Interesting posts from the statisticians. As a semi-layman. I'm comp sci so I have more mathmatics then the common man but I'm not a mathematician.
Here is the explaination of proability that I give to my players.
Assumption: Every number on any die has an equal probability of coming up.
Fact: A specific number from adding dice is not always equally probable. (See proof A)
Assumption: Game rules take that into account and so must be followed when rolling dice that will be added together.
Fact: A specific number from reading dice sequentially, i.e. percentile, is always equally probable. (See proof B)
Ruling: Different die combinations may be substituted for die that will be read sequentially, as long as the die continue to be read sequentially or only one die is rolled. The match must be exact. For example 25% = d4, 10% = d10.
Sides - % Chance of any given side
4 - 25%
6 - 16.6%
8 - 12.5%
10 - 10%
12 - 8.3%
20 - 5%
Proof A
When we add the dice together every side on either die has an equal perctent chance to come up so the probablity of the number occuring is a function of the number of combinations that a given number has such that:
x = the desired number
{y} = the set of all number combinations that can make that number.
n = the number of combinations in y
z = the total number of possible numbers rolled.
i = the toatal numer of possible rolls.
The probablity of a given number is: n * (100/z)
For example: We will roll 2d6
Possible numbers: 2 - 12 inclusive.
Possible combinations: 6 * 6 = 36
{y} for possible numbers
Num {y(d6,d6)} (Total)
2 {1,1} (1)
3 {1,2;2,1} (2)
4 {1,3;3,1;2,2} (3)
5 {1,4;4,1;2,3;3,2;} (4)
6 {1,5;5,1;2,4;4,2;3,3} (5)
7 {1,6;6,1;2,5;5,2;3,4;4,3} (6)
8 {2,6;6,2;3,5;5,3;4,4} (5)
9 {3,6;6,3;4,5;5,4} (4)
10 {4,6;6,4;5,5} (3)
11 {5,6;6,5} (2)
12 {6,6} (1)
Check: 1+2+3+4+5+6+5+4+3+2+1 = 36
There is a 100/36 = 2.78% (rounded from .7 repeating) chance of any particular combination appearing. Below are the odds of any particular number appearing.
2 = 2.78%
3 = 5.56%
4 = 8.34%
5 = 11.11%
6 = 13.89%
7 = 16.67%
8 = 13.89%
9 = 11.11%
10 = 8.34%
11 = 5.56%
12 = 2.78%
Check: (2.78*2)+(5.56*2)+(8.34*2)+(11.11*2)+(13.89*2)+16.67 = 5.56 + 11.12 + 16.68 + 22.22 + 27.78 + 16.67 = 100.06 (acceptable error occured from rounding)
Proof B:
When we read the numbers of the die sequentially every side has the same probablity of coming up and every number combination has exactly one set that can create it. Therefor the probability of any number coming up is the same.
x = the desired number
{y} = the set of all number combinations that can make that number.
n = the number of combinations in y (1 or 2)
z = the total number of possible numbers rolled.
i = the toatal numer of possible rolls.
The probablity of a given number is: n * (100/z)
For example lets roll 2d4.
z = {11,12,13,14,21,22,23,24,31,32,33,34,41,42,43,44)
If we premark a die as the tens roller then n has one combination for each number otherwise it has two but there are duplicates that way so we will assume premarked dice.
11 {1,1}
12 {1,2}
13 {1,3}
14 {1,4}
...
So our probability function degenerates to 1 * 100/z or just 100/z so no matter what dice you use the probablity stays the same.
