Posted Sun Jan 08, 2006 8:20 pm GMT by ambiguous1
Im new and have a question regarding kickers..
the board is K J 8 9 6
Player 1 has K 2
Players 2 has K 4
who wins the hand? I'm confused because I'm not sure if player 2 wins because he has a better kicker which is the 4..but since none of the players kickers( the 4 or the 2) beat the community cards, would it be split pot? After all, to win a hand one must have the best 5 cards which in this case the community cards are all higher then their kickers..please explain
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Posted Sun Jan 08, 2006 8:31 pm GMT by ComedyBee
Split pot.
Effectively the hand is a pair of Kings with J 8 9 as kickers. Because both of the held kickers are low than all the kickers on the board, they dont come into play and it is split.
Hope that helps bud.
CB
Posted Sun Jan 08, 2006 8:39 pm GMT by ambiguous1
so in order for one of them to win, the lowest kicker they'd need would be a 10 right?
Posted Sun Jan 08, 2006 8:42 pm GMT by Dat_Dude
| ambiguous1 wrote: | | so in order for one of them to win, the lowest kicker they'd need would be a 10 right? |
Actually, it would be a 7 as the lowest possible kicker to not split the pot.
If player 1 had a 7 instead of a 2 then his hand would be:
K J 8 9 7
And player 2's best hand would still be:
K J 8 9 6
Posted Mon Jan 09, 2006 10:52 am GMT by golddog
http://www.texasholdem-poker.com/forum/t3459/poker-faq--please-read-before-starting-a-new-thread
Posted Mon Jan 09, 2006 10:50 pm GMT by ambiguous1
usually in a case where a guy has a kicker in his hole card that doesnt play because its not higher then the community cards, will he usually check it alot near the end? because he knows that if his opponent has a kicker that will actually beat the board, then he loses. but if his opponent's kicker situation is just like him (the board plays), then its split pot.
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