So this week I was taking a look at coldblooded leadership. With this rule, whenever you take a leadership, you roll three dice and take the lowest two. Not surprising, this makes passing the leadership rolls much easier, but how much easier?
This was actually difficult for me to figure out with mathematics. The problem being, I can't just count up how many different ways there are for a person to roll a 6 on three d6 (33 different ways, fyi) but I had to count and differentiate the different rolls because 1,2,5 and 3,1,5 and 5,1,6 and 6,2,4 will all be the equivalent of a leadership roll of 6. I tried several different ways to make this easier on myself and I found that it was just a simple matter of listing out all 216 different ways you can roll three d6. Then I just went through them and figured out the sum of the two lowest dice.
So it turns out that the percentages for rolling 3d6 are as follows:
2 3 4 5 6 7 8 9 10
7.41% 19.91% 35.65% 52.31% 67.59% 80.09% 88.89% 94.91% 98.15%
2 3 4 5 6 7 8 9 10
7.41% 19.91% 35.65% 52.31% 67.59% 80.09% 88.89% 94.91% 98.15%
So there it is, most armies have only a 2.8% chance of getting insane courage, but coldblooded units have almost three times the chance of pulling it off.
For anyone that is interested, the possible different results of rolling 3d6 are:
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Maybe its a good thing that I go back to work in a week. I appear to have too much time on my hands.
These type of calculations all pretty much always involve writting out at least a short hand notation for all the possabilities.
ReplyDeleteFor example for result 2 I would think about it as needing 2 ones and the other die does not matter except if it is a third 1 So you get 1 for all 3 ones, then 5 (2-6) times 3 (whether the rejected die is the firt, second or third) giving 16 possible ways to get 2.
For a result like 7 I would first break it down by what 2 die results sum up then what the rejected die value is allowed to be and if 1,2,3 of the dice have different values for the 1,3,6 multipler.
7: 1,6 Other die must be 6 two dice with same value count as 3.
2,5 Other 5-6, 3 for the 5, 6 for the 6.
3,4 Other 4-6, 3 for 4, 6 for 5, 6 for 6.
Yields:3+3+6+3+6+6=27 out of 216.
Counting them just not as explicitly as you did. I did many of these same calculations for my Math Hulk series.
I tried it just like that to begin with, but I just kept confusing myself with it, so I went to the thorough method. In school I was known as a blunt force trauma mathematician so I guess I am keeping up with that!
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