11
$\begingroup$

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?

$\endgroup$

1 Answer 1

17
$\begingroup$

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

$\endgroup$
2
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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