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  #1  
Old 08-27-2006, 11:58 AM
Dire Dire is offline
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Default Annoying low draw problem

I've run into this situation alot lately: You're in a three way pot. You have nothing but the nut low draw. You know one of your opponents has a lock on the high and your other opponent is also drawing at the bare nut low.

You've got 16 outs to 1/4 so you'd need about 8:1 which you usually won't be getting, on the turn at least. So does that make this an easy fold? It's annoying giving the other guy a free shot at half of a decent sized pot but it seems like the only way.
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  #2  
Old 08-27-2006, 12:35 PM
Split Suit Split Suit is offline
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Default Re: Annoying low draw problem

fold. with no hopes to the top half of the pot, and knowing that ur drawing to 1/4 the pot, folding is best. something to keep in mind, you need the pot to be shown down 5handed on the river to breakeven/make money if you get 1/4erd.
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  #3  
Old 08-27-2006, 01:33 PM
Kuso Kuso is offline
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Default Re: Annoying low draw problem

[ QUOTE ]
fold. with no hopes to the top half of the pot, and knowing that ur drawing to 1/4 the pot, folding is best. something to keep in mind, you need the pot to be shown down 5handed on the river to breakeven/make money if you get 1/4erd.

[/ QUOTE ]

partially correct.

first, is this plo8 or lo8?

if plo8, it's almost always an easy fold if you are CERTAIN about your read. even if you're not, it's often a good fold. don't ignore draws to draws, though.

if lo8, you should usually fold if the pot is small. if there was a lot of money in pf AND the other low won't start jamming like an idiot, then you can potentially both call down profitably. note that this takes a very specific set of circumstances, but sometimes they happen. also don't undervalue counterfeit protection and (again) draws to draws.
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  #4  
Old 08-27-2006, 07:08 PM
Buzz Buzz is offline
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Default Re: Annoying low draw problem

Dire - Bear in mind that my degrees are not in mathematics and I do not fancy myself a mathematician.

Let's try to get straight that which you are stipulating.
<ul type="square">1. You have two opponents on the turn.
2. You are certain one of these opponents has the nut high.
3. You have no chance to win high.
4. Your have the bare nut low draw.
5. You are certain your other opponent also has the bare nut draw.
6. And that's it. Nothing more.[/list]Do I understand correctly?

I don't know how you can know these things, but I'll accept that you do.

Let me paraphrase a bit.
<ul type="square">1. You know your four cards. Let's make your hand
A[img]/images/graemlins/heart.gif[/img], 2[img]/images/graemlins/heart.gif[/img], Q[img]/images/graemlins/club.gif[/img], K[img]/images/graemlins/club.gif[/img].
2. You know the four cards on the board. Let's make them
7[img]/images/graemlins/diamond.gif[/img], 8[img]/images/graemlins/diamond.gif[/img], 9[img]/images/graemlins/diamond.gif[/img], T[img]/images/graemlins/heart.gif[/img].
3. You know two of the cards in one opponent's hand. Lets make them
T[img]/images/graemlins/diamond.gif[/img], J[img]/images/graemlins/diamond.gif[/img].
4. You know two of the cards in your other opponent's hand. Lets make them
A[img]/images/graemlins/club.gif[/img], 2[img]/images/graemlins/club.gif[/img].[/list]Does that fit the pattern of what you know?

If so, you know the whereabouts of 12 cards while the exact location of the other 40 is unknown.

Of the unknown cards, 16 will make the nut low for you and also for one of your opponents. The other 24 cards are misses.

The odds against you making low on the river, from your vantage point, are 24 to 16 (or 3 to 2). Thus on the 16 times when you win (tie), you need to win 3/2 chips to make up for the 1 chip you lose on the 24 times when you lose.

Let's digress a bit to the pot. Let's assume this is a limit game and make 1 chip = 1 big bet. It will cost you 1 chip to see the river, but then if you make your low, it will also cost you 1 more chip on the river. Let's assume, for now, no raises on the river. Thus if you make your low, both of your opponents will also put in one chip on the river.

So take one white chip and two blue chips to represent the bets on the river.

Then take a red chip and stack it on top of the first white chip. Those two chips are your total investment if you make the nut low.

Now add four more blue chips to make a total of four stacks, each with two chips. One of these stacks is the fourth of the pot you'll get if you make low. With these four stacks of two chips each, you're breaking even for your last two bets when you win.

But you really need to win at least one and a half chips when you win in order to make up for the chips you lose when you lose. You need to win in a 3/2 ratio.

So stack another blue chip on top of each of the four stacks. With this distribution, you'll be winning one chip, starting from the third betting round.

Not quite enough yet. We need a half chip more on each pile. Just use two blue chips to represent the four half chips. And it doesn't matter where you put them, because finally you're going to take away two blue chips to get back to what needed to be in the pot when the bet was up to you on the third betting round. Ten chips is the correct amount that needs to be in the pot when the action gets to you on the third betting round.

The pot odds you need on the third betting round are 10 to 1. I could probably write some equations to come to this conclusion, but I think stacking the chips makes it very clear.

I don't know where you got the 8 to 1, but I think it's wrong. I think you need 10 to 1. And we haven't considered the possibility that you'll get sixthed.

Could you be getting 10 to 1 whole pot odds on the third betting round? I suppose it's possible, though generally unlikely. More likely, in my humble opinion is that (1) your opponents don't have exactly the cards you have put them on, and (2) you actually do have some chance to win high. The guy you think also has the nut low draw could have the second nut low draw, or could have a set - and the guy you think has the nut high could have the same nut low draw as you.

But at any rate, if you're certain your two opponents have the hands you stipulated, then in order for you to call the single big bet on the turn, there need to be ten big bets (twenty small bets) in the pot and you need to have last position.

Buzz
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  #5  
Old 08-27-2006, 07:36 PM
Dire Dire is offline
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Default Re: Annoying low draw problem

2:1 was the [very] rough estimation of the low out odds. Getting a quarter need 4x as much -&gt; 8:1.
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