 PM #11 adept   Join Date: Jul 2009 Location: Canada Posts: 1,004 Re: Ante Only NLH Game Theory Question Of what I understand (OP will rectify if I'm wrong): 1) Once ante are posted, stacks are 100\$ 2) We're not trying to find the threshold hand in a range where we would consider open-folding. Rather, we're trying to find the threshold amount "x" in the pot (which presumably is the sum of each player's ante) where open-folding even one hand would be a mistake.
 06-30-2012, 06:29 AM #12 veteran     Join Date: Aug 2009 Location: Stanford, CA USA Posts: 3,325 Re: Ante Only NLH Game Theory Question cliffs x=110 for n=2 players, much more difficult to solve for more than 2 because the fact others have folded severely depletes outs for say 32o for CO making the problem much less trivial. So i need to know how many players have folded. The next deals mostly with 2 CO+btn ignoring others before as if they didnt exist. Push/fold poker studied only. You need to be producing a profit with even 32o nomatter how the btn reacts. However the btn cannot call you with anything unless x is big enough. So we need a situation where the equilibrium is you 100% and the btn some R. But R itself will be close to 100% anyway. The situation therefore from the btn ought to be that his hands vs 100% range result in at least equity of E>=100/(100+100+x). If we ignore the folds of others which is absolutely impossible to do in a serious game where the x would be so large that CO pushes everything (indicating that they received so terrible hands that now the range of btn is superrich in A,K and many outs of 32o are used) then we may get an answer with the above. But i doubt it applies in real life because the others would nt have folded so easily and if they did the rthe equity of 320 vs say 99-100% range is seriously less than 32%more like 20% 15% even. So look for something like 23o vs R with R very close to 100% which is equity ~ 32.3%. So basically 100/(200+x)=32.3% or x= 109.6. Notice that even if R=99% the fold equity is huge and the EV is 0.01*(x+100)+0.322*0.99*(100+100+x)>100 or x>107. Basically the R cannot even be 99% it has to be 100% when x=110. The idea is that both ranges the pushing and calling arrive to 100% almost the same time. This is why you can take 32o vs 100% and be confident this is it, which yields x~110 However notice that this is only a valid problem in 2 player game because due to the huge size of X the others before CO would absolutely adore to push with even 70% ranges etc meaning their hands are so terribly poor (if there are 7 others say) that it makes the range of btn rich in K,A and depletes most outs of 32o. Basically imagine the others for the X suggested here have to fold something like 70-80% ranges which is unimaginably rare with 7 of them say (1 in 5000 chance or worse). So if this happened say we have a situation where out 32o vs the random range is nolonger 32.3% but as low as 20%or less even (because for others to have received bottom20-15%ranges it means a lot of 2-9 rank cards in say 14 cards dealt to them meaning possibly of the other 6 3s and 2s left on the deck at least 4 are taken) So i ask again fora clarification how many others have folded otherwise the problem cannot be answered. In the 2 person case it seems x=110. You can see that for example you cant push 32o if x was like 100 and opponent had near 99% range as; 0.01*(100+x)+0.99*0.322*(200+x)>100 requires x>107. Basically 32o is a fold for some x<107 Notice that R is definitely close to 99% near the threshold because the next hand vs 100% after 32o is nearly 1% better in equity already so we cant take it out so easily. Basically the btn sees eg 42o giving vs 100% 33.2%. So in something like X=101 already much less than 110 above he has 100/(100*2+101)=33.2% so he can call. But see that 32o vs 100% which is 32.3% with x=107 say gives 100/(100*2+107)=32.5% making it a marginal fold. Basically i just gave an argument why near the transition R is 99% but not 98% or more. Last edited by masque de Z; 06-30-2012 at 06:35 AM.
 06-30-2012, 06:35 AM #13 veteran     Join Date: Aug 2009 Location: Stanford, CA USA Posts: 3,325 Re: Ante Only NLH Game Theory Question What i just did above is for the push fold situation. If you can study normal poker with betting anywhere from 1 to 100 and raises allowed etc the situation is back to a complex nearly as tough as poker itself problem. So i doubt it can be solved in a different context than all in push/fold sense. So i need it to be clarified here are we talking push fold poker or any kind of bet and raising allowed?
06-30-2012, 02:41 PM   #14
journeyman

Join Date: May 2012
Posts: 399
Re: Ante Only NLH Game Theory Question

Quote:
 Originally Posted by masque de Z are we talking push fold poker or any kind of bet and raising allowed?
When is x large enough to where you'd rather put in \$1 with 7 2 off than fold it.

06-30-2012, 04:34 PM   #15

Join Date: Jan 2006
Posts: 702
Re: Ante Only NLH Game Theory Question

Quote:
 Originally Posted by David Sklansky Everyone folds to you in the cutoff. There is x dollars in the pot and if you don't put in at least one dollar the button wins it automatically. You and the button both have \$100 stacks. How big does x have to be such that you should fold none of your hands?
I'm not really familiar with AONLHE. My understanding is the bring in is equal to the smallest chip denomination. If this is the case, then if the minimum bring in is \$1, that means that in a 9 handed game, the antes must make x be equal to a multiple of \$9 dollars. In the case of \$9, we already have a stack pot ratio of about 11.1. So it seems we are already in the territory of push/fold poker.

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