On ability scores and randomness

This post is about finding a more balanced way of determining abilities with random rolls. The random method described in the PHB (roll 4 six-sided dice, remove the lowest one, sum the rest or 4d6d1: roll 4 dice with 6 sides and drop the lowest 1 I will use this type of notation henceforth) provides a range of 3 through 18, whereas the standard array and the variant point-buy provide a range of 8 through 15. In this post, I provide another alternative. Many have been given before me, and without doubt, there will be many after me. The alternative I provide is 1d9+7. In my opinion, this will provide a good distribution of abilities.

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I'm doing this research because I hear many complaints about how the standard way of rolling for stats can create either very powerful PCs, or some very weak PCs. And that's something some DMs and players don't want to risk. Especially in 5e with its bounded accuracy, where ability modifiers remain fairly relevant, even in the late game. As such, I set out to find a way to roll for abilities that results in abilities that are more in line with the standard array, in order to provide an alternative so that you don't get crippled or overly powerful characters.

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There are a few criteria I required of the distribution. Firstly, it should be possible to roll with a standard set of dice: d4, d6, d8, d10, d12 and d20. This means that any integer a=(2^p)x(3^q)x(5^r) is viable to be used in rolling (just trust me on this one). Or, no prime numbers save for 2, 3 and 5. Or any multiples of these excluded prime numbers. (So no d7, d11, d13, d14...)

Secondly, I (initially) wanted a distribution that returned numbers between 8 and 15. Just like the standard array and point buy. I also looked at distribution returning numbers between 8 and 16, 7 and 15 (which are similar to 8 to 15, but minus one), 7 to 16, 6 to 16 and finally 5 to 16. I did this by determining the values of A and C as a function of B, if the dice rolls are described as AdB+C. If m is the minimim value and M the maximum value, then A+C=m and AxB+C=M. From this we get that A=(M-m)/(B-1) and C=(mxB-M)/(B-1), with the added stipulation that A is an integer (you can't roll half a die).

The third requisite is that the distribution of ability scores is good. Of course, this is a subjective term, so I'll describe what I regard as a good distribution. But first, some more explanation. I'll be using the term 'average ability scores' not as the average of the ability scores, but as a set of average ability scores. This assumes that the primary score will get the highest of the rolls, the secondary the second highest etc. So the normal 4d6d1 has average ability scores of {15.66; 14.17; 12.96; 11.76; 10.41; 8.5}. Similarly, I'll be using '50% ability scores' as a term denoting the set of (integer) ability scores that have a minimum of 50% chance of occurring. As such, 4d6d1 has 50% ability scores of {16; 14; 13; 12; 10; 9}. Using these two sets, I determined 4 criteria:

1. How close to the Standard Array are the average ability scores?
2. How close to the Standard Array are the 50% ability scores?
3. What is the point buy cost of the average ability scores?
4. What is the point buy cost of the 50% ability scores?

Two problems immediately arise. Firstly, there is no possibility for non-integer point buy scores. This is easily solved by using rounded values of the average ability scores (AAS). Secondly, it isn't possible to buy scores higher than 15 and lower than 8. However, only the point cost of 16 is needed, which I have set at 12. (The reasons for this would make this long post even longer.)

Next up is a way of determining how close a distribution's AAS and 50AS to the Standard Array (SA) are. I have done this via the residual square sum (RSS). That is, subtract each AAS or 50AS value from its corresponding SA value, square that value, and sum all six of them. This means that 4d6d1 has an AASRSS of 0.9418 and a 50ASRSS of 2. Both of these values are pretty darned good, and in fact the best of the examined distributions. For the final two criteria, I have determined the point buy cost, subtracted 27 and squared the result. This means that in all cases, lower is better.

The first four criteria (usable dice, range of 8 to 15, SA vs AAS and SA vs 50AS) then led to five viable means of rolling for abilities: 1d9+7; 2d12d1+4; 3d6d1+4; 4d4d1+4; and 11d2d1-4. These were then further reduced to the final 1d9+7 by the final two criteria.

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Of course, now remains the question of how exactly do I roll a d9, since I don't have any. For this, you'll need 2 d6s (or d12, but that requires more math). Each roll of the d6 should be divided by 2, and rounded up. This way, you effectively get a d3. From 2 d3s you can get a d9 distribution by doing 3x(1d3-1)+1d3. In fact, any d(MxN) can be made by rolling Mx(1dN-1)+1dM. This means the you are rolling 3x(ceil((1d6)/2)-1)+ceil((1d6)/2)+7 six times.

Any thoughts, criticisms and questions are always welcome. I'll tidy up the post if needed, as well as the Google Sheets document so that it's legible for others. As soon as AnyDice is up again, I'll link to that program as well.

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TL;DR: I have tried to find a new way of rolling for abilities so that you won't roll crippled or overpowered characters. This resulted in 1d9+7.