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Re: $27... another hand in push fold mode... AQ
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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. [/ QUOTE ] 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. [ QUOTE ] 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. [/ QUOTE ] 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. -- tjh |
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