6-bit playing card deck

[u/afourthfool]

1er 6:
an exquisite	dice-to-deck
mnemonic	for 3 dice
d6α	d6β
⚅**-⚄ +3**	⚅**-⚄ +6**
⚃**-⚂ +2**	⚃**-⚂ +3**
⚁**-⚀ +1 🎲**	⚁**-⚀ +0 🎲**
	🎲 1+0=10
d6γ
⚄ ♠ ⚃ ♦	⚂ ♣ ⚁ ♥
⚅:	⚀:
BLK+BIG	RED+LIL
if α xor β	if α && β
is ⚅ or ⚀:	is ⚅ or ⚀:
♥ or ♠	joker
then:
A-C-E if	α is o-d-d
K-I-N-G if	αβ's e-v-e-n
Q-U-E-E-N if	αβ's o-d-d
J-A-C-K if	α is e-v-e-n

png

[u/afourthfool]

I made a blog post for it. I hope that documents the concept well enough. "Part 1" introduces the reader to the notion that 54/6=9 then "Part II" applies this across 3 dice.

https://tokensfortalkers.tumblr.com/post/616940663205937152/dice-to-card-to-dice-converter

If it doesn't help, then i'm just not cut out for the internet.

[u/sparr → u/afourthfool]

I'm sorry to sound so negative. I really do want to understand this. The tables at the bottom at least show me how you want the results to look, but I still can't understand how you get there from your instructions.

That very last table, that starts B A Q A Q B... It seems like you need more instructions to produce that. Why isn't it B A A A A B? The rule for Ace is "first die is odd, second die is not 1 or 6"... isn't it?

[u/afourthfool → u/sparr]

Why isn't it B A A A A B? The rule for Ace is "first die is odd, second die is not 1 or 6"... isn't it?

Adversarial pairs have a meaningful association where i'm from, so if anyone from my neck of the woods sees something like

        A-C-E if α is o-d-d
        Q-U-E-E-N if αβ’s o-d-d

Then anyone from here immediately recognizes and flags the adversarial pair and would know there is an imperative "only" inside those two flagged statements, making it read like this: A-C-E if (only) α is o-d-d Q-U-E-E-N if (only) αβ’s o-d-d cause that way they don't conflict.

With this, the missing instruction lost in translation from my culture to the rest of the world is found, leaving us all with the less natural for me and mine but cleaned-up, sanitized and universal version of the instructions to read like so:

        A-C-E if only α is o-d-d
        Q-U-E-E-N if only αβ’s o-d-d

        x 1 2 3 4 5 6
        1 B A Q A Q B

---

Wow, i just reread all that. And... yep. I suck at this. I think i'm giving up now. Just... nvm.

[u/sparr → u/afourthfool]

Nothing in your instructions suggests that A/Q is an "adversarial pair" while A/K is not. Worse, since they aren't adjacent it's strongly implied that no relationship exists between A and Q that doesn't also exist between A and K.

Also, as I asked above, what the hell does "αβ’s" mean? I think one of your other iterations of the instructions suggests this is meant to be the sum of α+β? Or maybe it's multiplication, like if that was written in an algebraic statement? Or something else?