Do you have to play the board and beat the Q card even if your opponent has a higher kicker of 8?
2006-09-25 17:40:49 · 10 answers · asked by babysmilea 1 in Games & Recreation ➔ Card Games
tie i believe
2006-09-25 17:44:01 · answer #1 · answered by lawumun 3 · 1⤊ 0⤋
It is a split pot. In a case where the board is paired only one card is going to play as a kicker because both people have two pair, and in this case the Queen is higher than the second hole card of both players, so the Queen plays as a kicker for both players.
http://www.theultimatepokerguide.com/article.php/texas-holdem-poker-game-rules illustrates the rules of when certain cards play and how to read the board correctly, that may help you out in future situations like this.
2006-09-25 17:46:22 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The pot is split in this case. Each player has two pair (Kings and Tens), with a Queen kicker. The 3 in your hand, and the 8 in the opponents hand is discarded.
2006-09-25 17:46:16 · answer #3 · answered by Kevin A 1 · 0⤊ 0⤋
It's a split pot. You both have Kings and 10's with a queen kicker. The queen plays. Best five cards.
2006-09-25 17:44:25 · answer #4 · answered by MoMoney23 5 · 0⤊ 0⤋
Split pot. Only five cards are played. Two pair; kings and tens with a queen kicker for each. Either way playing a 10-3 and a 10-8 to the river with that board is a poor decision.
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2006-09-25 18:52:05 · answer #5 · answered by Poker Guide 2 · 0⤊ 0⤋
the hand is a draw since both play best 5 cards hence the hands are K K 10 10 Q .
2006-09-26 04:52:15 · answer #6 · answered by gproy2 1 · 0⤊ 0⤋
it is a split pot. 10 10 K K Q is the 5 cards you use. the 5 highest cards.
2006-09-25 17:43:12 · answer #7 · answered by cascade_waterfall04 2 · 1⤊ 0⤋
You both have two pair with Q kicker, pot is split
2006-09-27 06:07:18 · answer #8 · answered by Anonymous · 0⤊ 0⤋
Can't answer this without knowing the suits of each card.
2006-09-25 17:44:15 · answer #9 · answered by ValleyR 7 · 0⤊ 0⤋
a draw.
2006-09-25 18:49:45 · answer #10 · answered by johnny 1 · 0⤊ 0⤋