The following is a passage from Wikipedia on starting hands probability:

The 1,326 starting hands can be reduced for purposes of determining the probability of starting hands for Hold 'em—since suits have no relative value in poker, many of these hands are identical in value before the flop. The only factors determining the strength of a starting hand are the ranks of the cards and whether the cards share the same suit. Of the 1,326 combinations, there are 169 distinct starting hands grouped into three shapes: 13 pocket pairs (paired hole cards), 13 × 12 ÷ 2 = 78 suited hands and 78 unsuited hands; 13 + 78 + 78 = 169. The relative probability of being dealt a hand of each given shape is different. The following shows the probabilities and odds of being dealt each type of starting hand.

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The equation used to calculate the total possible number of starting suited hands and unsuited hands is 13 x 12 ÷ 2, but what I don't get and try to figure out is what does 12 and 2 represent in the equation?

Another thing I'm confused about is how the suit combinations for each unsuited non-paired hand is calculated?

2 Answers 2


When you are dealt a card, it has 13 possible ranks. When you are dealt a second card, for it not to be a pocket pair, that can be of any rank different to your first hole card - i.e. 12 different card ranks. For the division by 2, the calculations are for the number of combinations of different types of hands and a combination disregards order. If you take 13 * 12 you get each possible non-paired hand twice, e.g., AK and KA or 74 and 47, so you divide by 2 to count each combo only once. Thus the equation 13 × 12 ÷ 2 = 78.

For the 12 possible (suit combinations for each hand) for (unsuited cards nonpaired) hand shape equation, the 4 represents the number of suits and the 3 represents the number of suits any single suit can be paired with to give an offsuit combination. Four suits x three combinations each = 12 combinations total. The equation is not a fraction, the 1 in the equations is not a denominator. The symbol with brackets such as (4 1) is for the number of combinations of 4 things taken 1 at a time.

  • Just to add additional info, as you mention, "Combinations" DON'T consider the relative order of items, whereas "Permutations" DO consider the relative order of items.
    – Toby Booth
    Commented Mar 11, 2013 at 22:52
  • @TobyBooth - With combinations, order doesn't matter, with permutations, order does matter. But surely you know this. Are you using the word "consider" in a way I don't understand?
    – TTT
    Commented Mar 12, 2013 at 17:24
  • @TTT Oops, Typo. Just mixed them up ;) Thanks for catching that! Fixing now.
    – Toby Booth
    Commented Mar 12, 2013 at 19:02

Remember, don't get confused with the method used in applying combinatorics in the overall game of poker. What is being worked out here is the probability of starting hands being dealt to you, nothing else. As far as applicability to real poker, and the methods used, there are 16 combinations of AK off-suit, but only four combinations of AK suited. The same is true for all unpaired combinations, and this is very important to understand when applying combinatorics to poker. The best way, if you want to be clear and keep things easy to understand, is just write it out.

If we want to know the probability of being dealt a given strength starting hand, the above answer is correct. But technically, we want to know the probability of an exact starting hand being dealt, e.g. Js7h, suits included given that poker has the flop, turn, and river streets, to play or run out. Js7h is one single combo, the chance of being dealt this is therefore 1 in 1326. But the chance of being dealt any combo of J7 off-suit is 16 in 1326, because there are 16 combinations of J7o. To illustrate with the suited cards - there are four suits - spades, clubs, hearts, diamonds. So there can only be 4 combinations of J7 suited - J7 spades, J7 clubs, J7 hearts, J7 diamonds. The probability of being dealt Jd7d (J7 of diamonds) is the same as being dealt Js7h, 1 in 1326. But the probability of being dealt J7 'suited' isn't 16 in 1326, like it is with J7 off suit, instead it is 4 in 1326.

Hope this adds to/clarifies on the other post.

[Correction there are 12 not 16 combinations of AKo and J7o. There are a total of 16 combinations of AK or J7. Four of those combinations are suited, 12 are un-suited.]

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