• The VOIDRUNNER'S CODEX is LIVE! Explore new worlds, fight oppressive empires, fend off fearsome aliens, and wield deadly psionics with this comprehensive boxed set expansion for 5E and A5E!

D&D 5E Ability score generation if you REALLY like rolling dice....

DND_Reborn

The High Aldwin
So, just throwing this out there as I had the idea and tried it and it worked fairly well IMO.

Successive Four d20 Method

Your ability scores begin at 1 (you were born back then LOL). You roll four d20 against a DC equal to your current score plus one. If any of the d20 rolls succeed against the DC, your ability score increases by 1, and continue rolling the four d20 against the new score. If none of the d20 rolls succeed against the DC, you are done and the current ability score is what you have for that ability.

Obviously, you get the thread title now... you have to really enjoy rolling dice to use this method. But, I found it interesting and thought I would share. If you try it, please post your results.

Thanks for reading!
 
Last edited:

log in or register to remove this ad

Mercurius

Legend
Fun idea. Out of curiosity, what made you settle on four? Why not three or five?

p.s. How about a 20 increases it by 2, two 20s increases it by 3, three 20s turns it into an automatic 18, and four 20s gives you two 18s. Or something like that.
 

jgsugden

Legend
So, just throwing this out there as I had the idea and tried it and it worked fairly well IMO.

Successive Four d20 Method

Your ability scores begin at 1 (you were born back then LOL). You roll four d20 against a DC equal to your current score plus one. If any of the d20 rolls succeed against the DC, your ability score increases by 1. If none of the d20 rolls succeed against the DC, you are done and the current ability score is what you have for that ability.

Obviously, you get the thread title now... you have to really enjoy rolling dice to use this method. But, I found it interesting and thought I would share. If you try it, please post your results.

Thanks for reading!
I'd save time time by starting at 5.

I'd also add a fun quirk: you can choose to stop at any time. If you stop on a roll that includes one or more 20s, you get a minor perk tied to that ability score. It might be a minor (common) magic item, a small special ability, a circumstantial skill bonus, etc... These perks would be written on decks of perks (one for each ability score) and if you rolled more than one 20 when you stop, you'd draw multiple cards and select the one you want. So you might decide to stop on a 14 if you roll 2 20s, or you might continue to push your luck.
 

Jer

Legend
Supporter
I don't know why, but this reminds me of the buddy I had back in high school who wrote a BASIC program to generate characters according to Method 3 or Method 4 from the DMG. I bet he would have loved this idea.
 

DND_Reborn

The High Aldwin
Fun idea. Out of curiosity, what made you settle on four? Why not three or five?

p.s. How about a 20 increases it by 2, two 20s increases it by 3, three 20s turns it into an automatic 18, and four 20s gives you two 18s. Or something like that.
Two works really well for normal people with a DC of the current score (average is nearly exactly 10) instead of higher.

Four is a bit more forgiving for a bit better scores (which a lot of players want of course).

Five is even more forgiving, and really not too crazy if you want to do it.

As for the idea of natural 20's doing more, I am not a fan of that in general (which is why our critical damage is based on damage rolls now instead of on the attack roll). You could certainly do it, but it isn't my "thing". 🤷‍♂️

I'd save time time by starting at 5.
I actually like 3, but 1 allows for the really rare how crap I got a 2 DEX thing to happen! :D

I'd also add a fun quirk: you can choose to stop at any time. If you stop on a roll that includes one or more 20s, you get a minor perk tied to that ability score. It might be a minor (common) magic item, a small special ability, a circumstantial skill bonus, etc... These perks would be written on decks of perks (one for each ability score) and if you rolled more than one 20 when you stop, you'd draw multiple cards and select the one you want. So you might decide to stop on a 14 if you roll 2 20s, or you might continue to push your luck.
Also an option, but getting too fiddly for me, personally.
 
Last edited:

DND_Reborn

The High Aldwin
I don't know why, but this reminds me of the buddy I had back in high school who wrote a BASIC program to generate characters according to Method 3 or Method 4 from the DMG. I bet he would have loved this idea.
LOL, yeah I have an Excel macro which does it for me since I don't enjoy rolling dice that much! ;)
 

DND_Reborn

The High Aldwin
In our session yesterday a few players tried this method and here were the final results (in order) for three sets of ability scores:

17
16
14
13
11
11

16
14
13
11
9
8

15
14
11
11
9
9

One better set than the standard array, one close to it, and one worse than it. But, all three sets could have easily been rolled by doing 4d6 drop lowest.

I'll add that in one case the player rolled 1,1,3,5 when trying to get the increase to 5, so could have failed that and had a 4 LOL.

Anyway, it didn't take too long, just a few minutes, and it was actually sort of fun. So, if anyone else wants to give it a try, please share. :)
 

Three works really well for normal people (average is nearly exactly 10 with a fantastic distribution!).
Ehm, sorry, there's something I really don't get. Can you calculate how the average is nearly 10, using three dice?

Example, let's say my score went up to 10, now the DC is 11. So, I win against the DC rolling from 11 to 20. So... I lose when I roll three dice from 1 to 10.
Well, the probability for three dice to roll under 11 is: 50% x 50% x 50% = 12,5%

12,5% is not average, there's a great probability (87%) to win against the DC.

A fair (average) probability of 50%, for three dice rolled against DC, is when DC = 17.

On anydice.com I used this formula:
X: 17
output (1d20 < X) & (1d20 < X) & (1d20 < X)


The average, around "10" is found when we use only 1d20.
 

Mannahnin

Scion of Murgen (He/Him)
So, just throwing this out there as I had the idea and tried it and it worked fairly well IMO.

Successive Four d20 Method

Your ability scores begin at 1 (you were born back then LOL). You roll four d20 against a DC equal to your current score plus one. If any of the d20 rolls succeed against the DC, your ability score increases by 1. If none of the d20 rolls succeed against the DC, you are done and the current ability score is what you have for that ability.

Obviously, you get the thread title now... you have to really enjoy rolling dice to use this method. But, I found it interesting and thought I would share. If you try it, please post your results.

Thanks for reading!
Is there supposed to be a step that says you can keep re-rolling as long as you want, until all the d20s fail?
 

Dausuul

Legend
Ehm, sorry, there's something I really don't get. Can you calculate how the average is nearly 10, using three dice?

Example, let's say my score went up to 10, now the DC is 11. So, I win against the DC rolling from 11 to 20. So... I lose when I roll three dice from 1 to 10.
Well, the probability for three dice to roll under 11 is: 50% x 50% x 50% = 12,5%

12,5% is not average, there's a great probability (87%) to win against the DC.
This is one roll of a series. To get a score above 10, you must succeed on ten successive rolls (the first to go from 1 to 2, the second from 2 to 3, et cetera). If you fail any one of those rolls, you stop where you are.

The total chance to get above a ten is:

(chance to go from 1 -> 2) x (chance of 2 -> 3) x (chance of 3 -> 4) x ... x (chance of 10 -> 11)

Now, all that said, when I multiply it out, I get a 67% chance of getting 11 or better. When I calculate the probability of stopping at each individual result, then take a weighted average, it clocks in just shy of 12. So either the OP has a mistake in their math someplace, or I do.
 

Voidrunner's Codex

Remove ads

Top