Dice and Math-- Need math people!

Hodgie · Oct 16, 2007

The 4e threads talk a lot about reducing the number of dice rolled by having a few dice and then a flat plus. Even in mid level play I saw a lot of situations of players rolling 10-15 d6 multiple times per round.

My question is how many dice do you need to maintain a fair randomization? I am capable of picking a range, but I was wondering if any math folks could lay down some statistics to show that (for example) 5 dice would always be enough to represent 90% of all variance.

Thanks in advance.

Scribbler · Oct 16, 2007

The more dice you have the less random things get. If you have 1d6+3 you're equally likely to get 4, 5, 6, 7, 8, or 9. If you have 2d6 you're more likely to get a 7 than any other number, and you're not very likely to get either a 2 or a 12.

At the same time, more dice gives you a bigger range. You may be most likely to get a 7 with 2d6, but you can't ever get a 2 or 12 with 1d6+3.

So if by "less random" you mean "tending to get the same numbers," more dice makes it less random; if you mean "smaller range of numbers," less dice does that.

Hodgie · Oct 16, 2007

So I did some quick math on my own and found out that in many cases 2d6 seems to capture most of the variance.

For 4d6, 2d6+7 captures ~89% of all possible rolls (the middle 1156 of 1296 possibilities). It perfectly mirrors the average of 14.

For 5d6, 2d6+10 captures ~85% of all possible rolls (the middle 6557 of 7776 possibilities). It also reduces the average from 17.5 to 17.

For 6d6, 2d6+14 captures ~81% of all possible rolls (the middle 37,654 of 46656 possibilities). It perfectly mirrors the average of 21.

So the obvious (and intuitive) pattern is that 2d6 represents a diminishing portion of the total variance as total dice rolled increase. The question is: When is the representation insignificant?

Hodgie · Oct 16, 2007

Scribbler said:
The more dice you have the less random things get. If you have 1d6+3 you're equally likely to get 4, 5, 6, 7, 8, or 9. If you have 2d6 you're more likely to get a 7 than any other number, and you're not very likely to get either a 2 or a 12.

At the same time, more dice gives you a bigger range. You may be most likely to get a 7 with 2d6, but you can't ever get a 2 or 12 with 1d6+3.

So if by "less random" you mean "tending to get the same numbers," more dice makes it less random; if you mean "smaller range of numbers," less dice does that.

My intention is to transform the 20d6 breathweapon into [a few d6 + a number] while still being fair. I realize that I could just turn that 20d6 into 4d6 + 56 (which is the remaining 16 dice averaged to 3.5 and summed), but what I want to know is if those 4d6 can sufficiently represent the bulk of the variation you are likely to see.

I don't care if we never see a 20d6 breathweapon deal 120 damage because that wasn't going to happen anyways.

Elemmakil · Oct 16, 2007

Ideally, you need someone can calculate the standard deviation of

d6, for any positive integer n. Unfortunately, I'm not that person. Sorry.

Then, if the standard deviation of 4d6 is close enough to that of 20d6 (for example), your 4d6+56 works fine. "Close enough" is entirely subjective.

Has anyone taken courses in probability? Please?

-Elemmakil

thegreyman · Oct 16, 2007

I can, but my books are at home, as I am at work. Once I get home, I'll dig out my probability book and try to show my work.

eamon · Oct 16, 2007

Well, calculations with variance are simpler (for this kind of thing). Standard deviation is simply the square root of variance, and variance is just the average of the square deviation.

So, for a single d6, with average 3.5, the variance is [ (1-3.5)² + (2-3.5)² +(3-3.5)² +(4-3.5)² +(5-3.5)²+(6-3.5)² ] /6 which is exactly 17.5/6 and roughly 2.92 (and the std. dev. is thus about 1.71).

The variance of the sum of independent random variables - say, Var(2d6) - is the sum of the variance of the the independent random variables - say, 2*Var(d6), which is about 5.83.

So, by reducing the number of d6's you're always reducing "randomness".

Why then is it generally perceived that the more d6's you roll, the less random the result is?

Actually, the std. deviation rises as you add more d6's, but since the std. deviation is the sqrt of the variance, which is directly proportional to the number of d6's and thus the average roll, the std. deviation as a percentage of the average drops when you add more dice.

So at 1d6 the std. deviation is about half the average, but for 10d6 it's already less than a sixth of the average, and by 20d6 it's less than a ninth.

Since we're dealing with distributions that are roughly normal, about 70% of all die-roll's will be within one std. dev. and 95% within two.

Conclusion: if you're just using d6's and just adding them, you can't simulate higher-d6's increased std. dev. just by using fewer dice and adding them. However, you could multiply the d6's, or you could use dice with more sides.

And of course, it's questionable how important damage variance really is anyhow, since the effective randomness is largely determined not by these minor fluctuations in damage, but but the far more major factor of whether you actually hit at all (or, for spells, whether the opponent saves)

In game, that means that I'm not too worried about the reduced randomness of damage - any minor changes in d20 rules are liable to outweigh it. Frankly, you could probably just use average damage almost everywhere without big in-game effect, and leave the randomness to hit-n-miss (and criticals...) instead, especially once you're rolling more than a few damage dice per combat.

Darklone · Oct 16, 2007

I just wanted to say: I love my dice bags. The big ones. With hundreds of dice.

LonePaladin · Oct 17, 2007

The formula they're using to reduce the actual number of dice rolled is fairly straightforward: out of any set of dice rolled, they leave only one-third as actual rolls. The remaining two-thirds are averaged (and, sometimes, they round this off to the nearest multiple of 5 to make the math easier).

I've just started using this in my high-level campaign. Now, when the 19th-level archmage casts disintegrate, he rolls 13d6+88 instead of 38d6.

I've made a spreadsheet that has all of these dice figured up, for every size (from d4s up to d12s) and from one die to 40. I'm not sure how well this forum handles tables, but I might try to put those figures up on this discussion to save all of you the trouble.

LonePaladin · Oct 17, 2007

Here is the table. I may have to edit this a few times to get everything to line up right. Also, I'm excluding the first four lines, and assuming the 'abbreviated dice' idea won't come into play until at least five dice are being rolled for anything.

Rounding is entirely up to you; be aware that doing so will have results where two numbers have identical amounts (for example, if you round up to the nearest 5, the entries for 14d6 and 15d6 both read "5d6+35").

Code:

[FONT=Courier New][B]       d4       d6       d8        d10        d12[/B]
5    2d4+8    2d6+11   2d8+14    2d10+17    2d12+20
6    2d4+10   2d6+14   2d8+18    2d10+22    2d12+26
7    2d4+13   2d6+18   2d8+23    2d10+28    2d12+33
8    3d4+13   3d6+18   3d8+23    3d10+28    3d12+33
9    3d4+15   3d6+21   3d8+27    3d10+33    3d12+39
10   3d4+18   3d6+25   3d8+32    3d10+39    3d12+46
11   4d4+18   4d6+25   4d8+32    4d10+39    4d12+46
12   4d4+20   4d6+28   4d8+36    4d10+44    4d12+52
13   4d4+23   4d6+32   4d8+41    4d10+50    4d12+59
14   5d4+23   5d6+32   5d8+41    5d10+50    5d12+59
15   5d4+25   5d6+35   5d8+45    5d10+55    5d12+65
16   5d4+28   5d6+39   5d8+50    5d10+61    5d12+72
17   6d4+28   6d6+39   6d8+50    6d10+61    6d12+72
18   6d4+30   6d6+42   6d8+54    6d10+66    6d12+78
19   6d4+33   6d6+46   6d8+59    6d10+72    6d12+85
20   7d4+33   7d6+46   7d8+59    7d10+72    7d12+85
21   7d4+35   7d6+49   7d8+63    7d10+77    7d12+91
22   7d4+38   7d6+53   7d8+68    7d10+83    7d12+98
23   8d4+38   8d6+53   8d8+68    8d10+83    8d12+98
24   8d4+40   8d6+56   8d8+72    8d10+88    8d12+104
25   8d4+43   8d6+60   8d8+77    8d10+94    8d12+111
26   9d4+43   9d6+60   9d8+77    9d10+94    9d12+111
27   9d4+45   9d6+63   9d8+81    9d10+99    9d12+117
28   9d4+48   9d6+67   9d8+86    9d10+105   9d12+124
29  10d4+48  10d6+67  10d8+86   10d10+105  10d12+124
30  10d4+50  10d6+70  10d8+90   10d10+110  10d12+130
31  10d4+53  10d6+74  10d8+95   10d10+116  10d12+137
32  11d4+53  11d6+74  11d8+95   11d10+116  11d12+137
33  11d4+55  11d6+77  11d8+99   11d10+121  11d12+143
34  11d4+58  11d6+81  11d8+104  11d10+127  11d12+150
35  12d4+58  12d6+81  12d8+104  12d10+127  12d12+150
36  12d4+60  12d6+84  12d8+108  12d10+132  12d12+156
37  12d4+63  12d6+88  12d8+113  12d10+138  12d12+163
38  13d4+63  13d6+88  13d8+113  13d10+138  13d12+163
39  13d4+65  13d6+91  13d8+117  13d10+143  13d12+169
40  13d4+68  13d6+95  13d8+122  13d10+149  13d12+176[/FONT]