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Push tables in Kill Every One Push tables in Kill Every One

11-07-2010 , 09:45 AM
Do the push tables in Kill Everyone account for the card bunching/removal effect? How significant is this effect?

Example: If everyone folds to the button, in a 10-handed game, a player would be more likely to encounter AA from one of the two blinds than if the game was three handed..
Push tables in Kill Every One Quote
11-08-2010 , 03:29 PM
I would be interested int his too. Although the hand bunching theorem I don't belive holds as much weight as some people want it to because decent players are folding alot of brodway calrds and weak aces oop anyway. Also broadway rag combos are pretty common. So when it is folded around to the button we can say that the hand distributions are of the blinds are more likely to be stronger but the significance is hard to say. So I would think that if this changes the kill everyone shove ranges on the button it is going to be very very marginaly.

Alot of the theory behind the shove ranges is considering the whole range of oponenets and how likely they are to fold but also it is trecognizing that we NEED to double up and are looking for a spot to get our chips in with the most equity possible even if it is negative equity...

Would love to see some mathmatics of the bunching theory though. Its pretty interesting. I often talk about the bunching theory to players on my right who like to raise the button and cutoff alot espec. after I show them a big hand in the blinds or some1 else does. Just make a comment of oh of course I had or he had a big pair everyoen else folded all their little cards. Alot of times weak players will take the theory and run with it and stop opeining unopened pots in late position. And if they are smart enough to understand the flaws in the logic they will probably think I suck now and that's never bad imo.
Push tables in Kill Every One Quote
11-08-2010 , 04:04 PM
chance of being dealt at least one ace out of a normal 52 card deck: 4/52+4/51=0.155 so about 15%

lets assume, at a 10 handed table, that nobody would fold an ace, and all hands that do not contain aces are always folded by everybody (so everyone folds KK or QQ or whatever, and everyone always plays A2o). Your on the button, and the 7 hands that went into the muck did not contain an ace. Thats 14 cards that are not aces out of the deck.

so now, the chance of being dealt at least 1 ace is: 4/38+4/37=0,213 so about 21%

I'd say the difference is small enough to ignore, especially if you think about the fact that most people usually fold A2o UTG and usually play KK.

Last edited by woutman984; 11-08-2010 at 04:05 PM. Reason: fixed typo
Push tables in Kill Every One Quote
11-11-2010 , 05:57 PM
The tables do not take into account the fact that previous hands have folded. If I wanted to take that into account, I'd have to calculate a unique solution for shoving as SB when 2, 3, 4, 5... handed. Plus it also depends on what opening range each of the folders has. Too many possibilities and not going to happen in a book.

I wrote an article for the 2+2 magazine a few years ago that looked into the effect of prior folds. For jam/fold decisions, even in the worst case scenario it only makes a small difference. It's interesting to think about, but once you see the numbers you realize it's not worth worrying about.

Tysen Streib
Push tables in Kill Every One Quote
11-11-2010 , 08:43 PM
Quote:
Originally Posted by woutman984
chance of being dealt at least one ace out of a normal 52 card deck: 4/52+4/51=0.155 so about 15%
The math here is not correct. The probability of getting at least one ace is

4/52*3/51 + 2*48/52*4/51 = 14.9%

Another way

P(1 or 2 aces) = 1 - P(no aces) = 1 - (48*47)/(52*51) = 14.9%

Similar mistake was made in other part of the posting
Push tables in Kill Every One Quote

      
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