# How likely is it to get a complete set of commons from a single booster box?

I am looking at getting into MTG again. Thinking about just playing pauper and maybe some EDH. However it's kind of fun opening packs. My thoughts are I buy a box at the release of a set and just use it to build play sets of the commons. Sell off everything else except perhaps a couple cards that may work with EDH or something. Not asking if this is the cheapest way, just whether or not it would be viable to get a full common play set from a single box or maybe two.

• Kind of seems like a question for the math section - it's probability based on total commons in the set, commons in a pack, and packs in a box. – sirjonsnow Aug 3 '17 at 14:44
• @sirjonsnow Indeed, although we have had some other questions here that were basically math questions and nevertheless seem to have been received well enough by the community. – David Z Aug 3 '17 at 19:14
• Here is an example of the analogous question asked on Mathematics. Also, this BCG question is related, in that the underlying math seems to be the same, but the approach I took there is somewhat less practical for this case. – David Z Aug 3 '17 at 19:20
• You could probably just buy 5 Commander pre-cons for the 100\$ a booster box costs you. Cracking packs for cards is never the solution, Cracking packs to play limited is fine though. – Neil Meyer Aug 4 '17 at 9:29

The exact value is difficult to calculate. This is due to the print runs of the cards - the 74 commons have "print runs" of 5 to make the printing easier. There are hundreds of these, but the fact that no pack contains the same cards does complicate the calculation.

I did throw together a pack simulator to test this and it ran in the background. Below is the output after 10 million trials. The highest was 200 packs and the lowest was 45. On a different run (of about 22 million trials) I was able to get a playset in only 44 trials. This run I've let go for 35 million trials and it was only able to get that 45, although eventually I got so unlucky I needed 215 packs on one of them. The last 17 million trials have not changed the min and max, though. I don't expect them to be changing any time soon.

Thus I think it's safe to say you are never going to get a playset from a single booster box.

I modified it to check the number of boxes and found a very consistent: 2 boxes were sufficient 39.2% of the time, three boxes 58.8% of the time, and four boxes sufficient 2.0% of the time. Five boxes were only necessary on 0.03% of runs but that was consistent far far into the tens of millions of trials.

If you want to play it safe, buy 3 boxes - you'll have them all 98% of the time.

``````new max: 72 on trial: 2
new max: 77 on trial: 3
new max: 81 on trial: 4
new low: 72 on trial: 6
new max: 87 on trial: 9
new max: 88 on trial: 10
new low: 65 on trial: 21
new max: 92 on trial: 25
new low: 64 on trial: 31
new max: 95 on trial: 42
new max: 108 on trial: 64
new low: 61 on trial: 78
new max: 115 on trial: 80
new max: 116 on trial: 110
new max: 119 on trial: 146
new low: 60 on trial: 153
new low: 59 on trial: 162
new low: 58 on trial: 222
new low: 56 on trial: 296
new low: 55 on trial: 607
new low: 53 on trial: 744
new max: 140 on trial: 2928
new low: 51 on trial: 6183
new max: 156 on trial: 9309
new low: 49 on trial: 49366
new max: 162 on trial: 60593
new max: 170 on trial: 112123
new low: 48 on trial: 117347
hit trial: 500000
new low: 47 on trial: 830525
hit trial: 1000000
new max: 186 on trial: 1307797
hit trial: 1500000
hit trial: 2000000
hit trial: 2500000
hit trial: 3000000
hit trial: 3500000
new max: 195 on trial: 3563925
hit trial: 4000000
hit trial: 4500000
hit trial: 5000000
hit trial: 5500000
hit trial: 6000000
new max: 198 on trial: 6005659
hit trial: 6500000
new low: 46 on trial: 6614767
hit trial: 7000000
hit trial: 7500000
hit trial: 8000000
new low: 45 on trial: 8015142
new max: 200 on trial: 8333263
hit trial: 8500000
hit trial: 9000000
hit trial: 9500000
hit trial: 10000000
``````
• Oh man, on my 80 millionth trial (or so) it took 220 packs. Yikes! – corsiKa Aug 4 '17 at 0:42
• Wow, that's a details rundown. Thanks for doing that work. I dont think I would invest in 3 boxes to get commons. Again thanks for the rundown. – jeffwllms Aug 4 '17 at 2:32
• @nitsua60 In 39% of cases I bought two boxes (72 packs) and I had my entire play set. In the other 61% I had to go back to the store to buy a third box. 2% of the time, I had to go and get a fourth box, but very rarely did I have to buy a fifth. I mean, if I was a person buying stuff instead of a computer simulation. – corsiKa Aug 4 '17 at 3:51
• Yanno, on second read it makes perfectly good sense. I don't know why I was getting twisted up last time I read it. In an case, be sure you recycle those hundreds of millions of wrappers! =D – nitsua60 Aug 4 '17 at 3:56
• @nitsua60 I was planning to just sell the rares. You know how many planeswalkers you get opening 100,000,000 BOXES of magic cards? – corsiKa Aug 4 '17 at 5:11

Someone else might be able to do the math a bit better than me, but I will try to break it down for you. For this example lets use the recently released Hour of Devestation set.

Each Booster box you get will contain 36 packs.
Each Pack will normally contain 10 Commons, 3 Uncommons, 1 Rare, 1 Land.
However if you get a foil card it will take the place of one of your commons, so you may only get 9 Commons in a pack.
So one box of Magic cards will get you a bit less than 360 Commons.

In Hour of Devastation, there are 74 Commons not including Basic lands (so a full play set of Commons is 74 x 4 for 296 total cards). Since the cards you get from the packs are random it is unlikely (though I guess technically possible) for you to get a full play set of Commons from a single box.
Note: I really have no idea how to do probability calculations, so anyone who does feel free post an answer with the actual precentage chances. I am kind of intrigued by this now.

However, if you are willing to trade in your cards at your Local Game Shop, you should be able to swap any Rares or Uncommons you don't want for the Commons you are missing. As of writing the Commons in Hour of Devestation are between \$0.16 and \$0.11 a piece. Where as the Rares range from ~\$7 to \$0.30 (which you will have around 36 of) are and the Uncommonsrange from ~\$2 to \$0.15 (which you will have around 108 of). So you shouldn't have an issue converting any of the value you do get into a complete playset.

As a complete aside, if you do plan on doing this I would recommend getting some sort of binder or card box. There are a lot of cards in a booster box (550+ per box with tokens), and it can get pretty unwieldy pretty quick.

• Thanks. I was more looking for experience not an exact mathematical. Figured there would be people on here that crack boxes and just wondering more how things tend to work out for them. My guess is your right with thethe randomness a full pkayset of commons is unlikely. Might end up with 10 of one and 1-2 of another or something like that. – jeffwllms Aug 3 '17 at 15:11
• @jeffwllms That is possible to get 10 of one and none of another, but I don't think it is that likely. In my experience cracking boxes, there might be a couple commons that you get 5-6 of but you usually get at least 1 or two of each common. It really depends on how big the set is though, since every set is different. If you want a full play set of commons SocioMatt's answer is cheapest, but I find opening boxes to be fun and you may luck out and get an mythic rare that is useful for commander. – Malco Aug 3 '17 at 15:40
• thanks packs are fun to bust open and it is nice to possible get some EDH bonuses. – jeffwllms Aug 4 '17 at 2:20

Many e-tailers sell preorder and post-release packs of common playsets, which will guarantee you the full playset. Coolstuffinc has a pack that you can preorder for \$19.99 that includes a full playset of commons (link is for Ixalan). You can also do this through ebay vendors both before and after set release (link is for Hour of Devastation).

This solution also prevents you from having to spend a lot of time selling extra cards you don't want. Additionally, sometimes you won't make back your initial cost anyway, depending on set value and the rares you open. A listing of all expected box values can be found at mtg.dawnglare.com. If you look at Hour of Devastation, assuming you don't get an invocation, the expected value is just over \$83. If you buy a box at \$110 from your LGS, you probably aren't making that back by trading in cards you don't plan to use.

You best solution is probably to just buy the playset outright, and buy any uncommon or rare cards you want as singles.

• @scociomatt thanks for the info. I always came across the common/uncommon sets. And sometime after shipping you get up to getting within \$30 or so of a box. My reasoning for the question was more if there is a good chance of getting close to the same with the box the extra value for cards may be worth it. Many say booster boxes are worth it because you get a bunch of bulk of no value. But for those wanting to just do pauper there is a some value there. Anyway, thanks for the info and the links. – jeffwllms Aug 4 '17 at 2:24

The probability of getting 4 copies of each common is effectively 0.

Taking @Malco's numbers of 360 commons in a booster box and 74 commons in a set, I believe the probability can be computed as:

360 Choose 292 * (73/74)^4 * (72/74)^4 * ... (1/74)^4

(You are guaranteed to get 4 of one common, so it's not counted in the formula, so there are only 292 cards that you are trying to open).

The product of this is 2*10^-49, which is to say 0. Even though 360 Choose 292 = 3*10^74, the sum of fractions is around 10^-123.

That said, you will probably get pretty close to a complete playset, and can fairly easily trade or sell the rares to get the rest of the way.

• There is an important piece of information, though. Each set of 10 is composed of a limited number of runs of 5. Thus you get 72 sets of 5 cards, and each of those sets is guaranteed to be unique. They are further guaranteed that no two sets that contain the same cards are going to be in the same pack (because packs are guaranteed unique). Exactly how much this changes the math, I don't know. – corsiKa Aug 4 '17 at 0:07
• @corsiKa Well, that makes the math impractically more complicated. If I had to estimate, I would say that gets you up to about 10^-30 or 10^-20. It's still effectively 0. I see you went the route of answering how many packs does it take rather than what is the probability for a given number of packs. Where are you getting the details about the runs? I'd be curious to see more about their structure – Zags Aug 4 '17 at 12:59

There are many online shops that have bulk deals. Star City Games the shop I usually buy from has for 20\$US the gold collection with the following.

• 865 commons/uncommons
• 25 rares
• 25 foil commons/uncommons
• 5 foil rares
• 80 basic lands

If you want to get into pauper that would probably be a much better idea than buying maybe 6-8 packs for the same money. I have only bought their rare lots but they are not charlatans. They don't actively try and con you, they are not above putting some casual goodness in their lots.

(PS Unaffiliated)