Thank you for the kind words! Appreciate it

The Addition Method outlined by Michael Schell is something that you could look at, here: www.cribbageforum.com/YourCrib.htm#addition . This "quick and dirty" method can be used to evaluate tricky discards by adding the hand value together with the expected value of the two-card discard, to give a general idea of which discard option is better. So it's certainly easier than calculating the expected averages, but this method does require having a good understanding of average crib values. Something to consider

Thanks for the counting tip! I’ve read that before and will review again. The star method is an effective simple guide.

Thanks for the comment! I'm happy to clarify. We know there are 20,358,520 possible 6-card Cribbage hands. We also know, from doing the math, that 18,031,624 of these hands contain 3 cards of the same suit, or fewer. Since flushes in Cribbage require 4 cards of the same suit (at a minimum), that means that no flushes can be formed from any of these 18,031,624 hands. And therefore, suit is a non-factor in these hands. If suit is a non factor, then all that remains is rank (e.g. ace, two, three.. etc.) And the count of Cribbage hands, rank only, (i.e. unsuited) is 18,395: A-A-A-A-2-2 all the way to Q-Q-K-K-K-K. So these 18,395 unsuited hands effectively represent all 18,031,624 hands. Hope that helps. (All credit to the author of the "Cribbage Discards" book, from where I adapted this information.)

@@Cribbage_with_J Thanks for your reply. I followed all your combination calculations for getting to the number of hands with flush possibilities and subtracting these from the total number of hands possible to get the number of hands that contain 3 cards of the same suit, or fewer. However, I can't figure out what combinatorial equations you need to use to get to the 18,395 unsuited hands when suit is a non-factor. Is there a formula or set of formulae, similar to what you did in calculating the number hands with flush possibility, that can be used to equate to the 18,396? I see that if you divide 18,031,624 hands by 18,395 you get 980 but I'm not sure how you would get to that number (980) if you were trying to figure out what proportion of the 18,031,624 hands are basically equivalent once you factor out suit - e.g. AD, 3C, 5H, 6C, 8S, KD equals any hand with an Ace, 3, 5, 6, 8 and King.

Using your example of AD, 3C, 5H, 6C, 8S, KD, that hand cannot form a flush. Even if that hand contained one additional diamond (e.g. AD, 3C, 5D, 6C, 8S, KD) it still could not form a flush. Same with AC-3D-5S-6C-8H-KS. And AH-3S-5S-6D-8H-KC. These are all just variations of the same flush-less hand. The suits don't make any difference, because no flush is possible. So all of these hands could simply be represented in an unsuited way: A-3-5-6-8-K. I'm not sure what the combinatorial equation would be for this, (perhaps others might be able to comment) but in the absence of this, the logic is still sound: for any hand containing 3 or fewer cards of the same suit (which is 90% of all possible hands), no flush is possible. And thus, suit does not matter. Therefore, all of these 18,031,624 hands can be represented by an unsuited version.

@@Cribbage_with_J Thanks again for your reply. I follow everything you've said. I just wondered, going back to your original reply and your video regarding the table that goes from A-A-A-A-2-2 all the way to Q-Q-K-K-K-K in Myers book if there was some formula for calculating that there are 18,395 unsuited hands in this set or table. Naively, I was thinking there were 13 possibilities (Ace through King) for columns 1 to 4 in the book (e.g.13*13*13*13) and then something less than that for the 5th and 6th column because you can't have more than 4 of each rank in a hand but couldn't figure out any calculations which would give me the 18,395. Maybe I just have to lay out all the possibilities in Excel and count them up. Any idea of a more elegant way to get to this magic number?

Ah ok, I understand. In Myers' book, there is a table in the appendix section (page xxiii) where he shows the various combinations for all unsuited hand "types". For example, one type of hand is 4,1,1 ("four of one rank, one of another, and one of yet another") and the count of combinations for this specific hand type is 858. There are nine different hand types in total, and the sum of combinations for the nine hand types adds up to 18,395. Hope that helps!

I'm interested to know what you think the math should be