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

[DND_Reborn]

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

You mean this:

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.

 

[DND_Reborn]

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.

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.

I'll double check it later tonight hopefully. The 10 might have been from an earlier variant--like 3 d20 instead of 4...

Ok, the 10 was from 2d20 equal or above current score for average people. I'll update my later post.

For the record (FWIW, I did about a dozen different variants...):

If the DC is above the current score:

• 2 d20: avg 9.2
• 3 d20: avg 11.8
• 4 d20: avg 13.6

If the DC is the current score or higher:

• 2 d20: avg 10.2
• 3 d20: avg 12.8
• 4 d20: avg 14.6

 

[Lakesidefantasy]

I like it. I'll test it.

I like that we can adjust the power level by increasing or decreasing the number of d20s we roll.

I'm no mathematician but it seems to me that rolling a single d20 would be equivalent to rolling 3d6 for scores.

 

[DND_Reborn]

I like it. I'll test it.

I like that we can adjust the power level by increasing or decreasing the number of d20s we roll.

Cool! Please share a set of scores when you try different numbers of d20s.

I'm no mathematician but it seems to me that rolling a single d20 would be equivalent to rolling 3d6 for scores.

Well, not really, if you roll just a single d20 your score is going to be pretty low.

Remember, the way this works is you have to roll above your current score (or use the current score instead of above it for slightly easier...) each time.

So, you start at 1, and have to roll a 2 or better.
Then you are at 2, and have to roll a 3 or better.
Then you are at 3, and have to roll a 4 or better.
And so on.

Typically around 5 or 6 you start having a good chance of rolling too low and not getting your score increased.

 

[ragnodimaggio]

<quote> This is one roll of a series. </quote>
Right, I added the calculation for the series
Simply made with excel

DC	Probability to fail DC x 3 P (d1 >= DC, d2 >= DC, d3>= DC)​	P (pass DC) x P (previous passes)​	Probability to fail DC x 2 P (d1 >= DC, d2 >= DC)​	P (pass DC) x P (previous passes)​
2	0%	100%	0%	100%
3	0%	100%	1%	99%
4	0%	100%	2%	97%
5	1%	99%	4%	93%
6	2%	97%	6%	87%
7	3%	95%	9%	79%
8	4%	91%	12%	69%
9	6%	85%	16%	58%
10	9%	77%	20%	46%
11	13%	67%	25%	35%
12	17%	56%	30%	24%
13	22%	44%	36%	16%
14	27%	32%	42%	9%
15	34%	21%	49%	5%
16	42%	12%	56%	2%
17	51%	6%	64%	1%
18	61%	2%	72%	0%
19	73%	1%	81%	0%
20	86%​	0%	90%	0%

 

[Charlaquin]

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.

…So your maximum score in any ability is 5?

 

[Charlaquin]

I'm no mathematician but it seems to me that rolling a single d20 would be equivalent to rolling 3d6 for scores.

No, because 1d20 is linear, 3d6 is a bell curve. You’re much more likely to get a total between 7 and 13 (roughly one standard deviation from the mean) with 3d6, whereas with 1d20 every result is equally likely.

 

[Baldurs_Underdark]

It's a fun idea, but I would start the rolling at a score of 3, which is the lowest you can get with 4d6 drop one. Also, the initial rolls probably aren't so interesting.

…So your maximum score in any ability is 5?

You roll 4 d20, and if any of the 4 dice is a success, the score goes up one, and you roll 4d20 again for the same score. If that succeeds, it goes up one more and you roll 4d20 again. If that succeeds, it keeps going up one more. And at some point all 4 dice will not succeed and that score is done.

And then you go to your 2nd score, starting at 1 again (or as I propose, at 3). Keep rolling 4d20 until all 4 fail. Then onto your 3rd score.

 

[Mannahnin]

You roll 4 d20, and if any of the 4 dice is a success, the score goes up one, and you roll 4d20 again for the same score. If that succeeds, it goes up one more and you roll 4d20 again.

Yes, this the part he forgot to write in the OP.

 

[DND_Reborn]

Yes, this the part he forgot to write in the OP.

If that part wasn't obvious to you, you might as well click to Ignore my posts, because that is implying your scores could go up to 2 and then you would stop there.

Jeez...

It's a fun idea, but I would start the rolling at a score of 3, which is the lowest you can get with 4d6 drop one. Also, the initial rolls probably aren't so interesting.

Sure, I did that once as well. I've also suggested starting at 8 since that is the low of the standard array and point-buy.