Off-Topic Forum

Maybe this could have been posted in the general forum but I decided to err on the side of caution.

I’ve always wondered, how many packs of cards would I need to open (rip) on average to complete a basic set? I feel like there must be a formula for this that I could put into a spreadsheet but I don’t think I have the mathematical skills to do it. 

Just a disclaimer, I’ve searched the internet a couple of times without finding anything, although I probably just didn’t know enough to use the correct search terms. I’ve also searched this site and if the answer is already here, I missed it.

Oddly enough, I’ve worked with people who were math majors and when I ask them this, I’ve struggled to explain the issue to them in terms they can understand and have not gotten any results. For example, things usually stall when I say “well there are 792 cards in this base set and they come in packs of 17” and they say “divide 792 by 17” and I have to explain duplicates. They have the math in there somewhere but they definitely haven’t ripped packs with an answer like that.

To simplify things, we could assume each pack only has one card and the card has an equal chance of being any card in the base set. I’m aware that nothing is truly random and there are more of some cards produced than there are of others. For example, maybe each pack contains one all-star card. Ideally, I’d like the formula to be for X number of cards in a base set.

Clearly, trades or buying a complete set are the way to get around buying more and more packs of cards that will only have the cards you already have. I’ve just wondered about this in some form or another for at least four decades.

If your base set is 792 cards (1987 Topps baseball for example), your first “pack” of one card will be 792/792 chance of being a unique card but your next “pack” will have a 791/792 chance of being unique. From here, I can guess that it would take you 792/792 + 792/791 cards on average to get 2 unique cards of the set (a little more than 2 because your second card could be a duplicate of the first, though it isn’t likely). I threw this into a spreadsheet and had 792 rows with a column of just 792 (all the way down), a column of 792, 791, 790, 789 . . . down to 1 and in the third column, the first column divided by the second column. Then I got a total of the third column of about 5744 cars, which is 338 packs or 10.56 boxes. It feels right, but I don’t know if it is right and there must be a better way to write it.

Any idea if this is in the neighborhood of being correct? Is there a better way to write this (there must be)? Also, I’ve noticed from other sites that this sort of question generates passionate debate though I’m not sure why, let’s try to be civil.

If reading this made your brain hurt, thinking about it has done the same to me and I’m sorry for both of us.

Thanks!

This is actually an interesting question. I don't think I have the brain capacity right now to think of an answer, but maybe someday this could be answered and I would like to know!

The issue with this is the math changes with each pack you open.

I opened a pack Of Panani Contenders Football last night and there were 2 of the same card within the pack. A couple of different times. That variable of the company mismatching the packs changes the math.

I normally look at it as if I need 50% of the set still, the box / pack of cards should be  50 / 50 new and dup.

Lol yet another reason I am not a fan of Panini. I have opened a handful of blaster and hobby boxes of Upper Deck products, with no dupes in each box respectively, let alone in each packs. I can't say I wouldn't be upset if I were in your shoes

The main problem is the assumption that the variables of collation and distribution were constant.That's as far as my brain can go.

Math Major here - the answer is "it depends".

I have opened boxes upon boxes of cards over the years (both for myself as well as for shops). Some companies had GREAT collation within the box and others had terrible collation (even within packs). So to say "it will take x number of packs to get the 660-card set" is false for the next product. For instance, back in the 80s, Fleer had nearlt perfect collation in their wax boxes. Rarely (if ever) any duplicates. So with them, if you pulled 2 boxes out of a case, your chances were REALLY good at completing a set. Donruss, not so much. If there were 600 cards in the wax box, Fleer would net you 600 different cards. Donruss, you'd be lucky to pull 500 different. And 2 boxes from a case, may or may not net you a complete set.

So, in essence, there is no formula. It's all random. You could open 2 packs from the same box and have exactly the same cards in them. Losing battle trying to figure this out.

Well back in 1991 you could buy 2 boxes of upper deck and get a complete set. 3 boxes if you were unlucky.  Today. I would say it would take 4 boxes for a complete set but in 2021 I bought a box of bowman and got the paper set in one box. And about 50% of the chrome with duplicate refractors 

it's a game of chance. Just like how many licks will it get to the center of a Tootsie Roll Tootsie Pop.  The world may never know

This question is related to what's known in mathematics as the "Coupon Collector Problem".  Basically, those problems are typically set up as something like "You get one coupon per box.  How many boxes will you need to buy before you can expect to get all 10 coupons."  There are calculators and formulas available online that will get you that type of answer.  

A couple complications arise, as Dan suggested.  First, all of the solutions I've seen assume that each coupon is equally likely.  Things like short prints or double prints would drastically change the equation as do the modern inserts/parallels, and you would need to know the odds of getting each individual card.  The models I've seen in the past have also pretty much all assumed you were getting 1 coupon/card at a time.  You could get the expected number of packs by dividing the expected number of cards by the number of cards in a pack (i.e. if you expect to have to get 10,000 cards and they come 10 in a pack, you'd expect 1000 packs), but that assumes that we allow the possibility of 10 copies of the same card in a single pack--that everything is TRULY random.  As Dan said, collation within packs and within boxes varies from company to company and set to set.

There are approximate formulas out there for 1 card at a time, but they pretty much all assume true randomness (each card has no impact on the probability of the next card).  For a 792 card set, for example, the expected number of cards for a complete set would be approximately 5744 cards or  383 15-card packs.  Collation and the lack of true randomness likely makes that number wildly inaccurate, however.  I'd guess you'd expect to have to buy way more than that.

Anyway, in practical, real-world terms, there are likely too many unknown variables to really get a reliable answer.  If you want the formula for approximating it assuming true random samples, just google "Coupon Collector's Problem"..

Just because they print off the same amount of cards per player doesn't mean that they are distrubuted in an equitable manner. That's why you can get 5 - 10 of the same card for several players and none of another player. It's gone way beyond the luck of the draw. This is done on purpose so we keep buying.

163.5 packs @17-cards per required to complete a 792-card set cut from 6 standard 132-card sheets  ... or about 4 1/2 boxes @ 36-packs each, or 6 4/5 boxes @ 24-packs per box, or 11 2/3 Blaster boxes. These totals do not include double or short prints and, or inserts.                                                                                                                                                                                                                                                                                                                         H.H. Munro said "A little inaccuracy sometimes saves tons of explanation."