Home / Community / Forum / Support Area / Poker News /

Noodle thinking

Old
Default
Noodle thinking - Fri Oct 11, 2002, 09:41 PM
(#1)
Deleted user
ok blinds 200/100 with ante
i am UTG with AK and make a pot raise
i had about 9.5k before i raised

GEEZER raises pot from MP with just under 6k starting stack
leaving him with 1750$

putting geezer on QQ (he actually had JJ) i tried to work out wether i had the "noodles thinking " to put him all in

i thought about it and thougth about it and because i had made the raise the pot now gave me the odd to go all in am i correct?

if i was in MP and acting after geezer i would not have pot odds is this correct?

now my old thinking would be geezer was ahead of me and too fold
which was what geezer said i should have done when i typed QQ into chat telling him thats what i thought he had before the showdown

when no help came to either of us geezer won but was the play i made the "right poker move"
 
Old
Default
Fri Oct 11, 2002, 09:49 PM
(#2)
Deleted user
Quote:
when no help came to either of us geezer won but was the play i made the "right poker move"
In a way yes. You were likely getting the pot odds (no implied odds because I was all in).

In a way no. You increase your variance when you take the worst of it in terms of winning hands and there is a time value to money.

A nearly identical situation put me out a few hands later when I tried to isolate on an all-in raiser (who had AA to my JJ) and got overcalled by AK who hit K on the river. I think we were all "correct" there. If the "best going in usually wins" had prevailed I would have kept playing. Win one, lose one - break even NOT!
 
Old
Default
Fri Oct 11, 2002, 10:18 PM
(#3)
Deleted user
Further I am sure that the choice, particularly early in the tournament, is no different than it would be in a cash game.

So the correct move is to take the odds BUT it depends on your read being correct, which you can't always know.
 
Old
Default
Fri Oct 11, 2002, 10:42 PM
(#4)
Deleted user
geezer your correct if my read was wrong and i was up against AA or KK i would not have had the odds but from my playing off you i was sure you were on QQ

the fact it was still early in tourny and i would still have enough to play with also helped me decide ( i got the 3k up to 19k before busting out)
 
Old
Default
Sat Oct 12, 2002, 06:43 AM
(#5)
Deleted user
Excuse me, but I don't see where Ironside had the odds in this hand. Wasn't Ironside betting $6000 to win $6600 while a 43% dog? I think you are using the 11-10 odds quoted on these posts. Is that accurate for this hand? I've found Andrew Prock's Pokerstove useful for quick hand analysis. http://prock.freeshell.org/ Enjoy!

Regards, Den
 
Old
Default
Re: Noodle thinking - Sat Oct 12, 2002, 08:56 AM
(#6)
Deleted user
Quote:
ok blinds 200/100 with ante
i am UTG with AK and make a pot raise
i had about 9.5k before i raised
So you raised to, lets say, 950.

Quote:
GEEZER raises pot from MP with just under 6k starting stack
leaving him with 1750$
Strange raise from geezer.

Quote:
when no help came to either of us geezer won but was the play i made the "right poker move
Well, it depends on your read of the range of hands he had. If he turned his cards over and you saw that he had QQ, then I would have called due to getting just under 2-1 odds on a coin flip. However, not knowing for sure that he had QQ, I would have folded unless getting higher odds.

With those odds and situation, I would give geezer no chance of having a worse ace. The only range I put him on is QQ, KK, AA. Less than 2-1 is not good enough to call.

When would I have called with AK?

a) If the range of hands that I put him on meant he was just as likely to have a worse ace as a pair, then I would have called, almost regardless of odds.

b) If the range of hands I put him on included slightly worse pairs also, such as 99, TT, JJ, but no worse aces, then I would call getting just under 2-1 but fold getting slightly less.

c) If the pot odds were larger than they were, I would also have called as not only would I be getting higher odds that compensate me for those times against KK, but geezers range of hands would have widened slightly due to his stack pressure.

I am out of practice so the above about when I would have called may not be completely accurate, (particularly regarding point 'b'), but I think it is. This was the first time I thought about it in ages. The basic premise is correct anyway.
 
Old
Default
Sat Oct 12, 2002, 10:14 AM
(#7)
Deleted user
Just wanted to add:

The fact you have AK reduces the chances of being against AA, KK so this should be considered in the calculations when deciding whether to call.

Thinking about the actual example Ironside posted, getting just under 2-1. I would also think geezer's range of hands would include AK combinations also. So whether to call or not getting those odds against that range of hands is probably very close mathematically. If you could expand the range to include JJ also, then I think it is a call mathematically. With just the range of AA, KK, QQ, AK, then I think 2-1 or just over may be the needed, though I am just guessing.

I think I would call getting 2-1 against a typical opponent as I think the range that they would raise against an UTG opponent would include JJ and AK.

Thinking some more about point 'b' also. As the range of pairs increases, then the relative chances of being against AA, KK, goes further down, especially as you hold 2 of them. It therefore more or less becomes a case of what pot odds you want to take a coin flip with 2 overcards. 2-1 certainly, probably down to 3-2 would be correct for all but the best few players in the field.

Either way, what pot odds you need to play an all-in showdown against a certain range of hands can be worked out mathematically. Though, for the life of me, I cannot remember how. It involves counting card combinations and using weighted averages or something. Hopefully someone that knows how can post an example of how to do these equity calculations. Use this example :

A :h: K :s: v a range of QQ, KK, AA, AK. Also for AKo v same range. Thanks. In the calculations, assume that it is early in the tournament so other tournament factors are not relevant in the calculations. Some other examples would also be appreciated.

An example from later in the tournament, where the payouts ar efactored into the equity equation would also be good.

Situation for that example:

payouts:

1st 10k
2nd 8k
3rd 5k
4th 3k
5th 1k

5 players left, all have same size stack except for 1 shorter stacked player with 10k. Blinds 500/1000. Short stack not in the hand. 2 players in all-in confrontation AK v range of AA, KK, QQ, AK, (yes, I know range is likely to be wider at this stage, but just for simplicity and comparisons sake). Just under 2-1 pot odds in deciding whether to call the all-in. Any more info needed?

If you can't be bothered using my examples, then use any examples you wish
 
Old
Default
Sat Oct 12, 2002, 11:32 AM
(#8)
Deleted user
I may have been wrong about the pot odds on offer. They may have been much less than 2-1, so a fold is obvious. I think geezer would have folded Ak in this spot, and definitely JJ, so the range is definitely AA, KK, QQ in general.

Regardless of the odds being offered in this example, it doesn't affect anything I said in my above posts as I mention the pot odds figures in discussing the decision process.
 
Old
Default
Sat Oct 12, 2002, 01:39 PM
(#9)
Deleted user
the pot odds if i went all in (it was all in or fold) were around 11/10
i tried to work them out fully at the time but ran out of time

the main reason i put geezer on QQ first was that i knew he had a big PP to reraise me (from utg) due to the nature of his bet i knew he would proberly go all in no matter what but he hadn't hit the all in button meaning he wasnt 100% sure on his hand
A "good" player at the table made same read in the discussions afterwards

i used to always prefer getting all my chips in when i am infront and folding if behind.
and still working on this so if i was getting 2-1 pot odds its right to make the call? but 3-2 pot odds wouldn't be?


MARADEN because i had already put close to 1k in the pot i was betting 5k to win 6k which is 12/10 or close enough to 11/10
 
Old
Default
Sat Oct 12, 2002, 02:46 PM
(#10)
Deleted user
Quote:
...so if i was getting 2-1 pot odds its right to make the call? but 3-2 pot odds wouldn't be?
I thought I had explained all this in the above posts. Was it not clear?
If what I wrote in the above posts are not clear, then give some more exemples with your thinking and we will see if I can make it clearer.

When you only give geezer a range of AA, KK, QQ then 2-1 isn't really enough anyway. Though, if you were getting 2-1 then I think that the range would be just a little bit wider, so that the call would be right. But if getting 2-1 and the range is only AA, KK, QQ, then 2-1 isn't enough, probably.

However, if you knew for sure that it was QQ, say he turned them over and showed you, then 3-2 is enough- unless you are the best player in the field and then it may or may not be. Though even then, if that best player was a good cash player, they may still play it, and be correct to do so if you factor their cash expectation in the event of busting into the equity equation.

It is all about equity. The inputs to the calculation can change from a typical players, in certain exceptional circumstances.

For example, Some people may have an higher equity in the tournament in certain situations than a typical player would in the same circumstance. For example, a player may be deadly with a big stack early, but average with a medium stack. This may change the equity equation enough in working out whether to play an all-in showdown, so they may be mathematically correct to take lower odds etc.

Basically, like I said, the odds I would say that are needed are just a guess. It can be proven mathematically exactly what odds you need.
 
Old
Default
Wed May 07, 2003, 09:49 PM
(#11)
Deleted user
Another thread from the "Golden Oldies" that deserves a 2nd and 3rd look.

Shane.
 
Old
Default
Sun Jan 25, 2004, 08:06 AM
(#12)
Deleted user
bump for Key West
 
Old
Default
Wed Jan 05, 2005, 01:38 PM
(#13)
Deleted user
I started reading the Noodles posts because Aaron has bumped so many of them and I thought this one was worth a bump because I played a similar hand only I was the geezer.

9 left final table of a last chance event. UTG who I've known since I was 12 pot raises, we both have just over 200K and are co chip leaders. I put him on AK right way and I have QQ 3 seats away it folds to me and I go all in because I don't want him to see the flop and I'm pretty sure he'll fold. Folds to him and after 3 time buttons he folds. I tell him QQ and he confirms AK.

Could have been a risky play but I had a solid read and the Big One was on my wife's birthday so short of divorce I wasn't going to be able to play any how. :P

If Ironside hasn't departed yet it would be interesting if he or geezer have changed their opinions on how they played the hand out.

The part about how you play with a particular stack entering into the equation was also interesting to me.
 
Old
Default
Wed Jan 05, 2005, 02:25 PM
(#14)
Deleted user
200 + 100 + 10 x 25 = 550 in pot
Ironside puts in 200 + 750 = 950.
geezer has 6000
The call is 6000 - 750 = 5250 to win a pot of 550 + 950 + 6000 + 5250 = 12750. 5250/12750 = .41, so ironside needs a 41% probability of winning to justify call on Hand EV terms.

I assume geezer will play only AA KK QQ JJ here. Since ironside has AK, AA and KK are half as likely as QQ JJ. (How did I get that? 50 cards left in deck. 3 aces. prob 2 aces in a row is 3/50 x 2/49 = 6/2450. Same with kings. 4 queens left, so prob 2 Q in a row is 4/50 x 3/49 = 12/2450, twice as likely as AA or KK)
AK beats AA 7% of time
AK beats KK 30% of time
AK beats QQ and JJ 43% of time.

So, the probability of winning the pot, given my subjective beliefs that he holds JJ-AA and given that I hold AK is:
1/6 x .07 + 1/6 x .30 + 1/3 x .43 + 1/3 x .43 = .35.
(Where did I get 1/6, 1/6, 1/3, 1/3? Well, let x=prob geezer has AA. So prob KK also = x, prob QQ = prob JJ = because twice as likely as AA. x + x + + adds up to 6x. So, given that geezer holds AA KK QQ JJ and Ironside has AK, prob geezer has AA is x/6x = 1/6, etc.)
This is below .41. So I should fold.

Did ironside do the wrong thing by calling? No.

Why do I say no? Because my analysis is based on geezer holding JJ-AA whereas Ironside's is based on geezer holding QQ w/certainty (probability = 1). And since AK beats QQ w/prob .43, he believed he had odds.

Different beliefs, different answers. Probability is subjective. When people say they play by feel or intuition, that is what they mean. They've seen lots of situations and believe they have a very accurate assessment of the other player's cards. The players with the most accurate beliefs and the ability to do the math will crush everybody else in the long run, and probably the short run, too.

Let's just pretend for the moment that geezer actually held QQ instead of JJ. Because, Ironside was able to accurately narrow geezer's possible hands to QQ whereas I was able only to narrow his possible hands to JJ-AA, Ironside made a +EV call on a hand that I would have folded (a fold which, paradoxically, would have been mathematically correct given my relatively imprecise read on geezer's holdings). It is the accumulation of such small edges that separates good players from mediocre ones.

How does stack size fit in? Well, suppose geezer had only 3000 instead of 6000. Now, Ironside's call is 2250 and the total pot is 6750, so the required probability is 2250/6750 = .33 Now it is correct to call even if you put geezer on the better set of hands AA KK QQ JJ.

In general, the bigger the other player's stack relative to dead money, the closer and closer the required probability to justify a call gets to .50. That is why you should let 'er rip with weaker hands when the blinds and antes get big. You can have correct odds even with a hand substantially weaker than your opponent's.
 
Old
Default
Wed Jan 05, 2005, 02:48 PM
(#15)
Deleted user
Quote:
Originally Posted by Lloyd
So, the probability of winning the pot, given my subjective beliefs that he holds JJ-AA and given that I hold AK is:
1/6 x .07 + 1/6 x .30 + 1/3 x .43 + 1/3 x .43 = .35.
(Where did I get 1/6, 1/6, 1/3, 1/3? Well, let x=prob geezer has AA. So prob KK also = x, prob QQ = prob JJ = because twice as likely as AA. x + x + + adds up to 6x. So, given that geezer holds AA KK QQ JJ and Ironside has AK, prob geezer has AA is x/6x = 1/6, etc.)
This is below .41. So I should fold.
I'm pretty good at math but if you can do that while sitting at a table/computer off the top of your head god bless you.
 
Old
Default
Wed Jan 05, 2005, 02:59 PM
(#16)
Deleted user
Quote:
Originally Posted by marshfield
I'm pretty good at math but if you can do that while sitting at a table/computer off the top of your head god bless you.
Can I do this off the top of my head? Um, err, more or less, YES. :lol:

My great failing as a poker player is the beliefs part. Take my word for it--the math is easy when you are a wuss like I am and think everybody who raises has AA. ops:
 
Old
Default
Wed Jan 05, 2005, 03:05 PM
(#17)
Deleted user
For now I'm just taking one hand that I put somebody on and using that for my pot odds. Maybe next year I'll sign up for Dr. Lloyd's advanced poker calculus. :P
 
Old
Default
Wed Jan 05, 2005, 03:05 PM
(#18)
Deleted user
I used to score chicks in college with my mental calculation skills. Let's just say, my friends, that MANY were the panties that dropped when my one of my frat brothers would holler out, "Quick, Lloyd, what is 223 x 449?" :wink:
 
Old
Default
Wed Jan 05, 2005, 03:08 PM
(#19)
Deleted user
So did I but it was 24*36*24.
 
Old
Default
Wed Jan 05, 2005, 03:21 PM
(#20)
Deleted user
Quote:
Originally Posted by marshfield
So did I but it was 24*36*24.
That's easy, just factor out the 12's and it is 12 x 12 x 12 x 12= 20,736. Man, girls weren't very selective at your school if they were impressed by that!!! Perhaps the lack of selectiveness is related to the unusual 24 x 36 x 24 body shapes. Hmm....
 

Getting PokerStars is easy: download and install the PokerStars game software, create your free player account, and validate your email address. Clicking on the download poker button will lead to the installation of compatible poker software on your PC of 51.7 MB, which will enable you to register and play poker on the PokerStars platform. To uninstall PokerStars use the Windows uninstaller: click Start > Control Panel and then select Add or Remove programs > Select PokerStars and click Uninstall or Remove.

Copyright (c) PokerSchoolOnline.com. All rights reserved, Rational Group, Douglas Bay Complex, King Edward Road, Onchan, Isle of Man, IM3 1DZ. You can email us on support@pokerschoolonline.com