I innitially posted this in MtgSalvation, but I thougth I'd post it here as well.
From
this Magic Arcana, we know that there are 13 extra cards in Dark Ascension compared to a regular set. Since Innistrad had a regular rarity breakdown on one-sided cards (not counting basic lands) with DFCs being extra, I assume that will be the same in Dark Ascension. If that's correct, then there will be 13 DFCs in Dark Ascension.
We know that they made the odds of pulling each DFC out of a pack as close as possible to the odds of pulling a regular card of the same rarity.
In a regular small set they print rares and mythics in groups of 80, with 2 copies of each of the 35 rares and 1 copy of each of the 10 mythics. They'll probably also print the Dark Ascension DFCs in groups of 80 cards. That way they can ensure the odds of pulling a specific mythic out of a pack is close to the odds of pulling a regular mythic.
Small set mythics are printed once in a group of 80 cards and rares are printed twice in the same sheet. So, each DFC mythic should be printed once on a group of 80 cards and each DFC rare should appear twice.
Small set uncommons are printed once in every 40 cards. Since there 3 uncommons in a pack, the odds of pulling a specific one are 3 in 40 or 6 in 80. So, each DFC uncommon should appear 6 times in the 80 cards.
Small set commons are printed once in every 60 cards. Since there 9 uncommons in a Dark Ascension pack (as in Innistrad the DFC card will surely replace a common), the odds of pulling a specific one are 9 in 60, wich is the same as 12 in 80. So, each DFC common should appear 12 times in the 80 cards.
So, let's recap what are my conclusions and assumptions: 13 different DFCs printed in groups of 80 cards, with mythics appearing 1 time each, rares appearing 2 times each, uncommons 6 times each and common 12 times each. Now, let's see what are all the possibilities and then we'll see which ones are reasonable.
I'll put all the calculations inside a spoiler box to prevent the post from getting too big.
Spoiler:
Show
Possibility 1: 6 commons
This would ocupy 6*12=72 slots, leaving 8 slots for 7 cards.
a) If there is 1 uncommon, since uncommons are printed twice, then there are 78 slots occupied with 6 commons and 1 uncommon. The 2 remaining slots are not enough, since we need 6 more cards. Not possible.
b) If there are 0 uncommons, then there are 72 slots occupied with 6 commons and 0 uncommon.The only possibility that totals 13 cards is 6 commons, 0 uncommons, 1 rare and 6 mythics. (Since rares appear twice and mythics once, changing the number of rares and mythics would change the total number of cards).
Possibility 2: 5 commons
This would ocupy 5*12=60 slots, leaving 20 slots for 8 cards.
a) If there are 3 uncommons, then there are 78 slots occupied with 5 commons and 3 uncommons. The 2 remaining slots are not enough, since we need 5 more cards. Not possible.
b) If there are 2 uncommons, then there are 72 slots occupied with 5 commons and 2 uncommons, leaving 8 slots for 6 rares or mythics. The only possibility is 5 commons, 2 uncommons, 2 rares and 4 mythics.
c) If there is 1 uncommon, then there are 66 slots occupied with 5 commons and 1 uncommon, leaving 14 slots for 7 rares or mythics. The only possibility is 5 commons, 1 uncommon, 7 rares and 0 mythics.
d) If there are 0 uncommons, then there are 60 slots occupied with 5 commons and 0 uncommon, leaving 20 slots for 8 rares or mythics. Not possible, even they were all rares there would be 4 empty slots (if some of them were mythic there would be even more empty slots).
Possibility 3: 4 commons
This would ocupy 4*12=48 slots, leaving 32 slots for 9 cards.
a) If there are 5 uncommons, then there are 78 slots occupied with 4 commons and 5 uncommons. The 2 remaining slots are not enough, since we need 4 more cards. Not possible.
b) If there are 4 uncommons, there are be 72 slots occupied with 4 commons and 4 uncommons, leaving 8 slots for 5 rares or mythics. The only possibility is 4 commons, 4 uncommons, 3 rares and 2 mythics.
c) If there are 3 uncommons, then there are 66 slots occupied with 4 commons and 3 uncommon, leaving 14 slots for 6 rares or mythics. Not possible, since there would be at least 2 empty slots.
d) 2 or less uncommons is not possible, for the same reason that 3 uncommons is not possible.
Possibility 4: 3 commons
This would ocupy 3*12=36 slots, leaving 44 slots for 10 cards.
a) If there are 7 uncommons, then there are 78 slots occupied with 3 commons and 7 uncommons. The 2 remaining slots are not enough, since we need 3 more cards. Not possible.
b) If there are 6 uncommons, then there are 72 slots occupied with 3 commons and 6 uncommons, leaving 8 slots for 4 rares or mythics. The only possibility is 3 commons, 6 uncommons, 0 rares and 4 mythics.
c) If there are 5 uncommons, then there are 66 slots occupied with 3 commons and 5 uncommon, leaving 14 slots for 4 rares or mythics. Not possible, since there would be at least 6 empty slots.
d) 4 or less uncommons is not possible, for the same reason that 5 uncommons is not possible.
Possibility 5: 2 commons
This would ocupy 2*12=24 slots, leaving 56 slots for 11 cards.
a) If there are 9 uncommons, then there are 78 slots occupied with 2 commons and 9 uncommons, leaving 2 slots for 2 rares or mythics. The only possibility is 2 commons, 9 uncommons, 0 rares and 2 mythics.
b) If there are 8 uncommons, then there are 72 slots occupied with 2 commons and 8 uncommons, leaving 8 slots for 3 rares or mythics. Not possible, since there would be at least 2 empty slots.d) 7 or less uncommons is not possible, for the same reason that 8 uncommons is not possible.
Possibility 6: 1 common
This would ocupy 12 slots, leaving 68 slots for 12 cards.
a) If there are 11 uncommons, then there are 78 slots occupied with 1 common and 11 uncommons, leaving 2 slots for 1 rare or mythic. The only possibility is 1 commons, 11 uncommons, 1 rare and 0 mythics.
b) If there are 10 uncommons, then there are 72 slots occupied with 1 common and 10 uncommons, leaving 8 slots for 2 rares or mythics. Not possible, since there would be at least 4 empty slots.d) 9 or less uncommons is not possible, for the same reason that 8 uncommons is not possible.
Possibility 7: 0 commons
Not possible, since even if all the cards are uncommons there would be 78 slots occupied, leaving 2 empty slots.
CONCLUSION
Based on everything done before we conclude that (if my assumptions are correct) the Dark Ascension DFC rarity breakdown has to be one of the following:
6 commons, 0 uncommons, 1 rare and 6 mythics 5 commons, 2 uncommons, 2 rares and 4 mythics 5 commons, 1 uncommon, 7 rares and 0 mythics 4 commons, 4 uncommons, 3 rares and 2 mythics 3 commons, 6 uncommons, 0 rares and 4 mythics 2 commons, 9 uncommons, 0 rares and 2 mythics 1 commons, 11 uncommons, 1 rare and 0 mythics
Of these combinations, the only one that seems reasonable to me is the following:
4 commons, 4 uncommons, 3 rares and 2 mythics
There are 3 kinds of people in the world: Those who know math and those who don’t.
There are 10 types of people in the world: Those who understand binary, and those who don't.
I doubt there's 6 Mythics either (nor 4 Mythics), so if your calculations are correct then the most possible ones are [2,9,0,2] and [4,4,3,2]. So I kinda agree that [4,4,3,2] is most likely. 
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