D&D 5E (2014) Is Point Buy Balanced?

[AlViking]

Would you? I might be able to find my old spreadsheets from a decade ago. In there I generated all 54,000+ combinations. Would that help?

Also, I was talking with Alexa this morning and found that the number of permutations of the Standard Array is 6!=720. Can you confirm that?

Every possible option? Here you go ...

Spoiler: All options with 27 point buy. I think.

13, 13, 13, 12, 12, 12
13, 13, 13, 13, 12, 11
13, 13, 13, 13, 13, 10
14, 12, 12, 12, 12, 12
14, 13, 12, 12, 12, 11
14, 13, 13, 12, 11, 11
14, 13, 13, 12, 12, 10
14, 13, 13, 13, 11, 10
14, 13, 13, 13, 12, 9
14, 13, 13, 13, 13, 8
14, 14, 12, 11, 11, 11
14, 14, 12, 12, 11, 10
14, 14, 12, 12, 12, 9
14, 14, 13, 11, 11, 10
14, 14, 13, 12, 10, 10
14, 14, 13, 12, 11, 9
14, 14, 13, 12, 12, 8
14, 14, 13, 13, 10, 9
14, 14, 13, 13, 11, 8
14, 14, 14, 10, 10, 10
14, 14, 14, 11, 10, 9
14, 14, 14, 11, 11, 8
14, 14, 14, 12, 9, 9
14, 14, 14, 12, 10, 8
14, 14, 14, 13, 9, 8
15, 12, 12, 12, 11, 11
15, 12, 12, 12, 12, 10
15, 13, 12, 11, 11, 11
15, 13, 12, 12, 11, 10
15, 13, 12, 12, 12, 9
15, 13, 13, 11, 11, 10
15, 13, 13, 12, 10, 10
15, 13, 13, 12, 11, 9
15, 13, 13, 12, 12, 8
15, 13, 13, 13, 10, 9
15, 13, 13, 13, 11, 8
15, 14, 11, 11, 11, 10
15, 14, 12, 11, 10, 10
15, 14, 12, 11, 11, 9
15, 14, 12, 12, 10, 9
15, 14, 12, 12, 11, 8
15, 14, 13, 10, 10, 10
15, 14, 13, 11, 10, 9
15, 14, 13, 11, 11, 8
15, 14, 13, 12, 9, 9
15, 14, 13, 12, 10, 8
15, 14, 13, 13, 9, 8
15, 14, 14, 10, 9, 9
15, 14, 14, 10, 10, 8
15, 14, 14, 11, 9, 8
15, 14, 14, 12, 8, 8
15, 15, 11, 10, 10, 10
15, 15, 11, 11, 10, 9
15, 15, 11, 11, 11, 8
15, 15, 12, 10, 10, 9
15, 15, 12, 11, 9, 9
15, 15, 12, 11, 10, 8
15, 15, 12, 12, 9, 8
15, 15, 13, 10, 9, 9
15, 15, 13, 10, 10, 8
15, 15, 13, 11, 9, 8
15, 15, 13, 12, 8, 8
15, 15, 14, 9, 9, 8
15, 15, 14, 10, 8, 8
15, 15, 15, 8, 8, 8

 

[Lakesidefantasy]

There are only 65 valid 27-point arrays, so it shouldn't be too difficult to determine which ones match those rules. I'll take a quick look.

Edit: Of the 65 valid arrays, only 20 meet these rules.
1) No odd stats after racial/floating/background stat adjustments. Practically, this means either one odd stat in the array (for +2/+1) or three odd stats (for +1/+1/+1).
2) Highest stat is 16+ after stat adjustments.
3) Second highest stat is 14+ after stat adjustments.

14, 14, 12, 12, 11, 10
14, 14, 12, 12, 12, 9
14, 14, 13, 12, 10, 10
14, 14, 13, 12, 12, 8
15, 12, 12, 12, 12, 10
15, 13, 12, 12, 11, 10
15, 13, 12, 12, 12, 9
15, 13, 13, 12, 10, 10
15, 13, 13, 12, 12, 8
15, 14, 14, 10, 10, 8
15, 14, 14, 10, 9, 9
15, 14, 14, 11, 9, 8
15, 14, 14, 12, 8, 8
15, 15, 11, 10, 10, 10
15, 15, 12, 10, 10, 9
15, 15, 12, 11, 10, 8
15, 15, 12, 12, 9, 8
15, 15, 13, 10, 10, 8
15, 15, 13, 12, 8, 8
15, 15, 15, 8, 8, 8

Thank you this great!

So, 45 of the 65 Standard Array combinations fail satisfy the aforementioned criteria.

If we were to use those criteria to evaluate the 65 Standard Array combinations as either effective or ineffective then roughly 2/3 of them would be lacking in efficacy. Some more than others I would assume. I wonder which are the worst? And, how would we even evaluate that?

Furthermore, there are 720 permutations of each of those 65, some of which would be ineffective for specific builds.

 

[FrogReaver]

That is a pretty cool autocorrect/swipe typo.

Thanks, and had I read your post earlier I wouldn't have had to talk to Alexa about it.

Definitely was autocorrect. Should have stated ‘with replacement’.

It’s quite amusing though.

 

[Lakesidefantasy]

Every possible option? Here you go ...

Spoiler: All options with 27 point buy. I think.

13, 13, 13, 12, 12, 12
13, 13, 13, 13, 12, 11
13, 13, 13, 13, 13, 10
14, 12, 12, 12, 12, 12
14, 13, 12, 12, 12, 11
14, 13, 13, 12, 11, 11
14, 13, 13, 12, 12, 10
14, 13, 13, 13, 11, 10
14, 13, 13, 13, 12, 9
14, 13, 13, 13, 13, 8
14, 14, 12, 11, 11, 11
14, 14, 12, 12, 11, 10
14, 14, 12, 12, 12, 9
14, 14, 13, 11, 11, 10
14, 14, 13, 12, 10, 10
14, 14, 13, 12, 11, 9
14, 14, 13, 12, 12, 8
14, 14, 13, 13, 10, 9
14, 14, 13, 13, 11, 8
14, 14, 14, 10, 10, 10
14, 14, 14, 11, 10, 9
14, 14, 14, 11, 11, 8
14, 14, 14, 12, 9, 9
14, 14, 14, 12, 10, 8
14, 14, 14, 13, 9, 8
15, 12, 12, 12, 11, 11
15, 12, 12, 12, 12, 10
15, 13, 12, 11, 11, 11
15, 13, 12, 12, 11, 10
15, 13, 12, 12, 12, 9
15, 13, 13, 11, 11, 10
15, 13, 13, 12, 10, 10
15, 13, 13, 12, 11, 9
15, 13, 13, 12, 12, 8
15, 13, 13, 13, 10, 9
15, 13, 13, 13, 11, 8
15, 14, 11, 11, 11, 10
15, 14, 12, 11, 10, 10
15, 14, 12, 11, 11, 9
15, 14, 12, 12, 10, 9
15, 14, 12, 12, 11, 8
15, 14, 13, 10, 10, 10
15, 14, 13, 11, 10, 9
15, 14, 13, 11, 11, 8
15, 14, 13, 12, 9, 9
15, 14, 13, 12, 10, 8
15, 14, 13, 13, 9, 8
15, 14, 14, 10, 9, 9
15, 14, 14, 10, 10, 8
15, 14, 14, 11, 9, 8
15, 14, 14, 12, 8, 8
15, 15, 11, 10, 10, 10
15, 15, 11, 11, 10, 9
15, 15, 11, 11, 11, 8
15, 15, 12, 10, 10, 9
15, 15, 12, 11, 9, 9
15, 15, 12, 11, 10, 8
15, 15, 12, 12, 9, 8
15, 15, 13, 10, 9, 9
15, 15, 13, 10, 10, 8
15, 15, 13, 11, 9, 8
15, 15, 13, 12, 8, 8
15, 15, 14, 9, 9, 8
15, 15, 14, 10, 8, 8
15, 15, 15, 8, 8, 8

Thanks! Are there 65 combinations there?

 

[AlViking]

Thanks! Are there 65 combinations there?

Yep. I could make a decent character out of any of these, although I'd still want to choose placement. The last one in the list of 3 15s and 3 8s isn't that far off of my barbarian.

It would be interesting to know if those people who don't like "cookie cutter" characters or who want to be surprised would have the same complaint if they used this array. Random number between 1 and 65 and then randomize placement.

 

[Lakesidefantasy]

Yep. I could make a decent character out of any of these, although I'd still want to choose placement. The last one in the list of 3 15s and 3 8s isn't that far off of my barbarian.

It would be interesting to know if those people who don't like "cookie cutter" characters or who want to be surprised would have the same complaint if they used this array. Random number between 1 and 65 and then randomize placement.

We don't talk about Fight Club here.

 

[EzekielRaiden]

Would you? I might be able to find my old spreadsheets from a decade ago. In there I generated all 54,000+ combinations. Would that help?

If you can import it into Google Sheets and DM me a link, there should be an easy way to check for PB validity. I'd just set up functions which calculate the PB cost of each array, and then a column which checks to see whether the PB sum equals 27.

Also, I was talking with Alexa this morning and found that the number of permutations of the Standard Array is 6!=720. Can you confirm that?

Since there are 6 different values and order matters, yes, that's correct. If there were duplicates (e.g. if two stats were both 12 or something) then we'd need a slightly more complicated formula.

It would also differ if we weren't picking all six values. The proper formula for permutations without replacement is n!/(n-r)!.

Worth noting, the difference between permutations (order matters) and combinations (order doesn't matter) is simply whether you don't or do divide by r!. Since we're choosing all six values from the list, there is only one combination, as order doesn't matter. If there were 7 values, then there would be six different combinations of six elements, because that's equivalent to choosing which one of the values to exclude.