Upgrade your account to a Community Supporter account and remove most of the site ads.
Enchanted Trinkets Complete--a hardcover book containing over 500 magic items for your D&D games!

Reply to thread

Message: <blockquote data-quote="EzekielRaiden" data-source="post: 9824268" data-attributes="member: 6790260">It is reasonably balanced. The pursuit of absolute perfection is rarely worth the price paid--but pushing toward better is often still worthwhile.In order: Yes. Usually no. Usually no."Balance" does not mean removing the possibility of bad/incorrect/worse/less-effective choices. Rather, it means that applying basic reasoning is likely to lead to choices that are generally pretty good. Being +1 or +2 at everything generally isn't as good as being +3 at one or two things, +2 at a couple others, and being lower at anything else. Or, more simply: What things power the stuff you do a lot? Make those numbers bigger first.These two statements can be addressed together. Do you have at least one stat you can increase to 16? Do you have at least one other stat of 14 or better? Have you avoided odd scores as much as possible?If the answer to all of these questions is "yes", then you have a fairly good set. If the answer to any single one of them is "no", then it's okay, but you're accepting some slight reduction in power. If the answer is "no" to more than one question, you might want to re-evaluate. The last question is pretty loose, because it has to be; with 27 point-buy, it is possible to have all even-numbered stats and 1 point left over, meaning you may need to accept one odd stat.If you care enough for an actual numerical evaluation of how many PB options conform to this setup, I can do so. Generating the total number of valid PB slates will be a slight pain, but not too difficult.Combinations with repetition would be (16+6-1) choose 6 = 21 choose 6 = 54624, so yes, your number is correct. And yes, the number of permutations is 16^6 = 16777216. Combos-with-rep, where you have n options and pick r from them, is always (n+r-1) choose r, and permutations is always n^r.</blockquote>

Verification