Evening, I've got a question I've been wrecking my head about--namely which two pair wins when one has top and minimum pair while the other players has second and third pair.

To paint you a picture, let's say the board is (Qc)(Kc)(3d)(4d)(8h)

Player 1 has (Qh) (8h)

Player 2 has (Kh) (3c)

In this case, would player 2 win due to him having the highest pair? This is what makes sense to me, but my friend keeps arguing that it should be a split because both players have one over the other.

I've been trying to google an answer to this, but either my wording is too vague or there simply isn't any answer to this due to this being such a simple question.

Anyhow, I appreciate any answer I might get.

  • To stand on Lee's excellent answer: The other 'home game rule' like your split pot scenario is a tied hand where they go to the sixth card to determine the outcome. It looks something like: Player1 has K5o, and Player2 has K10o. If the board is AAJKK they will argue that player2 has the best hand because the 10 beats the 5. But, this obviously would be a six card hand, and the rule is the best 5 card hand wins.
    – RobW
    Commented Apr 20, 2022 at 22:34

1 Answer 1


First, highest pair wins. KKxxz always beats QQyyz, regardless of the second pair. It is only when the high pair is identical that the second pair matters, so QQ99x beats QQ77y. Only when both pairs are identical does the fifth card play: KK88J beats KK885.

Not the answer you're looking for? Browse other questions tagged or ask your own question.