How to balance the number of cards for each rarity

galva94
2/2 Zombie

Posts: 123
Favorite Card: Cryptic Command
Favorite Set: Kamigawa
Color Alignment: Blue, Green

How to balance the number of cards for each rarity Jul 23, 2020 17:46:02 GMT

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Post by galva94 on Jul 23, 2020 17:46:02 GMT

This is not exactly a question about a card, but I hope this is the proper place to post.

My problem is the following: I would like to make a set and print it out, but I am not sure of how to balance the number of cards with respect to rarity. Let me explain with an example.

Let say I have a set with 101 commons, 80 uncommons, 53 rares and 15 mythic, which is more or less the standard set size. Let's consider the mythic to be as frequent as the rares for sake of semplicity (i.e. lets say we have 68 rares). To print this set and play it with friends as it would be an official set I would have to print out... ready..? 19040 cards! (If I get the math right!). And I would have to print out: 136 copies of each common, 51 copies of each uncommon and 20 of each rare and mythic. (Actually i considered 100 unique commons instead of 101, again for sake of semplicity). Math nerds can check that 136*100 + 51*80 + 20*68 = 19040, and with this pool we can create 1360 booster packs with the right amount of cards or each rarity.
Of course there's only a small issue: there is no way that this crazy amount of prints will ever be used!

One way this numbers can be taken down is to adapt the size of the set (i.e number of cards per rarity) such that they are "less coprime" in some sense, and more similar to the booster pack ratio 10:3:1. Maybe later I'll try to figure out some optimal sizes, but I doubt we can go under 1000 cards. Ideally I would like to print something like 336 cards (ideal for a 8 people draft of 14 cards/pack).

The number of commons, uncommon and rares for my purpose can be very flexible, so I can accept a solution that contemplate an unusual number of cards per rarirty, if it works!

An approach that I thought about is to print like 3 copies of each common, 2 copies of each uncommon and 1 copy of each rare/mythic with a set of 80c*3copies+36u*2copies+24r*1copy = 336 cards. In this way all cards are played in a 8-player draft: infact

10 * 24boosters = 80 * 3copies
3 * 24boosters = 36 * 2copies
1 * 24boosters = 24* 1copy

The problem here is that the proportons are completely gone! 101 : 80 : 68 is VERY different from 80 : 36 : 24 (especially the gam between commons and uncommons!)

I am really curious to know how you guys tackled this problem, because searching online I haven't found much! Again, I don't care if the solution is not perfect, as long as it works with small numbers.
Sorry for the math, but I decided to post it because maybe someone will find it interesting.

2wb
0/0 Germ

Posts: 47

How to balance the number of cards for each rarity Jul 28, 2020 21:29:37 GMT

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Post by 2wb on Jul 28, 2020 21:29:37 GMT

In an 8 player pod, statistically, you see 240 commons, 72 uncommons, 21 rares, 3 mythics.

You have a ~1% chance of seeing any common -- each pack is collated in normal printings to have at least one card of each color and not have duplicates.

So, for a given card, it has ten circa 1% chances to show up in a pack -- or a bit under 10% chance to see it in a given pack.

Thus, you'd only have to print 4 of a common to not run out of them, more likely than not.

But that's not too scaleable to the other rarities, so I went and asked people who are smarter;

cajun helpfully suggested multiplying things out --

So you have "one sheet of mythics". For each mythic, there are two of each rare.

That, together, comboes out to 121 packs (15 of which have mythics and 106 of which have rares).

In 121 packs, you need 363 uncommons and 1210 commons.

That works out to 12 of each common (round up) and 4.5 of each uncommon. Keeps the ratios intact (though obviously you have the issue of needing a half-a-card, which would be best done by just bumping it to 5 each uncommon).

Far cry from 19,040, it's only 1,733.

According to CyberChronometer, you'd need 6 of each common, 3 of each uncommon, 2 of each rare and 1 of each mythic. The math that led them to that was not posted, however; but you can see where it's similar --

Because you're not making 121 packs, only 24, you don't actually need to have every one of those commons/uncommons (you still need the rares/mythics because there's just too few so there's no better way to proportion them); and at that small of a number (almost 1/6th) the variance difference between 1,212 commons and 606 commons is small enough that it shouldn't matter; and the same applies to the 400 uncommons scaling down to 240.

Which leaves you needing 967 cards printed.

At numbers below that, your card count will start negatively affecting randomness -- again according to CyberChronometer, you can go down to 4 of each common (404) and two of each uncommon (160) while still preserving 90% accuracy to the correct randomness.

I would, personally, say that that 685 number is the lowest one you can get in a reasonable capacity.

Board!, Auxano, db2c40

galva94
2/2 Zombie

Posts: 123
Favorite Card: Cryptic Command
Favorite Set: Kamigawa
Color Alignment: Blue, Green

How to balance the number of cards for each rarity Jul 31, 2020 13:24:48 GMT

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Post by galva94 on Jul 31, 2020 13:24:48 GMT

Doing some math I figured out thw following formulas. Let's define this quantities:

c = number of unique commons in the set

u = number of unique uncommons in the set

r = number of unique rares in the set

m = number of unique mythics in the set

For example (c,u,r,m) = (101, 80, 53, 15).

C = printed copies of each common

U = printed copies of each uncommon

R = printed copies of each rare

M = printed copies of each mythic

For example (C,U,R.M) = (6,3,2,1). This is our unknow.

The direct formulas are:

M = 1

R = 2

U = 3 * (m + 2*r) / u

C = 10 * (m + 2*r) / c

of course, to be rounded down. Let's provide for examples: (c,u,r,m) -> (C,U,R,M)

(60,40,35,10) -> (13,6,2,1)

(101, 80, 53, 15) -> (11,4,2,1)

(101, 60, 53, 15) -> (11,6,2,1)

(60,40,45,0) -> (7,3,1,0)

(101,80,68,0) -> (6,2,1,0)

(101,60,68,0) -> (6,3,1,0)

I suggest any user to do the calculations again to take care of the decimal part as they desire.

Now, we can also find some inverse formulas. Let's say we want to print no more than 4 copies of each commons for budget reason. How can we construct a set of the right side to keep the proportions? Because some discrete math is involved, the formulas are not super easy, but this is what I've found (I hope that's correct!). This time I considered only sets with no mythics (or if you want, sets where mythics are found with the same frequency as rares)

r = t / R

u = 3*t / U

c = 10*t / C

What is "t"? It's an arbitrary parameter that decides the set size (while keeping the proportions). For example if I want to print C = 4 or each common, U = 2 or each uncommon and R = 1 or each rare, I can plug in these values and try a value of t, e.g. t = 24:

r = 24 / 1 = 24

u = 72 / 2 = 36

c = 240 / 4 = 60

This leave me with 60 unique commons printed in 4x, 36 uc printed in 2x, and 24 rares in 1x. The total numer of prints will be 60*4 + 36*2 + 24 = 240 + 72 + 24 = 336 prints.

Of course the proportions are warped with respect to the standard 101:80:53+15 but that's expected: you cannot have both proportions right while keeping low the number of cards!

This last example seems to be particularly convenient for a 8 player draft, because all cards are destributed, and they are in the exact rarity required. I would say it's a good compomise between the proportion 10:3:1 and the proportion 101:80:68.

Anyway I'm curious to read more about the math done by CyberChronometer, if someone can link me the topic