$27... another hand in push fold mode... AQ

[tjh]

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Here is my thinking process and tell me where i am wrong


If there are 4 hands in a row where it is push/fold then we know with the 100/200blinds we are getting to that stage with 7 still in.


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Perhaps, seems to me though that the all-in stage is usually controlled by one or two persons when it starts with this many players. A read on the pusher would be usefull here. If they have been the aggressor then you can expand the pushing range.

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I need to find a hand soon and I may not see a better one than this for a while.


I only have 1200 chips left and after the next few hands will only have 900 left so i need to look for a chance to double up.

I put him on a range of any pair, Any ace with 9+ kicker, KQ and maybe KJ.

that makes me a conflip agisnt 10 hand
A big dog to 3 pairs
A dog to Ak
A fav to 5 other hands.

I think this makes it a true coinflip decision.


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Without taking into account that you have an Ace and A Queen.

There are 6 ways of getting any pair.
So any pair hands Villain has 78 possibilities.

7 other hands that can be made 16 diffferent ways for a total of 112 ways.

So Villain is playing one of these 190 possibles.

AA KK QQ you are behind. 18 hands.
AK you are behind 16 hands

You are a fav to AJ AT A9 KJ KQ. 80 hands

Coinflip to the rest AQ an 22->jj 76 hands

Behind 34, ahead 80, coinflip 76.

Looks like a wee bit better than a coinflip.

Of course you having AQ cuts down on the probability of all the ace hands so you may still be correct. I just wanted to post soome numbers to show that AK is about three times as common as AA. The non-paired hands are easier to get so when you think villain has AA or AK he is 3/4 of the time going to have AK.

How do you do this calculation in real life...
I just weight the calling range towards the non-paired hands. One non-paired hand equals about 3 paired hands in terms of frequency.

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tjh

jmiller and tjh... thanks, i learned something today, greatly appreciated