How many packs of Mean Streets of Gadgetzan cards would I need to open, on average, to get 2x of all common cards and 1x of all rare cards? Additionally, how many different epics and legendaries would I get from those packs?
You are likely to hit the 2x of each common before you meet the 1x of each rare requirement. I won't bog you down with detailed mathematical formulae, but if you're interested, you can read more here: https://en.wikipedia.org/wiki/Coupon_collector%27s_problem
http://hearthstone.gamepedia.com/Card_pack_statistics cites a card being common as approximately a 71% chance. There are 49 unique commons in MSOG. If you were to open 60 packs, you would be expected to open [0.71*(60*5)] = 213 commons total. Looking at the coupon collector chart, the number of commons we need, on average, to have 49 unique commons is 220. We're not there yet, but we're close.
Rares, however, are substantially rarer, with most studies showing around a 23% chance for a given card. This means, with 60 packs, you're expected to open 69 total rares. There are, however, 36 rares in MSoG. So you're opening just over the amount of commons you need if you could just select the ones you wanted from a pool once you turned over a "common" rarity, but you're not even opening the amount of rares you need if you could select from a pool. Looking at the chart, the number of tries we need to have a complete set of 72 rares is 151. We're about 40% of the way there.
So, how many packs do we need to open to get a complete set? Solving for x, where 220 is the absolute maximum number of cards we need for two of each common, we can say [0.71*(x*5)] = 220, which after some algebra equates to 62 packs needed for 2 of each common. For rares, it's [0.23*(x*5)] = 151, which equates to 132 packs.
Thus, you need 132 packs to, on average, open one of each rare and two of each common through sheer discovery alone. But you'll have a ton of extra commons rares if you do this. That's a ton of dust worth of extra cards. As rares are 100 dust each and commons are 40, it's obviously substantially more cost effective to use dust to craft the last few cards.
EDIT: I made a obvious math error by assuming to collect a playset of 2x cards would be double the under needed to collect 1x. That's obviously not the case, as you would have collected duplicates of many cards before you finally found the last card you need, so only in the very worst case scenario would it be double.
TL;DR - Check out this amazing source on the data needed to get the expansion. This article goes into detail about the rarities in Hearthstone packs and the amount of packs you need to open ON AVERAGE to get a certain percentage of the expansion. It spikes at the end because of the amount of dust and low percentage to get all of the legendaries.
- 50 packs ≈ 50% complete
- 100 packs ≈ 75%
- 150 packs ≈ 85%
- 200 packs ≈ 90%
- 380 packs ≈ 100%
According to an article of PCGamer, the average result from the pre-order will give you 91 Commons (93% of all MSG Commons),46 Rares (64% of Rares), 11 Epics (20% of Epics) and 3 Legendaries (15% of Legs).
We come to the answer of the question how much you and I should invest for each stereotype proposed.
Scenario 1: competitive only: Average 175 Packs, $180 investment (500 coins left)
Scenario 2: competitive and fun, no crap: Average 240 Packs, $237 investment
Scenario 3: completionist (non golden): Average 380 Packs, $375 investment (2,500 coins left)
Scenario 4: completionist (golden): approx. 1,000 Packs, $930 investment Note: Brute forcing 155,600 dust for all golden cards would cost you 1,440.74 packs. Nevertheless, in 1,000 packs you get an average of 35 golden Legendaries and 95 Golden Epics which should be fine taking into account duplicates and dusting
In 1,000 packs you can only expect 3.5 golden Legendaries and 9.5 Epics, therefore crafting most of it seems the only validstrategy of obtaining a full golden set – which very few will ever do
using the cards odds and the collector problem from A. McDaniel's answer and correcting his results to account for the actual number of cards from MSoG, we get
49 unique commons will be collected in 220 tries there are 5 cards in a pack and a 0.7165 that a card will be common that gives us 61.4 ≈ 62 packs 36 unique rares will be collected in 151 tries there are 5 cards in a pack and a 0.2284 that a card will be rare that gives us 132.2 ≈ 133 packs
In my question, I'm asking for 2 times the common. It appears, as per the results in this article http://www.jstor.org/stable/2308930?seq=1#page_scan_tab_contents, that subsequent sets of a collector problem take n loglog n tries to complete (49 loglog 49 in our case) which amounts to 12 more packs, for a total of 74
This is still well below the number of packs for rares, so 133 is the right number