For example, it would take X packs to have 2 of every common, Y packs to get 2 of every common and rare, etc.
Ideally, I'd like to know how many packs to get 2 of every common, 2 of every rare, 2 of every epic, however many legs to be expected to come with that # of packs, and then 1600 dust for each leg to finish the playset
I've read a few places that it takes a couple hundred packs to get most of a set. From my experience I usually open 100-150 packs and have around 6000 dust saved up which gets me all the cards I want and most of the set minus a few legendaries and epics.
If you need n cards of a rarity, you need to open about n log(n) + 0.577n cards of that rarity on average (0.577 is roughly the Euler-Mascheroni constant). Logarithms have base e = exp(1).
For 98 commons, this relates to 98 log(98) + 0.577*98 ~= 506 commons opened (assuming 4 per pack [actual number will be a bit less due to packs with more than one rare or better]), so about 127 packs or so.
Most of the time I have a full set of commons by 90 packs though I find.
EDIT: Actual number is lower than this because the above assumes that 2 copies of a card at each rarity below legendary are distinct, but they aren't, I will think about this (and check wiki to see if they cover this case there).
You can get a complete Set with roughly 200 to 250 packs of an expansion. 250 to be sure. You're not going to open all the cards, but you'll have enough dust to craft the remaining missing cards.
Are You sure? Im knda too lazy to even think how to check that and im at work atm too, but this numbers seems way too low imho :P
@edit: taking into account there is like 5% per pack to get leg, u would have 10-12 legs (guess the acctual number would be lower tho) and i dont think You would have enough dust to craft even rest of the legs :P
If you need n cards of a rarity, you need to open about n log(n) + 0.577n cards of that rarity on average (0.577 is roughly the Euler-Mascheroni constant). Logarithms have base e = exp(1).
For 98 commons, this relates to 98 log(98) + 0.577*98 ~= 506 commons opened (assuming 4 per pack [actual number will be a bit less due to packs with more than one rare or better]), so about 127 packs or so.
Most of the time I have a full set of commons by 90 packs though I find.
EDIT: Actual number is lower than this because the above assumes that 2 copies of a card at each rarity below legendary are distinct, but they aren't, I will think about this (and check wiki to see if they cover this case there).
It covers this in the wiki article.
Expected number is
n log(n) + n log(log(n)) + a wee bit (formula for collect m of each is n log(n) + (m-1)log(log(n)) + a wee bit
so for n = 49 distinct commons collecting 2 we have about
49 log(49) + 49 log(log(49)) ~= 258 commons opened
So that makes a big difference (258 vs. 506).
EDIT: The wee bit error term may also be of the order n*Euler-Mascheroni constant, so I should probably add 0.577 * 49 ~= 28 or so to the number perhaps.
EDIT2: It's at least 0.577 * 49 higher since that is part of the formula for collecting the first 49, getting the extra duplicates takes about 49 log(log(49)) more (plus another error, smaller than 0.577 * 49 I expect).
Assuming that holds true, I came out way ahead. Counting 1 of 2 cards as .5 (i.e. 1 epic instead of the 2 I gave .5 points instead of 0 or 1. This also isn't including goldens or dust). In 78 packs I got this:
Anybody done the match on this?
For example, it would take X packs to have 2 of every common, Y packs to get 2 of every common and rare, etc.
Ideally, I'd like to know how many packs to get 2 of every common, 2 of every rare, 2 of every epic, however many legs to be expected to come with that # of packs, and then 1600 dust for each leg to finish the playset
https://www.reddit.com/r/hearthstone/comments/6sdu9w/a_beginners_shopping_guide_to_the_frozen_citadel/
I've read a few places that it takes a couple hundred packs to get most of a set. From my experience I usually open 100-150 packs and have around 6000 dust saved up which gets me all the cards I want and most of the set minus a few legendaries and epics.
https://en.wikipedia.org/wiki/Coupon_collector's_problem
If you need n cards of a rarity, you need to open about n log(n) + 0.577n cards of that rarity on average (0.577 is roughly the Euler-Mascheroni constant). Logarithms have base e = exp(1).
For 98 commons, this relates to 98 log(98) + 0.577*98 ~= 506 commons opened (assuming 4 per pack [actual number will be a bit less due to packs with more than one rare or better]), so about 127 packs or so.
Most of the time I have a full set of commons by 90 packs though I find.
EDIT: Actual number is lower than this because the above assumes that 2 copies of a card at each rarity below legendary are distinct, but they aren't, I will think about this (and check wiki to see if they cover this case there).
For a collection made from 200 packs opened
Assuming that holds true, I came out way ahead. Counting 1 of 2 cards as .5 (i.e. 1 epic instead of the 2 I gave .5 points instead of 0 or 1. This also isn't including goldens or dust). In 78 packs I got this: