Texas Hold 'em is a form of poker where each player has 2 "pocket" cards, which they combine with 3 (or more) community cards from a pool of 5 to form their hand, which are then scored under poker ranks.

Poker cards are scored on the following hands, from best to worst.

  • Straight flush (5 consecutive cards of the same suit)
  • Four-of-a-kind (4 cards of the same rank, and one other card)
  • Full House (3 cards of the same rank, and 2 cards of another rank)
  • Flush (5 cards of the same suit, not consecutive)
  • Straight (5 consecutive cards, not suited)
  • Three of a Kind (3 cards of the same rank, 2 cards of other different ranks)
  • Two pair (2 cards of the same rank, 2 cards of another rank, and 1 card of yet another rank)
  • One pair (2 cards of the same rank, 3 cards of different other ranks)
  • High Card (5 cards of different ranks, neither suited nor consecutive)

For the purposes of this puzzle:

  • Players are required to select the highest possible hand. If they have 2 spades as their pocket and there are 3 spades in the community cards, they cannot form any hand worse than a flush.
  • All hands of the same type are equivalent. E.g. all straight flushes are treated as the same hand, even ace-high royal flushes.

Is it possible to form a set of community cards and 9 sets of pocket cards so that all 9 kinds of scoring hands are formed? If not, what is the maximum number of different kinds of hands that can be formed?


1 Answer 1


I believe the maximum number of different hands is 7

Because 4 of a kind requires a pair on board, which eliminates High Card, and requires both of the other two available cards of that value, eliminating Three of a Kind. (An on board pair turns any other valued triplet into a Full House)

Here is one such example:

Clockwise Straight Flush, Flush, Straight, Four of a Kind, Full House, Two Pair, Pair
Hold' Em Hand

  • 12
    $\begingroup$ Also, if you try leaving out the quads, the full house still requires a paired board, making "high card" impossible, so you can't get to 8 that way either. $\endgroup$
    – Bass
    Apr 27, 2023 at 20:00
  • 1
    $\begingroup$ One can easily prove that 8 is not possible. First, if the board is not paired, then neither 4 of a kind nor full house can be achieved. Second, if the board is paired, one can't make high card and only one of 3 of a kind and 4 of a kind can be achieved. Finally, if the board contains either 2 pair or 3 of a kind then both single pair and high card can't be achieved. $\endgroup$
    – quarague
    Apr 30, 2023 at 17:49

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