# Two cards system

Ben system specifies 2704 images for cards, so I would call the smaller Ben card system modified. Ben card system is 16-13-13 (consonant-vowel-consonant), while the modified system uses 13-8-13. I use h/blank as one of my 13 consonants which is not one of the 16 Ben selected. My number-letter conversions are different from Major/Ben because I feel my conversions are more intuitive.

My system is in the making after the suggestions above. So:
I have my base of two cards system. Spades are mostly 40 sth numbers, clubs 60 sth numbers, hearts 20 sth numbers, diamonds 2 - 14 numbers.
Every type of card (queen, 1, 5, ace, etc.) gets assigned a different vowel or combination of vowels.
Spades are mirrors of hearts, cause hearts reversed look kinda like spades. Diamonds then have to be mirrors of clubs.
Me being minimalistic so:
52 x 51 x 1/2 = 1326
Queen, king, joker are so special that for now they get excluded. (actually they get excluded cause there are 10 digits, and thirteen cards of every suit)
Cardsâ combinations of different types (clubs - diamonds, spades - hearts etc.) get assigned first digit as follows (remembering that first card is always black because the reverse is first card being red):
clubs-hearts 9 _ _ me being minimalistic because my hearts from two card system are 20 sth numbers, clubs 60 sth, they get the highest digit
clubs - diamonds 8 _ _ diamonds are worth less in my two card system so they get one less than clubs diamonds = 9 - 1 = 8
clubs clubs 6 _ _ also -10 cause 611, 622, 633, 644, etc. otherwise get counted and there is only one 1 of spades for example
spades-clubs 5 _ _ obviously spades are worth less than clubs in my two card system so they get a lower value, poor spades
Sum = 800 (-20), me being minimalistic
From 1000 numbers remain 000, 100
So now special queen, king, joker time.
Because they are so special, they get to combine with themselves first, after that with the remaining (50-12) 40 cards.
How many combinations are there?:
2 (two black suits to start the pair) x 3 (three different types of cards, jokers, queens, kings to choose from) x 4 (second card so the suit doesnât matter at all) x 3 (three different types of cards again) - 6 (because 1 cardblack spades joker and 2 card black spades joker doesnât exist as the deck has only one black spades joker, the same applies to different figures here, in total 6) = 66
Spades are lower in value in my two card system so they get 000 series and not 100 series, also king, queen, joker feel special about their status, so they have more space to choose from:
Spades beginning from this trio: Joker 000 -033, Queen 034 - 066, King 067 - 099
Clubs: Joker 100 - 133, Queen 134 - 166, King 167 - 199.
Sum = 866 (-20)
Now what remains are combinations of king, queen, joker with the peasants (ace, which gets to be 1 in this system is also a peasant, 2,3,4,âŚ,10)
Number of combinations:
2 x 3 x 40 x 2 (because king, queen, joker can also take the second card place) = 480.
Sum = 866 - 20 + 480 = 1326, proving no pair got omitted nor counted twice.
So far vowels are unnnecessary. Now they will be for the remaining pairs of cards.
What remains is to invent 480 new images (now vowels for every type of card come in handy) and practice associations. This system seems pretty clear for me for now. Intriguingly, the combinations of king, queen, joker with the remaining 40 cards must include pairs beginning with red cards (x2) because the reversed black cards variants are also lacking.
I am thinking about including ace or maybe even 10 in the royal family so that they can live happily together in 000 and 100 series numbers, however, I am not convinced. I might change it.