Are Spectre or Veteran packs more cost effective?
Solution 1:
Spreadsheet
- Google Doc Version (Last Updated Apr 8, 2012)
- Download (Excel 2007+) (Last Updated Apr 8, 2012)
Notes
- According to this answer, values do not increase if you already have all of the items in that category, which should mean that these will remain accurate throughout.
- The values in the table are estimates and are not actual values. These values are used to predict what is likely better, but without the actual values, this is the best that can be done.
- Ultra-rare items are set to 0 for Veteran Packs based on the answer provided by rare candy here. I've seen disputes of this, however, so there is a chance that this is not the case. If anyone can show a screenshot of getting an ultra-rare from a veteran pack, that would be helpful.
- Formula used to calculate values is
Cost / Total Percentage
which gives the estimated cost per item level for each pack.
Buying with Credits
For obtaining uncommon items, buy a Veteran Pack. With my estimates, Premium Veteran packs offer little to no value in terms of uncommon, but some value in terms of rare items. Spectre packs have a much lower value in terms of uncommon items (based on my estimates) but have a significantly higher value for rare items. Premium Spectre packs, by my estimates offer slightly better value for rare items, and significantly better value for ultra-rare items.
Buying with (MS/Bioware) Points
Again, given my estimates a Veteran Pack and a Premium Veteran Pack offer little to no difference in terms of value. Also, as above, rare value is significantly higher in Spectre packs, but Uncommon value is also significantly lower. Premium Spectre packs are, by my estimates, a better value for both rare (slightly better) and ultra-rare (significantly better) items.
Summary
Buy Veteran packs to obtain Uncommon items, and Premium Spectre packs to obtain rares and ultra-rares. If a Premium Spectre pack is not available, I would recommend a standard Spectre pack for rare and ultra-rare items.