Tuesday, April 11, 2017

How Many Packs Does It Take To Make A Set? #TheHobby #Math #MyHeadHurts

(source: https://www.pinterest.com/adamwolfe13/pastime-card-app-people-driven-interaction-design/)

Sometimes, I get these weird thoughts in my head like, "How many packs would someone have to buy if they wanted to complete the 1981 set of Fleer Baseball?" And, so my fingers begin the ADD-inducing trek through the interwebs.

The problem with the seemingly innocuous question is that most of the answers I found are not based in reality. For example, most sites provide a scenario in which there are a finite number of cards per set (which is true) and those cards are sold in packs of ONE card (for the most part, not true)1.  Still, others try to use the "Coupon Collector's Problem" in order to try to provide an answer2.

I did manage to find one place where the authors were on the right track: The Paninimania study on sticker "rarity."3 In the study, it's concluded that the following formula will calculate how many packs of cards are needed for a set number of cards sold in packs of multiple cards:

Now, I do not understand the complexities of such things, but if I read the study as best I could, the formula above still presents a real world problem: the packs have no duplicates in them. While that would generally hold true today (pack collating has matured since the early days), in days gone by, each pack could very well have duplicates.

The study also compared buying individual packs versus buying boxes to see which would yield "better" results. That is, would it be more efficient to buy random packs versus buying several boxes. The problem, again, is that they had to set the rule that each BOX contained NO duplicates. Okay, in the real world, that's just not happening.

As an aside, the study does a really job in explaining why certain stickers (or cards) appear to be rare when, in fact, it's all a result of probability and calculation.

So, dear readers, is there a formula that DOES take into account that there may be duplicates? If not in a single pack, then surely in a box of X number of packs. I don't know.

I did find a java program that supposedly calculates such things, but I could not get it to run. During the compile, I get an error about the ArrayList. I assume I do not have my java JDK installed correctly or something. No idea on that one.4

In a nutshell: I tried to find a program that would help me calculate the number of packs needed (estimated) to pull enough cards to complete certain sets. Why? I dunno, for fun. Well, this quickly dropped the fun part... Haha!



  1. I wrote a draft of a post on this a long time ago and never finished it, inspired by the 2009 Google Code Jam.

    https://code.google.com/codejam/contest/188266/dashboard#s=a&a=2 (That should link to the "Contest Analysis" of Problem C)

    That one, again, assumes no duplicates in each pack. But, each pack is not related to each other pack, the way packs within a box usually contain different cards, assuming decent collation.

    If you allow for any combination of duplicates, like the possibility all 12 cards in a pack are the same Bip Roberts card, it just devolves to the coupon collector problem you mentioned, divided by the number of cards per pack.

    1. That is very cool! Yeah, the problem is that older packs could have two dupes in my experience, but not usually more than that. Notbsure how that affects things. Hahah!

  2. An interesting problem professor. My answer. Just buy a complete set. It's cheaper and a lot less work. I know. I'm kidding.

    1. Hahaha, actually, that was the conclusion drawn in most of the scenarios I found!