View Full Version : New Take on bell curve rolling

Dairian

2009-01-08, 03:00 PM

Ive never really been happy with rollign d20s, the range is just too great. But at the same time, the conventional bell curve rolling methods(3d6, 2d10) have too narrow a range, so ive decided in my games to roll a d12 and a d8. this makes rolling a 9-13 equally likely, meaning that you can't just assume an 11 like with the other methods.

I just wanted to se how you guys liked this method, any feedback would be appreciated.

Samurai Jill

2009-01-09, 09:36 AM

It's a fresh approach, certainly.

CarpeGuitarrem

2009-01-09, 01:21 PM

Very interesting. Lemme see how this plays out...(using "a" for a roll of 10, "b" for a roll of 11, "c" for a roll of 12)

02: 11

03: 12, 21

04: 13, 31, 22

05: 14, 41, 23, 32

06: 15, 51, 24, 42, 33

07: 16, 61, 25, 52, 34, 43

08: 17, 71, 26, 62, 35, 53, 44

09: 18, 81, 27, 72, 36, 63, 45, 54

10: 19, 28, 82, 37, 73, 46, 64, 55

11: 1a, 29, 38, 83, 47, 74, 56, 65

12: 1b, 2a, 39, 48, 84, 57, 75, 66

13: 1c, 2b, 3a, 49, 58, 85, 67, 76

14: 2c, 3b, 4a, 59, 68, 86, 77

15: 3c, 4b, 5a, 69, 78, 87

16: 4c, 5b, 6a, 79, 88

17: 5c, 6b, 7a, 89

18: 6c, 7b, 8a

19: 7c, 8b

20: 8c

Holy cow, that's a nice curve. I like it.

Catattack

2009-01-11, 04:02 PM

Or you can use 1d8 + 1d8 + 1d6 - 2.

That has the same range as a d20 but with more curve.

Or use 1d8 + 1d8 + 1d6 + 1d20 - 2 round down.

It gives a curve still a bit random, more like a banana curve than bell.

:smallsmile:

Daracaex

2009-01-12, 12:36 AM

Or you can use 1d8 + 1d8 + 1d6 - 2.

That has the same range as a d20 but with more curve.

Or use 1d8 + 1d8 + 1d6 + 1d20 - 2 round down.

It gives a curve still a bit random, more like a banana curve than bell.

:smallsmile:

I think the point is to generate a bell curve, not make rolling into a second-grader's math test.

Khanderas

2009-01-12, 07:00 AM

Very nice and I like it as this evens out the random quite a bit. Quite intresting as we get the same number of combinations for results 9-13 (not to be confused with the same % chance to get a 9, 10, 11, 12 or 13).

I note there are no "1" rolls, but that is problebly for the best.

WinterSolstice

2009-01-12, 07:10 AM

wouldn't a d8 and d12 mean the possibility of a score of 20 before racial modifiers?

Khanderas

2009-01-12, 07:14 AM

wouldn't a d8 and d12 mean the possibility of a score of 20 before racial modifiers?

It's 1d12 + 1d8 instead of 1d20

As in attackrolls, skillchecks and so on. Not the 3d6 for starting stats.

a d20 have 5% to land on any number. But the more (and smaller) dice involved the greater the chance of getting a number in the middle. Something I think is good, since that puts more "worth" in getting the skill up as opposed to just have a flat 5% to score a nat 20.

Edit Idonnohowmany: How would you handle a "take 20" here ? since the chance to get a 20 in this system is not 5%, but 1,04%

averagejoe

2009-01-12, 07:42 AM

(not to be confused with the same % chance to get a 9, 10, 11, 12 or 13).

Am I missing something. I'm pretty sure that is what it means.

(I should probably be completely sure, but I'm not completely sure about anything when I've been up all night.)

Khanderas

2009-01-12, 09:01 AM

Am I missing something. I'm pretty sure that is what it means.

(I should probably be completely sure, but I'm not completely sure about anything when I've been up all night.)

Well I don't feel mathematically fit (slept bad, but drank coffee, so percived reality may differ from actual reality, / feel awake but problebly not at 100%) but I thought that since the chance to roll.... say a 5 on a d12 differs from the chance to roll a 5 on the d8. I thought that it would problebly scewer the results slightly.

For example, 2d10 max result 20.... chance of that is (1/10)*(1/10) = 1%

1d12 + 1d8 max result 20, chance of that 20 is (1/12)*(1/8) = 1,04166666... %

Pointlessly small difference anyways. And I may still be wrong, but that is the reason I thought that while there are the same amount of combinations for the numbers 9-13, those combinations might not have the same chance to pop up. Problebly wrong... because it feels wrong when I look at it again and it problebly is a straight

(nr of possible combinations for nr X / total combinations possible)*100

for the % chance for any given number.

Studoku

2009-01-12, 09:23 AM

Well I don't feel mathematically fit (slept bad, but drank coffee, so percived reality may differ from actual reality, / feel awake but problebly not at 100%) but I thought that since the chance to roll.... say a 5 on a d12 differs from the chance to roll a 5 on the d8. I thought that it would problebly scewer the results slightly.

For example, 2d10 max result 20.... chance of that is (1/10)*(1/10) = 1%

1d12 + 1d8 max result 20, chance of that 20 is (1/12)*(1/8) = 1,04166666... %

Pointlessly small difference anyways. And I may still be wrong, but that is the reason I thought that while there are the same amount of combinations for the numbers 9-13, those combinations might not have the same chance to pop up. Problebly wrong... because it feels wrong when I look at it again and it problebly is a straight

(nr of possible combinations for nr X / total combinations possible)*100

for the % chance for any given number.

Each combination is based on a d8 and a d12 roll, so it has a 1/(8*12) : a 1/96 chance of rolling.

Khanderas

2009-01-12, 10:12 AM

Each combination is based on a d8 and a d12 roll, so it has a 1/(8*12) : a 1/96 chance of rolling.

Yeah. that is how it works, but my low-on-sleep-high-on-coffee brain tricked me into overcomplicating it.

Dairian

2009-01-12, 07:15 PM

How would you handle a "take 20" here ? since the chance to get a 20 in this system is not 5%, but 1,04%

I dont allow taking 10 or 20.

ShneekeyTheLost

2009-01-12, 07:22 PM

Reminds me of a dice system I was considering using in a game I am contemplating actually writing up.

Basically, ranks in a skill determine how many dice you get to roll to overcome difficulty. Kind of like the WW system, in that it uses a number of d10's, only instead of each die trying to overcome a number, the total has to overcome a number. So if the DC was a 40, then you'd need at least 5-6 ranks to be able to have a chance in heck of hitting it, but at 8-9 ranks, it wouldn't be too difficult.

Also reminds me of the 'other' way of doing percentiles: 11d10-10.

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