I'm so glad I found your channel.... But, that "if you remember back to math class.." comment cracked me up. I took algebra 4 times..LOL So, I can spit the quadratic formula or even pythagorean theorem. It wasn't a 'I cant do it' thing... its was a 'I dont like school thing.' This was early 80's also. Love what you are doing.
Your videos are helpful and clarify some strategies. I wonder if there is an easy way to quickly calculate or approximate expected averages when playing ? It’s a lot of mental math otherwise 😢
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
Very interesting video...thank you ! I am trying to learn as much as I can about this fantastic card game and your channel helps! I'm finding out that Cribbage is very addicting once you get a grasp of his to best score points...card combination values etc. Subscibed!
I just got a crib flush last night and I almost couldn't believe my eyes. In cribbage, I find flushes to be alluring because they are relatively strong in Poker, but I always remind myself... I'm not playing Poker!
At time code 7:33 I missed how you got from 18,031,624 non-flush hand possibilities to only 18,395 unsuited hands, Could you please explain? Thanks. Great work by the way and very interesting analysis.
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!
Pretty sure you are double counting your 5 card flush hands and triple counting your 6 card flush hands with the way your are doing the math. Either way, interesting video.
I was thinking the same thing. When using the choose function you’re counting how many hands have *at least* 4 cards of the same suit. Within that ~20 million hands you’ll already have any hands that happen to have the other one or two cards in hand also match the same suit included in your count. Doesn’t change the argument much at all, just pushes you closer to a 1/10 chance for a flush but the same point still stands. Great video!
@itselith Appreciate your kind words and similar to the other comment I left, I would like to understand where you think the math is incorrect. So here's a breakdown: 6-card flushes: We calculated the number of ways to get 6 cards all from the same suit. These are entirely distinct and cannot overlap with any hands with fewer than 6 cards of the same suit. 5-card flushes: We calculated the number of ways to get 5 cards from the same suit and 1 card from a different suit (39 choose 1, which is choosing one card from the other three suits). Since we're explicitly selecting 1 card from another suit, there is no overlap with the 6-card flush hands. These hands are distinct as well. 4-card flushes: Here, we select 4 cards from one suit and 2 cards from other suits (39 choose 2). Again, this cannot overlap with the 5-card or 6-card flushes, because we're specifically choosing 4 cards from one suit and the rest from different suits. [*If the calculation for selecting the 2 cards from the other suits was 48 choose 2, now this would introduce the possibility of selecting cards of the flush suit. But using the 39 choose 2 calculation mitigates this error and any potential overlap] Each case-4, 5, or 6 cards of the same suit-describes a distinct combination, and the calculations were structured in a way that prevents double counting. So, the math is correct, and there's no double counting of hands.
@@Cribbage_with_J Ah, I was getting confused with how the choose functions were interacting to prevent double counting, you're right. My brain tried to shortcut by doing [13 choose 4] and then [48 choose 2] which would allow the 4, 5, and 6 card flushes to be counted in one equation but ends up overcounting because you've lost the guarantee that the [4 choose 1] is only counting the examples with matching suits for the 5 and 6 card flush cases. You could try to divide all that stuff off if your goal was one equation but it would get messy really fast, the way you did it is much simpler for the same result. Sorry for the confusion!
I'm so glad I found your channel.... But, that "if you remember back to math class.." comment cracked me up. I took algebra 4 times..LOL So, I can spit the quadratic formula or even pythagorean theorem. It wasn't a 'I cant do it' thing... its was a 'I dont like school thing.' This was early 80's also.
Love what you are doing.
Your videos are helpful and clarify some strategies. I wonder if there is an easy way to quickly calculate or approximate expected averages when playing ? It’s a lot of mental math otherwise 😢
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.
Very interesting video...thank you ! I am trying to learn as much as I can about this fantastic card game and your channel helps!
I'm finding out that Cribbage is very addicting once you get a grasp of his to best score points...card combination values etc. Subscibed!
Thank you for the kind words! Appreciate it
I just got a crib flush last night and I almost couldn't believe my eyes. In cribbage, I find flushes to be alluring because they are relatively strong in Poker, but I always remind myself... I'm not playing Poker!
Some weird audio glitches throughout, including a glaring one which I assure you is a THANK you 🙏😬
At time code 7:33 I missed how you got from 18,031,624 non-flush hand possibilities to only 18,395 unsuited hands, Could you please explain? Thanks. Great work by the way and very interesting analysis.
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!
Pretty sure you are double counting your 5 card flush hands and triple counting your 6 card flush hands with the way your are doing the math.
Either way, interesting video.
I'm interested to know what you think the math should be
I was thinking the same thing. When using the choose function you’re counting how many hands have *at least* 4 cards of the same suit.
Within that ~20 million hands you’ll already have any hands that happen to have the other one or two cards in hand also match the same suit included in your count.
Doesn’t change the argument much at all, just pushes you closer to a 1/10 chance for a flush but the same point still stands. Great video!
@itselith Appreciate your kind words and similar to the other comment I left, I would like to understand where you think the math is incorrect. So here's a breakdown:
6-card flushes: We calculated the number of ways to get 6 cards all from the same suit. These are entirely distinct and cannot overlap with any hands with fewer than 6 cards of the same suit.
5-card flushes: We calculated the number of ways to get 5 cards from the same suit and 1 card from a different suit (39 choose 1, which is choosing one card from the other three suits). Since we're explicitly selecting 1 card from another suit, there is no overlap with the 6-card flush hands. These hands are distinct as well.
4-card flushes: Here, we select 4 cards from one suit and 2 cards from other suits (39 choose 2). Again, this cannot overlap with the 5-card or 6-card flushes, because we're specifically choosing 4 cards from one suit and the rest from different suits. [*If the calculation for selecting the 2 cards from the other suits was 48 choose 2, now this would introduce the possibility of selecting cards of the flush suit. But using the 39 choose 2 calculation mitigates this error and any potential overlap]
Each case-4, 5, or 6 cards of the same suit-describes a distinct combination, and the calculations were structured in a way that prevents double counting. So, the math is correct, and there's no double counting of hands.
@@Cribbage_with_J Ah, I was getting confused with how the choose functions were interacting to prevent double counting, you're right.
My brain tried to shortcut by doing [13 choose 4] and then [48 choose 2] which would allow the 4, 5, and 6 card flushes to be counted in one equation but ends up overcounting because you've lost the guarantee that the [4 choose 1] is only counting the examples with matching suits for the 5 and 6 card flush cases.
You could try to divide all that stuff off if your goal was one equation but it would get messy really fast, the way you did it is much simpler for the same result. Sorry for the confusion!