I was hoping someone could help with the math for the following variant scenario: game would play out just as Hold'em but all 7 cards would be live with expanded hand rankings. The expanded rankings would include the following along with standard rankings:

  1. High card
  2. 1 pair
  3. 2 pair
  4. Trips
  5. 3 pairs
  6. Straight
  7. Flush
  8. 2 trips
  9. 6 card straight
  10. Full house
  11. 7 card straight
  12. 6 card flush
  13. 7 card flush
  14. Four of a kind
  15. Straight flush
  16. four of a kind plus pair
  17. four of a kind plus trips
  18. 6 card straight flush
  19. 7 card straight flush

Intuition only puts them in that order. Could someone please help with the correct order from a mathematical perspective?

  • Isn't this completely opinionated? What makes a 7-straight beat a full house? Why are two sets over a flush? What about the royal flush, wouldn't it beat quads? For starters, there is no "one definite rule" to hand rankings; house rules such as sets losing to two pair has been used in open tournaments before (rationale being two pair uses more cards in the hand than sets and thus more chops will happen instead of letting the kicker battle overwhelm most situations, and also that sets being strong puts a lot of pocket pair battles into a game of luck where even AK hitting both can lose easily)
    – Unihedron
    Oct 17, 2022 at 12:59
  • @Unihedron I am looking for the rankings based on probability, not opinion. Yes, a royal flush beats quads as seen in numbers 14 and 15. I have never played with house rules that don't follow standard hand rankings.
    – SDH
    Oct 17, 2022 at 14:17
  • 2
    Poker hand strength isn't decided by probability, but the rule-makers. Hell, if it really were, then "high card" would beat "one pair" since it's more common to have only one pair as compared to none at all: en.wikipedia.org/wiki/Poker_probability#7-card_poker_hands You'd better calculate and put down the percentages into your question if that's what you want to work with.
    – Unihedron
    Oct 17, 2022 at 20:54
  • 1
    (Also take note of how, by adding new hands, you are removing probability from lower hands being made since some of them become new, upgraded hands of higher rank, which in some ways results in inbalances when comparing new hands like quad-trip "large full house" to a measly trip-pair "small full house")
    – Unihedron
    Oct 17, 2022 at 21:03

1 Answer 1


I'm curious about how these would rank out mathematically, too. I have most of these figured out using sets, but the straights, and flushes are difficult for me to figure out, especially when you consider 5, 6, and 7 card versions. For just sets (pairs P, triples T, and quads Q) the ranks are as follows:

1P (63258624) = 13x6x12C5x(4^5)
2P (29652480) = 13C2x(6^2)x11C3x(4^3)
0P (28114944) = 13C7x(4^7)
1T (6589440) = 13x4x12C4x(4^4)
1T1P (3294720) =13x4x12x6x11C2x(4^2)
3P (2471040) = 13C3x(6^3)x10x4
1Q (183040) = 13x12C3x(4^3)
1T2P (123552) = 13x4x12C2x(6^2)
2T (54912) = 13C2x(4^2)x11x4
1Q1P (41184) = 13x12x6x11x4
1Q1T (624) = 13x12x4

The 1T2P, 1Q1P and 1Q1T hands all lack names, but, I think I'd call them Duplex, Mansion, and Castle, respectively.

Now, what I do know is which hands the Straights, Flushes, and Straight Flushes (here I'll just say SF) are special cases of:
5SF: 0P, 1P, 2P, 1T
6SF: 0P, 1P
7SF: 0P
Ex. 5-Straight, 2 Pair: A, A, K, K, Q, J, 10.
5-Straight, 1 Triple: A, A, A, K, Q, J, 10.
5-Straight, 1 Pair: A, A, K, Q, J, 10, 2.
5-Straight, 1 Pair: A, K, Q, J, 10, 2, 2.
5-Straight, No Pair: A, K, Q, J, 10, 3, 2.
6-Straight, 1 Pair: A, A, K, Q, J, 10, 9.
6-Straight, No Pair: A, K, Q, J, 10, 9, 2.
7-Straight: A, K, Q, J, 10, 9, 8.

Now, this creates three different sets of ranks: one where SF hands can have Pairs and Triples using the same cards, one where only 5SF and a Pair is a unique rank, and one where 5SF with a Pair is just a 5SF with two kickers at the same rank. Also possible, and arguably easier to remember is for 7 card hands to play as 7 card hands, meaning all 7 cards must be the same Suit for a Flush, or in a 7 card sequence for a Straight, or, ignore some of the actual ranking theory and make 5,6,7 card SF hands all rank the same as 5SF hands but the 6 and 7 card variants are higher in their respective ranks. These create 5 alternative ranking models, 3 have additional ranks and two do not. The 3 with additional ranks have an additional 21, 9, and 6 ranks respectively. The other two have no additional ranks. The 5,6,7 SF are all the same rank form is closer to regular 7 card games using the best 5 card hand principle, while the 7SF only form is closer to actually playing Poker with a 7 card hand (and arguably easier to calculate probabilities of) The probabilities for the 7SF only variant is easy enough to calculate Straight (131040) = 8x(4^7) - 32, between 1Q and 1T2P.
Flush (6832) = 13C7x4 - 32, between 1Q1P and 1Q1T.
Straight Flush (32) = 8x4, above 1Q1T.

Using regular best 5 card hand rules, pulled straight from Wikipedia, we have

Straight = 6180020, between 1T and 1T1P.
Flush = 4047644, between 1T and 1T1P.
Straight Flush = 41584, Above 2T, below 1Q1P.

Mind, I did not subtract any of these from purely Pair, Triple, and Quad based ranks, so there is a double count here with these not pulling their possibilities from 0P, 1P, 2P, and 1T hands. To do this, I would have to calculate with all 21 additional possibilities to find where I need to subtract how much. Advantage to this more exhaustive method: I can then produce rankings for all 5 possible systems (duplicates allowed, no duplicates, no duplicates but 5SF pair, best 5, only 7).

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