16

This question already has an answer here:

  • The table cards were A,K,K,8,3.
  • All four suits were on the board.
  • Player one had A,8.
  • Player two had A,3.

Is this a split pot?

marked as duplicate by GendoIkari, Nij, Toon Krijthe, TheThirdMan, doppelgreener Jul 23 '17 at 9:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

25

Yes, it is a split pot. Both players have AAKK8. You always make your best 5 card hand from the cards available. From Wikipedia:

A hand always consists of five cards. In games where more than five cards are available to each player, the best five-card combination of those cards must be played. Any cards not included in the hand do not affect its ranking. For example, if player A holds 3♠ Q♦ and player B holds 3♣ A♣, and five cards 4♣ 5♦ 6♦ 7♠ 10♥ are available to both players, the players hold equally ranking 3-4-5-6-7 straights despite the fact that the player B's ace ranks higher than the player A's queen.

  • Wow 20 upvotes. I guess i should start asking and answering poker questions. – user1873 Jun 22 '13 at 3:10
5

The best hand in either case would be two-pair with 8 as a kicker (A-A-K-K-8); it makes no difference if the 8 came from the table or from the hole cards. Since suit also doesn't matter, the hands would tie.

So yes, it would be a split pot.

0

If the table were A, K, K, 7, 3, the A-8 hand would win. The same would be true if the table were say, A, K, 8, 7, 3. Then the A-8 holder would have Aces and eights with a K kicker, while the A-3 holder would have A, A, 3,3, K, a weaker second pair.

But the actual table gives the A-3 holder A, A, K,K, 8, same as the A-8 holder. That leads to a split pot.

The 8 on the table "counterfets" (nullifies) the effect of the superior 8 kicker held by the fist hand. At the same time, the second pair of kings means that the A-8 does NOT beat A-3 with the better second pair.

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