Theres a pocket 9s with a 9 on the flop, other player has a Q in pocket and 2 queens on the flop. Who wins and why? Is it just a matter of higher card?
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The best five card hand wins.
In this case, player one has 9 9 9 Q Q or a full house, 9s over queens.
Player two has Q x Q Q 9 or trip queens (3 of a kind where two of them are on the board), assuming the other hole card is not a nine.
A full house beats three of a kind, so player 1 wins. Note that even though the queens are the higher three of a kind, the two queens that contribute to player two's three of a kind actually give player one a better hand.
However, if by some chance player two's hole card was a nine, then they would have a better full house, ( Q 9 Q Q 9 = Q Q Q 9 9) queens over nines and would win.
Also, this is also assuming neither player improves on the turn or river, as all 5 community cards must be dealt before there is a showdown in Texas Hold Em.
Now if you want to ask about the same card (as your question title seems to imply) it doesn't matter if you make three of a kind with one hole card and a paired board, versus a pocket pair and a third card on the board. In other words, 9 x with a board of 9 9 y is the same hand as 9 9 with a board of 9 x y. However, the pocket pair is worth more in terms of equity because nobody else shares the pair on the board (and thus another player cannot have a full house if the board isn't paired) and it is less likely that another player also holds a nine.