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I have the following constellation: Player 1 has a full house, and player 2 has 2 three of a kind. I'm aware that because of the five card rule, player 2 also has a full house. But the question is, doesn't player 2 have a higher full house since player one has Aces full of kings, while player 2 has Kings full of Aces, which is higher (since the pair is relevant I believe) ?

It seems that most equity calculators consider this to be a draw, but I would argue that it is not.

Player 1: 2♥ 8♠

Player 2: 4⋄ K♥

Table: A♣ A⋄ A♠ K♠ K⋄

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    Player 2's king has been counterfeited.
    – user1934
    Oct 19, 2015 at 0:14

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They both have aces full of kings. The pot would be split. Aces are always higher than kings in holdem, except when played low as part of a A2345 straight.

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  • Thanks for the clarification, for some reason I though the pair was relevant rather than the triple to decide which full house is higher. Would you the agree that if the table is ['AC', 'AD', 'AS', 'KS', 'KD'] the only way to escape a draw would be to have a four of a kind? The reason I'm asking is I don't see how there are 45 possibilities for the opponent to win as described here: propokertools.com/simulations/…
    – Nickpick
    Oct 18, 2015 at 23:04
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    The 45 wins are Ah2c, Ah2d,...AhKc, AhKh, and KhKc, each of which makes quads.
    – Lee Daniel Crocker
    Oct 19, 2015 at 2:44
  • @Nicolas even if the pair were relevant (which is not the case), it's irrelevant here because each player has the same best hand: AAAKK. Player 2's K does not improve his hand at all and while you can find KKKAA in player 2's hand, player 2's best (and therefore played) hand is still AAAKK.
    – mah
    Nov 9, 2015 at 17:39

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