Posted Wed Nov 02, 2005 4:43 pm GMT by TwoBob
Two players in a showdown and they hold the following:
Player 1: A K J 9 2 (Hole cards were A 2)
Player 2: A K J 9 4 (Hole cards were A 4)
(The community cards were K J 9).
Who wins?
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Posted Wed Nov 02, 2005 4:54 pm GMT by twofotisx
Assuming KJ9 are the only cards on the board- or you were playing 5 card draw and the players both showed the mentioned hands, the person with the 4 outkicks the person with the 2.
Posted Wed Nov 02, 2005 5:01 pm GMT by galderon
| TwoBob wrote: | Two players in a showdown and they hold the following:
Player 1: A K J 9 2 (Hole cards were A 2)
Player 2: A K J 9 4 (Hole cards were A 4)
(The community cards were K J 9).
Who wins? |
In hold'em, that scenario is impossible. What are the two unnamed community cards? They cannot be unused, no matter what they are. If there's a 2, player 1 has a pair, if there's a 4, player 2 has a pair, if they're both 3's, they'll both have a pair of threes, and anything above a 4 will be used instead of their lower hole cards.
Most "who wins" questions are answered here: http://www.texasholdem-poker.com/forum/t3459/poker-faq--please-read-before-starting-a-new-thread
Posted Thu Nov 03, 2005 3:08 am GMT by traz
| galderon wrote: |
In hold'em, that scenario is impossible. What are the two unnamed community cards? They cannot be unused, no matter what they are. If there's a 2, player 1 has a pair, if there's a 4, player 2 has a pair, if they're both 3's, they'll both have a pair of threes, and anything above a 4 will be used instead of their lower hole cards.
Most "who wins" questions are answered here: http://www.texasholdem-poker.com/forum/t3459/poker-faq--please-read-before-starting-a-new-thread |
yea he's right, the situation is impossible in normal hold em. Either i'm interpreting it wrong, or you're playing another game heh.
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