杭州交通信息网公制英制美制螺纹样板:DnD stat creation probability...
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I am an avid DnD player and I have always wondered; given the standard stat creating method, (rolling three 6 sided dice and adding them up), what is the probability function of getting a specific stat, such as 13, or 17?
Is there such a function to express this probability? This obviously relates to ideas of partitions, particularly, unique partitions of 3 numbers 6 or below. If anyone comes up with a general function you will be my personal hero/chosen one.
Umm..yeah I know all this. Doing it manually is trivial. However I am looking for a general function based on the number of dice, how many sided, and what value you seek the probability of getting (by adding the dice up). Call these variables n, s, v.
The function then would be the coeffeciant of x^v of the polynomial(x+x^2+x^3+....+x^s)^n. Realize you cannot use the binomial theorem directly to figure out the coffeciants (you can use the binomial theorem repetedly however).
Up to a certain point (around the middle of the polynomial), the coeffecials are of this form, (n-1)C(v-1) (where C is the choose function or binomial coeffeciant). However I have been unable to determine what happens around the middle.
well, it's not hard.
for example, there r only 3 possible combinations for u to come to the number of 17 :
566, 656, 665
the probability for u to get this is the sum of individual:
3*(1/6^3)=0.0139
which means that the probability for u to get 17 is only 0.0139
similar for other values.
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the probability density function of the sum of several random variables can be calculated by the convolution of each of their PDF.
now the PDF of each of the 3 random variables is the delta function. u can make a 'convolution sum' of them to get the result.
well, i m going to take an exam tomorrow so i don't have time to deal with ur problem. maybe i will have after the exam.
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the problem becomes more complex when u r considering n dice.....
well, it's not easy to find the PDF of the random variable. but it doesn't mean that there is no method. u can solve this problem in 2 ways, but each of them may require the help of the computer.
the first way is, as i have already mentioned yesterday, to do the convolution.
suppose: f(n)=1/6 when n=1, 2, 3, 4, 5, 6, and is 0 for any other ns.
then the PDF u want will be:
f(n)*f(n)....f(n) (totally m )
'*'means convolution, and 'm' is the number of dice. it's not very easy to calculate in this ay.
the other method is to use the characteristic function.
let
f(x)=∑1/6*δ(x-n) where n=1,2,3,4,5,6
ψ(ω)=∫f(x)exp(jωx)dx
where the integration region above is (-∞,∞), and j stands for the imaginary unit.
then the PDF u want will be:
g(x)=(1/2π)*∫{[ψ(ω)]^m}*exp(-jωx)dω
where m is the number of dice.
u can use a computer to calculate it.
when m is very large, the value should be 3.5n accroding to the large number therom, and will be gaussion distributed accroding to the central limit therom
我帮你翻译一下（I help you to translate）
DnD stat 创作可能性...
????0 -??????? 14? 23??
这到目前为止stumped 我的一个有趣的问题,
我是一个热中DnD 球员并且我总想知道; 给标准stat 创造方法, (辗压三6 个被支持的模子和把他们加起来), 什么是得到具体stat 的可能性作用, 譬如13, 或17?
有是这样作用表达这个可能性吗? 这与分开想法明显地关系, 特别, 3 第号6 的独特的分开或下面。 如果任何人产生一个一般作用您将是我的个人英雄选上的一个。
it's ok, you know the char's stat is base on 4D6, so you ran roll 4 dices, use the biggest 3 one. It may produce a 18. Anyway it seem rarely but it fact it is.
And one point is you must roll 4d6 6times have no stop to build a char.