How to rank a new type of Hold 'Em Poker hand (hard)

Question

I am trying to figure out how to rank a new type of poker hand, henceforth called a 'Block'.

A Block consists of 6 cards, and so is not possible in regular Poker - only Hold'em.

Each Block consists of 2 sets of 3 consecutive cards, looping from Queen to Ace and ignoring Kings. Each Block uses the same numbers, in different colours (red and black).

For example: A '2' Block would consist of Ace, 2 and 3 of any combination of spades OR clubs, and A, 2, and 3 of hearts OR diamonds. In short, a red set and a matching black set.

A Looped Example: A 'Q' Block would consist of J, Q and Ace of (spades and/or clubs), and J, Q and Ace of (hearts and/or diamonds).

My brain melts when I attempt to calculate the odds, but here is my best attempt:

• So, there are 12 Blocks, each of which can be made in a bunch of ways (64?) - so 768 ways to make a block.
• A web search tells me there are 133,784,560 ways that 7 cards can peel from the top of a 52 card deck.
• 133,784,560 / 768 = 174,198

... So, if I'm at all in the region of correctness, the chances of forming a Block are very, very slim. Like, only 4 or 5 times more common than a straight flush, and slightly more rare than 4 of a kind (going off of https://en.wikipedia.org/wiki/Poker_probability#7-card_poker_hands).

So here are my questions:

1. How would a Block rank versus regular Hold 'Em hands? Aka, did I do the above math right (I almost certainly did not)?

2. If we add that Kings are wild, just for this particular hand, how much easier does achieving a Block become? [Trying to calculate this makes my eyes jitter - we can't reuse the kings but could use more than 1; brain hurty.]

... Thanks y'all.

Answer 1

Your calculation is almost correct; you're missing two points. The first point is that you also need to count the possibilities for the seventh card, which could be any of the $46$ remaining cards. So the number of possible "Block" hands is closer to $12\cdot 2^6 \cdot 46=35328$. The second point is that some of these hands can form a "Block" in two ways; specifically, this happens when the seventh card is the same rank as any of the other six cards. So the exact count is $$ 12\cdot \left(40\cdot 4^3 + 3 \cdot 4^3\right) = 33024, $$ where the first term counts the hands consisting of a Block and a seventh card of a different rank (these are two-pair hands), and the second term counts the hands consisting of a Block with an extra card in one of the three ranks (these are full houses). This gives a probability of $$ \frac{33024}{133784560}\approx 0.0247\%; $$ so a Block should beat a (non-royal) straight flush (being slightly less probable) and lose only to a royal flush.