Need number-crunching help, please.

[RangerWickett]

I'm working on an alternate rule system for skills, where instead of simply increasing DC, some skills require multiple successes. I mostly started this so that opening a lock would require you to spend a few rounds and get everything right, but the core rules already have similar rules in the form of, say, Climb checks each round instead of one Climb check to scale the whole cliff.

So, I know there are a lot of mathematically-minded folks around here, and I need help knowing what combinations of 'number of successes' and 'DC' are fair at various levels. Anyone willing to help?

As a basic starting point (I'm early in my work), let's imagine there are five levels of skill mastery to worry about:

Beginner +0
1st-level +5
Low-level +10
Mid-level +15
High-level +30

And let's say that some actions might require from 1 to 10 successful checks, each of the same DC - let's say DC 10, 15, 20, 25, and 40 for examples. Can someone give me a vague sense of what the success rate array would look like?

Thanks.

 

[Thanee]

The problem is... that highly depends on the skill modifier.

If skill modifier is +9 and base DC is 15, then the chance to beat the base DC is 75%, adding +5 to the DC reduces this to 50% and adding an additional roll instead changes the likelihood of success to 75% x 75% = 56,25%.

If the skill modifier is +12 instead, the chance to beat the base DC becomes 90%, adding +5 again reduces it in a linear fashion (by -25%) to 65%, but adding an additional roll does not work in a linear fashion and turns the DC into 90% x 90% = 81%.

So, basically, being very close to the DC with the skill modifier turns additional rolls into a much better deal than raising the DC, while being farther away turns this around, of course.

At an average difference between skill and DC of ~10, the +5 ~ one additional roll translation works well enough, but if you start adding more rolls, the additional DC steps need to be much smaller than +5 (as a guesstimate it's +5 / total number of additional rolls added successively for each additional roll; i.e. +5 for one, +5+5/2 for two, +5+5/2+5/3 for three, +5+5/2+5/3+5/4 for four, etc).

Problem there is, despite the flattening curve you quickly reach the point, where the DC is beaten automatically with the multiple rolls, or the DC is basically unbeatable with the single roll. It's only a small window, where this correlates well.

Bye
Thanee

 

[Steverooo]

Beginner vs. DC:10

Required Number of Successes:
1 10/20 ~ 50%
2 10/20 x 10/20 ~ 25%
3 10/20 x 10/20 x 10/20 ~ 12.5%
4 10/20 x 10/20 x 10/20 x 10/20 ~ 6.25%
5 10/20 x 10/20 x 10/20 x 10/20 x 10/20 ~ 3.125% (3 1/8%)
6 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 ~ 1.5625% (1 9/16%)
7 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 ~ 0.578125%
8 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 ~ 0.390625%
9 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 ~ 0.1953125%
10 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 x 10/20 ~ 0.09765625%

As you can see, randomness isn't the PCs' friend... In this case, since he has no adds, nd needs 10+ to succeed, he succeeds half of the time, and needing more than one success halves his chances, overall, for each additional one needed.

In other cases, with different numbers, the chances will be different, but the adding of more tasks decreases the chances of overall success terribly, when you have a less than 100% chance... After that, it doesn't really matter, as you basically can't fail.

You can construct similar tables for beginner vs. DC:20, Master vs. DC:10, whatever, and do likewise.

Shadowrun had a "number of successes" system, in each of the first three editions, which handled it a different number of ways (I haven't seen the fourth edition, but it probably does, too). Very different Non-D20 mechanics, though...

 

[Steverooo]

1rst-level vs. DC:10

Required Number of Successes:
1 15/20 ~ 75%
2 15/20 x 15/20 ~ 56.25%
3 15/20 x 15/20 x 15/20 ~ 42.1875%
4 15/20 x 15/20 x 15/20 x 15/20 ~ 31.6%
5 15/20 x 15/20 x 15/20 x 15/20 x 15/20 ~ 23.7%
6 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 ~ 17.8%
7 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 ~ 13.3%
8 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 ~ 10.0%
9 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 ~ 7.5%
10 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 x 15/20 ~ 5.6%

This one's not so bad, as the 1rst-level PC only needs a 10, and adds +5, succeeding on rolls of 5+. He has a better chance... Oops! Actually, he succeeds 16/20, not 15, so my math is off.

[EDIT Actually, the first one, above, is wrong, too... The Beginner will succeed on 10-20, which is 11/20, not 10/20!... Okay, I quit! I can't do math, tonight!

 

[Yair]

Table: Chance for Success
Req.   Normal       3        5      10
2 	95% 	99.88% 	99.62% 	98.04%
6 	75% 	89.65% 	75.64% 	37.07%
11 	50% 	50% 	22.66% 	1.93%
16 	25% 	10.35% 	1.29% 	0.004%
20 	5% 	0.12% 	0% 	0%
The Req column is the required number on the d20. Each number's column is 
the number of successes required (in each case, three failures ends the attempt.)

My math skills are poor, but my copy-paste skills are decent. See here.

What is it about people and wheels?

 

[Thanee]

My math skills are poor...

How exactly can the chance of success raise when you add additional rolls against the same number!?

Bye
Thanee

 

[Yair]

How exactly can the chance of success raise when you add additional rolls against the same number!?

That what happens when you don't pay attention...
It takes three failures to fail the test.
Which of course means it isn't quite the same thing the OP was seeking...

 

[cmanos]

Alrighty, I forget which system has this. I think it may be Shadowrun or Star Wars (the non-d20 versions). Certain tasks require certain number of successes to complete the task. Mind you, these are systems where you can get multiple successess in a round.

So, say you need 10 successes to complete something, say, climbing a cliff. In the system I refer to, which i can't remember what game it is for, you'd roll a certain number of dice equal to your skill, and very possibly adding in dice for your attribute, and each success you rolled would be one closer to the 10 you need.

How about this. Instead of needing a number of successes, you need to net a total of 10 points over the DC.

So, lets take something with a DC of 10. You need a net of 10 to succeed.

First roll - 14. 14-DC10 = 4 net successes. You are 4/10s of the way there
Second roll - 13 13-DC10 = 3 net successes, total 7 successes. 7/10 there

Now, Certain actions could have the possibility of setting you back if you fail. Say climbing a cliff. You don't succeed, but you catch yourself and don't slide all the way down the cliff.

Rolling a number equal to the DC would indicate you made no progress. Rolling a number less than the DC would mean you made NEGATIVE progress. You slid down the cliff. You wiped away some of those net successes.

Lets take our example above

Third Roll - 8 8-10 = -2. You slid. you had 7 successes but you just lost 2. 5/10 of the way there.

And then you could always add in some catastrophic failure that would mean you have to start again, no matter what you had towards success. Say you roll 5 under the DC. All successes are wiped away and you need to start the task again. Other consequences may depend on the actiuon you are attempting.

 

[Steverooo]

That was Shadowrun, although you got no dice for stats, in most cases, in the editions I played...

 

[punkorange]

Althought it may not be what your looking for you could raise the DC to say 30. Give the PC 3 rolls, add the results of the three rolls and compare to the DC. Almost like craft. For example.
Open Lock DC 15 (40) If you get less than a 15 on any of your 3 rolls then you fail, otherwise add your rolls. If you can beat 40 in 3 rolls it's a success, otherwise the lock is too difficult for you to pick.