Complicated Math: Dice and Probability - Downloaded from the Wizards Community, all coding removed by Solauren

Complicated Math: Dice and Probability

Post/Author/DateTime Post
#1
Novacat
May 03, 2012 4:13:02 So I'm tinkering with an RPG system, and I'd like to have a system where task resolution is accomplished by rolling X d6s, and then keeping the best 2 and adding them together. The problem is, I don't know how to determine the average result of 3d6 ignoring the lowest die, other than to brute-force it by charting out every possible combination (6*6*6 = 216 possible combinations).

The averages that I need are:

3d6 ignoring the highest
3d6 ignoring the lowest
4d6 ignoring the two lowest
5d6 ignoring the three lowest

If any of you math gurus know what kind of algorithm I can use to find these, that would be appreciated, cause I'd rather not manually average out 7,776 combinations of dice.
#2
lokiare
May 03, 2012 9:35:15 You multiply the number of sides with the number of dice used repeatedly subtracting one and then multiplying again until you count out the number of dice.

Well a chart is a good way to spot trends:

for 2d6:

2     3     4     5     6     7     8     9     10     11     12

1,1 1,2   2,2   2,3 3,3 3,4 4,4 5,4   5,5    6, 5   6,6

2,1 3,1    3,2 4,2 4,3 3,5 4,5   6,4    5,6

             1,3   1,4   2,4 2,5 5,3 6,3   4,6

                     4,1   1,5 5,2 2,6 3,6

                             5,1 1,6 6,2

                                    6,1

It doesn't matter how many dice you have, they will all look like that in general with the two outside ranges having the lowest chance and moving up while going toward the center.
#3
Novacat
May 03, 2012 10:20:24 I'm sorry, I don't understand. Could your provide an example?
#4
lokiare
May 03, 2012 14:15:44
I'm sorry, I don't understand. Could your provide an example?

Ok, to roll a 12 on 2d6 you have a 1 in 36 chance. To roll an 11 on 2d6 you have a 2 in 36 or 1 in 18 chance. Basically your chance doubles going toward the middle range (for 2d6 that's 7). This is because more number combinations means more chances of getting that number (for 7 there are 6 different combinations of dice rolls that can equal it).

For adding dice in, dropping the lowest it lowers the chance of getting numbers on the low spectrum. This is because in order to drop a number it has to be lower than or equal to the lowest number needed. So there is only one combination of rolls to get a 2 (1,1,1). As opposed to five combinations to get 12: (1, 6,6), (2, 6, 6), (3, 6, 6), (4,6,6), (5,6,6), (6,6,6). So you would have a lop sided pyramid diagram then. With the left side the same, but the right side barely going down.

Post/Author/DateTime	Post
#1 Novacat May 03, 2012 4:13:02	So I'm tinkering with an RPG system, and I'd like to have a system where task resolution is accomplished by rolling X d6s, and then keeping the best 2 and adding them together. The problem is, I don't know how to determine the average result of 3d6 ignoring the lowest die, other than to brute-force it by charting out every possible combination (666 = 216 possible combinations). The averages that I need are: 3d6 ignoring the highest 3d6 ignoring the lowest 4d6 ignoring the two lowest 5d6 ignoring the three lowest If any of you math gurus know what kind of algorithm I can use to find these, that would be appreciated, cause I'd rather not manually average out 7,776 combinations of dice.
#2 lokiare May 03, 2012 9:35:15	You multiply the number of sides with the number of dice used repeatedly subtracting one and then multiplying again until you count out the number of dice. Well a chart is a good way to spot trends: for 2d6: 2 3 4 5 6 7 8 9 10 11 12 1,1 1,2 2,2 2,3 3,3 3,4 4,4 5,4 5,5 6, 5 6,6 2,1 3,1 3,2 4,2 4,3 3,5 4,5 6,4 5,6 1,3 1,4 2,4 2,5 5,3 6,3 4,6 4,1 1,5 5,2 2,6 3,6 5,1 1,6 6,2 6,1 It doesn't matter how many dice you have, they will all look like that in general with the two outside ranges having the lowest chance and moving up while going toward the center.
#3 Novacat May 03, 2012 10:20:24	I'm sorry, I don't understand. Could your provide an example?
#4 lokiare May 03, 2012 14:15:44	I'm sorry, I don't understand. Could your provide an example? Ok, to roll a 12 on 2d6 you have a 1 in 36 chance. To roll an 11 on 2d6 you have a 2 in 36 or 1 in 18 chance. Basically your chance doubles going toward the middle range (for 2d6 that's 7). This is because more number combinations means more chances of getting that number (for 7 there are 6 different combinations of dice rolls that can equal it). For adding dice in, dropping the lowest it lowers the chance of getting numbers on the low spectrum. This is because in order to drop a number it has to be lower than or equal to the lowest number needed. So there is only one combination of rolls to get a 2 (1,1,1). As opposed to five combinations to get 12: (1, 6,6), (2, 6, 6), (3, 6, 6), (4,6,6), (5,6,6), (6,6,6). So you would have a lop sided pyramid diagram then. With the left side the same, but the right side barely going down.