Combinatorics (suit combinations)

    • nchris7
      Joined: 09.09.2011 Posts: 4
      Hey everybody,

      I was reading an article about Poker probability (Texas hold 'em) on Wikipedia ('em)) and I didn't quite understand everything. In the first section it has a table of probabilities for pocket pairs, suited and off-suited cards. What I don't understand is the difference between the first and the last row in the third column, so the "Suit combination for each hand".

      I guess I know what is the difference between 4C2 (4 over 2) and 4C1 x 3C1, though it was a couple years since calculus :) . The latter also takes into account the order.

      I know it must be correct because the cards are either pairs, suited or off-suited non pairs and the combination adds up to 1326.

      By the way, I calculated the pocket pairs this way: 52C1 + 3C1. This approach is easier for me but I'd like to understand the other one, too.

  • 5 replies
    • IngridN
      Joined: 02.03.2011 Posts: 12,162
      Hi nchris7,

      Welcome to our community.

      I've moved your thread into the relevant board.

    • EuanM
      Joined: 07.05.2011 Posts: 531
      Hey nchris7,

      Thanks for posting!

      "Suit combinations for each hand", simply refers to the amount of combinations for that type of hand, so:

      Pocket Pair = 6*
      Suited Cards = 4**
      Non-paired, Un-suited = 12***

      *For any PP, you can have one of any six combos in terms of the suit. For Pocket aces, for example, we can only ever have one of these:

      A:spade: A:heart: , A:spade: A:diamond: , A:spade: A:club: , A:heart: A:diamond: , A:heart: A:club: , A:diamond: A:club:

      Any pocket aces you ever get dealt, with be one of the above as shown or with the suits in reverse, hence six combos.

      **For any cards we get which are suited, we have four combos because we have four suits.

      ***For any card we are dealt which will be off-suit & not pocket pairs, we have 12 combos for each of those, because either of the two cards can be of any suit, and as they are not pairs or suited, we don't have any "blockers" as such for a higher amount of combos. Let's take this example:

      72o - We can have the seven in any suit, with the deuce in any suit, apart from the same suit. These are all the possible combinations of 72o we can ever have, because in our analysis we are looking specifically at un-suited, non-paired cards, there are simply more combos as a higher number of cards meet our criteria for analysis:

      7 :spade: 2 :heart: , 7 :spade: 2 :diamond: , 7 :spade: 2 :club:
      7 :heart: 2 :spade: , 7 :heart: 2 :diamond: , 7 :heart: 2 :club:
      7 :diamond: 2 :spade: , 7 :diamond: 2 :heart: , 7 :diamond: 2 :club:
      7 :club: 2 :heart: , 7 :club: 2 :diamond: , 7 :club: 2 :spade:

      4C2 = 6 Either card can have either suit, but never suited as we are looking at PP's only, therefore there are less (6) combos than with Off-suit, non-paired cards.

      4C1 over 3C1 = 12 Neither card will be the same and neither card will have the same suit, because we are only looking at hands which are unsuited, non-pairs, so our hand will always be off-suit and both cards always different, because this criteria applies less restrictions on the range of our calculation, the amount of combos increases. If we say the hand we are looking at HAS to be suited, or HAS to be a pocket pair, there are far less manners in which these cards can combine, hence less combos.

      It's easier to get your head round if you work with one specific hand example, as shown above.

      Does this explanation answer your question? Let me know if you need something more in-depth!

      All the best

    • nchris7
      Joined: 09.09.2011 Posts: 4
      Hey l33tsoz,

      Thanks for the detailed answer! I understand now how it works.

    • nchris7
      Joined: 09.09.2011 Posts: 4
      I have another problem about the same topic (that's why I didn't create a new one)

      There's another table which contains probabilities and odds of being dealt various other types of starting hands. (Just below the one I mentioned before) I calculated all the possibilities and I got everything right. I used always this 'new technique', to separate the suit and the hand combinations.

      I have a question about this type: Any 2 cards with rank at least X

      I divided the problem into two sub-problems.
      Any PP with rank at least X + Any not-PP with rank at least X

      So, let's say X = Q.
      1st sub-problem: 3 PP x 6 (suit combination) = 18
      2nd sub-problem: 3 not-PP x 16 (suit doesn't matter) = 48
      # of good cases: 18 + 48 = 66
      Probability of starting hand: 66 / 1326 = 0.0498

      After calculating this for X = J, T, 9 and I tried to generalize the formula.
      # of PP with rank at least X = 15 - X (X being 14, 13, 12 and 11 for A, K, Q, J)
      # of not-PP with rank at least X: using the formula for sum of integers I came up with this:
      K = 1, Q = 3, J = 6, T = 10, 9 = 15, 8 = 21, ...
      I assigned numbers from 1 to 12 to the cards from K to 2 and then N being the value N*(N - 1)/2 gives the # of not-PP.

      This formula seemed to be okay. When I wanted to calculate the possibility of having a starting hand when both cards are above 7, so X = 8, I assumed that it's gonna be more than 50% because there's more cards above 7 then below that (including 7). But I got this:

      X = 8
      # of PP: 7
      # of non-PP: 21
      Any PP with rank at least 8: 7 x 6 = 42
      Any not-PP with rank at least 8: 21 x 16 = 336
      # of good cases: 42 + 336 = 378
      Probability of starting hand: 378 / 1326 = 0.2851

      The 50% threshold with this formula is between 5 and 6. (47.5% and 58.8%)

      Where did I make a mistake in the calculation or the thinking?
    • EuanM
      Joined: 07.05.2011 Posts: 531
      I think your math is perfectly fine here:

      If X=Q

      1st problem seems ok, we have 18 combos of PP's above & including QQ.

      2nd problem, seems a bit strange.
      We have six non-PP hands which are QQ+. (KQs, AQs, AKs, KQo, AQo, AKo)
      12 combos for each off-suit hand + 4 combos for each suited hand = ( 3 x 12 )+( 3 x 4 ) = 48

      # of hand combos above + including QQ+ = 66
      Probability of being dealt a combo from this range = 66 / 1326 = 0.0498 = 4.98% = approx 1/20 times.

      Let's say X=8 (Only counting hands which are 88+, A8s+, K8s+, Q8s+, J8s+, T8s+, 98s, A8o+, K8o+, Q8o+, J8o+, T8o+, 98o)

      So this exact range:

      # of PP: 7 (8,8 - 9,9 - T,T - J,J - Q,Q - K,K - A,A)
      # of non-PP: 21 suited hands, each with 4 combos + 21 off-suit hands each with 12 combos (21 x 16 is ok)
      Combos of PP's: 7 x 6 = 42
      Combos of non-PP's: 21 x 16 = 336
      Total combos: 42 + 336 = 378
      Range = 88+, A8s+, K8s+, Q8s+, J8s+, T8s+, 98s, A8o+, K8o+, Q8o+, J8o+, T8o+, 98o

      Probability of being dealt a combo from this range = 378 / 1326 = 0.285 = 28.51%

      As you can see, your calculations are correct.

      The mistake comes from the assumption that the above range accounts for 50% or more of your starting hands combos, which is not true. As you can see on the grid in the attached screenshot, it does not cover half of the total starting combos and therefore cannot be a correct assumption.

      It's at this point, I recommend downloading our Equilab if you havn't done so already. This is the download link.
      It makes the visualization much more straightforward, in addition to this, it lets us do the calculation in a different way.

      We want to calculate the top 50% of our starting hand range? (As you want to know the 50% of the combos on the top-side of the range grid so we must go from here)
      We know that there are 1326 combos of starting hands, so the range we want to find includes 1326 / 2 = 663 starting hand combos.

      Suited cards in general, account for a smaller percentage of the top of our range than off-suit cards because they have less combos, pocket pairs account for the lowest percentage. Because the top of our range includes more suited cards than off-suit cards, the bottom of our range is perceived (in our minds or on the grid) to be physically larger than it actually is in terms of the starting hand itself.

      Here is the grid again, with the top 50% of our range, excluding all PP's for purposes of a more accurate perception.

      As you can see, Off-suit hands account for less of the grid than the suited hands, but there are actually more off-suit combos of course. This image is the same in our minds when we think "There are 13 cards, so every card valued about 7 must be more than 50% of the hands."

      There are two things we need to note about this thought process:

      1. It's incorrect
      2. We are trying to define the top 50% of our range, simply because we want to know the 50% of the cards with the highest values. (Remember we also have the bottom of our range)

      Why is it incorrect?

      It's incorrect because we make the assumption that each hand has the same amount of combos to be dealt within a defined range, on the back of the assumption that we are not taking suits into account, as we are only using numbers. This skews our perception of the starting hands possibilities and should be avoided.

      I hope this helps, this stuff can be confusing at times. Always think in combos!

      Now, We need to determine what we want to find out:

      Do we want to know if this range (88+, A8s+, K8s+, Q8s+, J8s+, T8s+, 98s, A8o+, K8o+, Q8o+, J8o+, T8o+, 98o) accounts for 50% or more of our starting hand possibilities?
      Simple answer is no, it includes 28.51% of our starting hand possibilities, as this particular range, has 378 combos, which is nowhere near 50% x 1326.

      Do we want to know which hands are included in the top 50% of our range & why these hands represent the top 50%?
      Here is the top 50% of our range: 33+, A2s+, K2s+, Q2s+, J4s+, T6s+, 96s+, 86s+, 76s, 65s, A2o+, K5o+, Q7o+, J7o+, T7o+, 98o

      Why do these hands represent 50% of our range?
      Because these are the 50% of combos of starting hands which will hold the most value in the long-term. Keep in mind that we shouldn't play anywhere near 50% of our starting hands, which is why you see trash (K5o, J7o, etc..) hands in the above range I gave.

      Do we want to know the number of combos of hands which are included in the top 50% of our range?
      That's easy it's 50% of 1326 = 663 combos

      Just a quick thing on the back of the last question, we cant have 663 (50%) combos in our range, we need to have 660 (49.77%) or 664 (50.08%). Your number of combos will always be an even number as no starting hand has less than four combos + are all multiples of two. We can only have an odd number if we include a single specific combo,

      There is one combination of 6 :heart: 5 :diamond: , but 12 combos of 65o + 4 combos of 65s, hence a defined range would only include an odd number of combos if we ourselves include a specific starting hand AND the suits for each card.

      I hope this helps, let me know if you have any other questions.