#1
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Harrington on Hold \'em Hand 4-9
pp. 165-168
Question regarding the flop action. With the pot at $170, "Action: You actually bet $100. Player E calls. The big blind and Player A fold. The pot is now $370. You bet too little. If Player E is on a club flush draw, he got the proper pot odds to call." Player E got 2.7-to-1 pot odds on a 4-to-1 shot. Perhaps this could be accounted for by implied odds, but then, in the segment on the river, when Hero makes trip kings, "Against most opponents you should bet about $250 here [into a pot of $370]. To call, he would have to put in $250 for a pot of $620, about 2.5-to-1 odds. He's more than 4-to-1 to hit his flush, so it's a blunder for him to call if he knows what you have." Can someone explain this incongruency to me? Thanks. |
#2
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Re: Harrington on Hold \'em Hand 4-9
On the flop you are 2to1 to get the flush by either the turn or river. On the turn you are 4to1.
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#3
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Re: Harrington on Hold \'em Hand 4-9
This seems counter to most of the advice in the book, I will have to look at it when I get home, he usually says to bet 1/2 to 3/4 of the pot.
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#4
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Re: Harrington on Hold \'em Hand 4-9
[ QUOTE ]
On the flop you are 2to1 to get the flush by either the turn or river. On the turn you are 4to1. [/ QUOTE ] But the opponent is not guaranteed to see a river just by calling the flop. Getting 2-to-1 on a flopped flush draw is not enough. It might just be a typo that I'm blowing out of proportion, though [img]/images/graemlins/smile.gif[/img] |
#5
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Re: Harrington on Hold \'em Hand 4-9
I took Harrington's point to be that giving an opponent 2.7 to 1 is just enough to justify a call given the implied odds, while 2.5 to 1 is just short of enough to justify a call even with the implied odds.
He makes the point somewhere that he believes you need to bet at least two-thirds the pot in order to deny your opponent the proper odds to call on a draw. If you apply that standard to the above example, you'll see his point. |
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