The Math

Cube Composition

I spent a little time fiddling with different deck distributions, but landed on:

• 6 decks of each single color (30 in total)
• 2 decks of each color pair (20 in total)

for a total of 50 decks.

The important thing to me was ending up in a situation where if you were dealt any two decks at random, you’d end up in 2 or 3 colors most of the time¹. Enter the hypergeometric distribution²! With the configuration above, I determined the probability of every distribution of single- and dual-colored decks:

So if you draw 2 random decks, you’ll get a 1-color deck and a 2-color deck slightly less than half the time, and of the rest of the possible draws, you’ll get two 1-color decks more than twice as often as two 2-color decks.

This is encouraging, but it’s only part of the story. I want to get from here to the probability distribution of the number of colors in the combined deck. To get there, I first needed to calculate the probability of each color count given the deck distribution. I’m sure there’s a clean formula for this somewhere, but there were few enough cases that I just wrote out explicit formulas for each case in terms of the deck counts and plugged them into the spreadsheet. The results:

Next up is brushing the rust off some elementary probability and determining that for any given color count:

Pr(n colors) =   Pr(n colors && 2 single)
               + Pr(n colors && 1 single, 1 dual)
               + Pr(n colors && 2 duals)

             =   Pr(n colors|2 single) * Pr(2 single)
               + Pr(n colors|1 single, 1 dual) * Pr(1 single, 1 dual)
               + Pr(n colors|2 duals) * Pr(2 duals)

And fortunately we’ve got all those values in the tables above. Asking the spreadsheet nicely to do all the arithmetic for us yields:

Looks good! So almost half the time you’ll end up with a 2-color deck, and most of the time when you don’t you’ll be 3-color. Combined, you’ll be in 2 or 3 colors about 89% of the time, and in the extreme cases of 1 or 4 about 11% of the time.

So how did I arrive at the original numbers? Well, I followed the extremely scientific method of “plug all of this into a spreadsheet with the 1- and 2-color deck counts as parameters, then fiddle with them until I find a combination that both had a distribution I like and doesn’t involve me building hundreds of decks”.

Lands

As a baseline, I’d like every final deck to have about 38 lands. Since the primary purpose of the colorless supplement is to provide universal color fixing and utility artifacts, I found a set of 10 lands for it that I felt almost any deck could benefit from. That leaves ~28 between the two half-decks, so each half-deck will ultimately contain 14±1 lands (13 lands for half-decks with very low curves, 15 lands for half-decks with very high curves).