Posted Sun Oct 26, 2003 3:20 pm GMT by matrix
Okay I have a question:
in Texas Hold'em - if the Flop is a straight and everyone of course gets to play the straight - who wins the person with the highest kicker, like an Ace or the person with a pair or even two pairs?
Thanks for any answers.
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Posted Sun Oct 26, 2003 4:07 pm GMT by karasz
the pot gets split big guy...
so if:
player aa
player kk
player qq
player jj
player aj
player kq
player 33
player 88
player 27
player 29
and the board is 10jqka
doesnt matter the pot gets split evenly to each remaining player... and usually nobody bets cuz you cant beat that hand...
Posted Sun Oct 26, 2003 4:08 pm GMT by Geno
In TEXAS HOLDEM, you aim to make the best 5-CARD hand from the 7 available to you (the community cards and the two u hold...).
If the board shows a straight, the only way the pot can be won outright is if a player holds 1 or 2 cards making a higher straight, all pairs, kickers etc are meaningless bcos the 6th card simply does not count. For example:
Board shows: 6,7,8,9,10
Player 1 holds AA
Player 2 holds KQ
Player 3 holds J3
Even though Player 1 holds a pair and Player 2 holds nice high cards, Player 3 wins bcos he can make the straight 7,8,9,10,J whereas Players 1 and 2 simply have wot's on the board - 6,7,8,9,10. If no1 held a Jack, the pot would be split between all remaining players. Note that a 5 also means nothing as it would be part of a LOWER straight.
NB. If a player held Q,J in this situation, he would win outright as he holds a straight 8,9,10,J,Q which is 'the nuts' hand in this situation assuming there is no flush on. Bear that in mind too that if 3 or more of the community cards are of one suit, some1 may well steal the pot with a flush which always beats a straight of course!
Hope that helps, I think I've written it clearly enuff 8)
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