There are quite a few questions and answers here on specific holdings in poker and their frequency. I am more generally interested in the number of distinct 5-card hands that are possible in 7-card poker. That is, seven cards are drawn from a standard deck of 52, but only the best 5-card combo of all the possible combos in that 7-card draw counts.
For example: The draw is KKK87654, and contains also six cards of the same suit. That holding has 1) Three of a kind; 2) a straight; and 3) two flushes. But because the flush beats the other holdings, only the higher flush counts. So I don't want to count the lower flush, the trips, and the straight, as it would be double counting.
Another example is AT98642. This is an Ace-high holding, with kickers T986. This hand is equivalent to AT98632, because the lowest two cards are ignored in the evaluation.
By equivalence classes I mean:
- Straight Flush (including Royal Flush)
- Four of a Kind
- Full House
- Flush
- Straight
- Three of a Kind
- Two Pair
- One Pair
- High Card
So, to summarize the question: how many distinct hands are there for each of these categories, and how to calculate them?
The exercise of counting all possible 5-card holdings in 5-card poker is relatively straight forward. There are ${52 \choose 5} \sim 2.6$ million holdings, but only 7642 distinct hands.
The problem gets trickier for 7 draws, because we need to discount "kickers" that may overturn the original hand, or because some holdings are not possible at all. For example, a 7-high holding is impossible a draw of 7 distinct cards will necessarily include at least one card higher than 7. I am stuck with the combinatorics of these cases.
Note that there are $52 \choose 7$ holdings possible ($\sim$ 133 million), but because of the restrictions referred to above, the number of distinct 5-card holding is supposed to reduce to just 4824.