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.
2006-08-12 12:32:14 · 6 answers · asked by Gary W 1 in Games & Recreation ➔ Card Games
Best 5 card hand wins. In this case it would be two pair with an Ace kicker. KKQQA would beat KKQQ8
2006-08-12 14:21:51 · answer #1 · answered by HokiePaul 6 · 0⤊ 0⤋
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.
2006-08-13 03:17:03 · answer #2 · answered by sincity usa 7 · 0⤊ 0⤋
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
2006-08-12 20:02:48 · answer #3 · answered by JOHNSHAFT005 2 · 0⤊ 4⤋
Ace wins with K K Q Q A, other player has K K Q Q 8, he is
outkicked by the Ace.
2006-08-12 19:37:37 · answer #4 · answered by ? 2 · 0⤊ 0⤋
There's no such thing as "three pair". The A3 hand winds because his Ace outkick's the opponent's 8.
2006-08-13 12:07:25 · answer #5 · answered by tesla_man_1999 2 · 0⤊ 0⤋
THe other person. 3 pair.
2006-08-13 09:43:24 · answer #6 · answered by centreofclassicrock 4 · 0⤊ 5⤋