Texas holdem question [duplicate]

If the community cards are Q,,Q,10,10,2 and one players hold cards are a 2 how do you decide the winner if no other player can match any of the community cards?

• just a hint 3 pair is a six card hand, only five cards count, so if the board has two pair, the best hand Q,,Q,10,10 X, X= the largest kicker a player may have, whom ever has the best kicker wins, if two player have the same best kicker the pot splits.
– Jon
Commented Oct 15, 2022 at 0:05

If no one else has a Q, 10, or pair higher than 10, then the player with the single highest fifth card wins. Poker hands have exactly five cards, no more, no less. So if player 1 has, for example, 9-2, the best five-card hand he can play is Q-Q-10-10-9, while player 2's K-5 wins with Q-Q-10-10-K. Under no circumstances do a sixth or seventh card come into play.

Insufficient information. The precise answer is winning kicker wins the hand, and if all kickers are void due to being under the board, then it becomes a chop.

Board: Q♥Q♣T♥T♣2♥

Hand: 2♣3⋄, best 5 cards makes Q♥Q♣T♥T♣3⋄, Two pair with kicker on 3

Hand: 4⋄6⋄, best 5 cards makes Q♥Q♣T♥T♣6⋄, Two pair with kicker on 6. This player wins!

As "three pair" is not possible with 5 cards, the best two pairs are kept, then the highest card as kicker.

Another example: Q♥Q♣2♥2♣T♥, this time anyone without a Q, 2 or 10 would either have flush (two hearts) or nothing but two pair, but T♣3♣ would beat A⋄K⋄ (Queen Ten two pair > Queen Deuces two pair)

Another example demonstrating the chop:

Board: Q♥Q♣T♥T♣9♥

Hand: 2♣9⋄, best 5 cards makes Q♥Q♣T♥T♣9⋄, Two pair with kicker on 9

Hand: 4⋄6⋄, best 5 cards makes Q♥Q♣T♥T♣9♥, Two pair with kicker on 9

Both players win and they chop the pot!