Posted in Poker March 2nd, 2010
Texas Holdem?
In Texas holdem, there is a pair of Kings showing and a pair of Queens showing and a four. One person has an Ace and a three, and the other person has a pair of eight. Who wins? Is it the person with two high pair with the Ace kicker or is it the person with a pair of eights in their hands? Thanks for your help.
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6 Comments on this post
Posted by the authority March 2nd, 2010 at 2:56 am
Ace wins with K K Q Q A, other player has K K Q Q 8, he is
outkicked by the Ace.
Posted by JOHNSHAFT005 March 2nd, 2010 at 3:30 am
I would think the person with the pair of 8′s would win because it would be three pairs to two pairs and an ace or
K K Q Q A 3
vs
K K Q Q 8 8
Posted by HokiePaul March 2nd, 2010 at 3:40 am
Best 5 card hand wins. In this case it would be two pair with an Ace kicker. KKQQA would beat KKQQ8
Posted by sincity usa March 2nd, 2010 at 3:50 am
If the board is k-k-q-q-4 and you have an A-3 and the other player has a pair of 8-8s in the hole then the A-3 would win. you both have kings and queens, but you have an ace kicker and he only has an 8 kicker, he got counterfeited. You can only play you best 5 card hand out of the 7 cards available.
Posted by centreofclassicrock March 2nd, 2010 at 4:13 am
THe other person. 3 pair.
Posted by tesla_man_1999 March 2nd, 2010 at 4:45 am
There’s no such thing as “three pair”. The A3 hand winds because his Ace outkick’s the opponent’s 8.