Are there some common rules on which cards to discard immediately?
That's pretty hard to say because it really depends on your build and playstyle. But the cards that cost 5-10 mana should in most cases be removed without thinking. Unless they're some really powerful high quality cards in which case it could be smart to keep one but only if your other cards seem good.
I think it's not worth risking early game by keeping high cost cards unless you're sure you can pull it off.
Besides the mana cost, which factors should I take into account making this decision?
Spell cards:
Some spell cards are more effective late game, even if they cost low amount of mana. Rather remove them unless you feel like they might come in handy early game. Keep good early game spells, especially those that cost low amount of mana and are able to destroy enemy monsters.
Minion cards:
I'd suggest keeping good taunt cards that are up to 5 mana cost. Taunts are very useful and you never know will you get one when you'll really need him later.
Low cost minions that let you draw more cards in any way should be kept as well. Quite useful.
In the end there's no magical formula which will tell you what to keep since you can't know what you're going to get next by drawing. Sometimes you remove cards you don't need early game just to get them replaced with other un-needed cards.
This is according to my own experience. I have 257 arena wins and I like to keep stats. I searched previously but couldn't find an official source for this.
1st, 10th, 20th, 30th pick in Arena are always rare or higher: I call this Rare Turn, the others are normal turns.
At any normal turn, you can get one of these sets:
- 3 free
- 2 free 1 common
- 1 free 2 common
- (possibly 3 common as well didn't pay attention, can't verify this)
- 1 Rare turn ( % not sure I would guess about 5% - 10%)
In a normal turn roughly you have a 60% chance of getting a free and 40% chance of getting a common card.
At any rare turn, you can get one of these sets:
- 3 rare
- 3 epic
- 3 legendary
I got a legendary about every 10 games and since we have 4 rare+ picks I assume the chance of getting a legendary in a single rare turn is about 2%
I got an epic about every game so I assume the percentage for that is about 20%
TL;DR:
Normal Turn:
Rare Turn:
- Rare+ : 4 times
- Epic : 20%
- Legendary : 2%
I don't have a source for this its all according to my personal experience. These numbers are most likely wrong, but should give the general idea. More data is needed for better estimates.
In a normal arena deck you will have something like this:
- 15 Free
- 10 Common
- 4 Rare
- 1 Epic/Legendary
Of course if you are lucky you could get 4 legendaries, 10 rares, etc..
Best Answer
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.