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:
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!