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#11
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putting your opponent on a statistically-distributed range of hands based on combinatrics [/ QUOTE ] How can I find the exact statistics of a hand? I've always just been guessing. |
#12
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[ QUOTE ] putting your opponent on a statistically-distributed range of hands based on combinatrics [/ QUOTE ] How can I find the exact statistics of a hand? I've always just been guessing. [/ QUOTE ] It is just a guessing game - sorry if I made it sound like an exact, mathematical science. It's really a combination of math and psychology. After Villain's turn raise, for instance, my thought process was first considering the past play in the hand. I decided that (by far) the most likely hands he would raise preflop and play this strongly would be a high overpair or 88. There's 3 unseen eights, so his possible winning starting hands could be 8[img]/images/graemlins/club.gif[/img]8[img]/images/graemlins/diamond.gif[/img], 8[img]/images/graemlins/spade.gif[/img]8[img]/images/graemlins/diamond.gif[/img], or 8[img]/images/graemlins/spade.gif[/img]8[img]/images/graemlins/club.gif[/img]. That's 3 hands. Now there's 4 unseen aces, kings, and queens, for instance. This is where the reads affect the statistical distribution: they determine what overpair hands I would put him on. Some players wouldn't play any overpairs this strongly, some would just play aces, some play aces or kings, etc. Let's assume he'll only play aces like this. There's four unseen, so that's a total of 6 combinations of losing hands he could play. This again is a simplification as there's a slim chance, for instance, that his cat is walking across his keyboard and he's playing 72o here. But in game time we need to neglect the stuff that's not significant enough to make a huge difference. So I see 3 hands beat me, and I beat 6 hands. I figure I'm ahead at least 1 time in 3. And that's enough to make me want to push it. |
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