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【博德之门3】优势/重骰收益究竟多少?

2023-08-11 20:01 作者:半萌不萌  | 我要投稿

今天让学生帮忙打工算了一下,有错误会更正

太长不看版:

dn骰子的期望为%5Cfrac%7Bn%2B1%7D%7B2%7D%20 (地球人应该都知道吧)

如d20的期望为10.5


重骰1/2的专长(如巨武器战斗)带给每个dn骰子的收益为%5Cfrac%7Bn-2%7D%7Bn%7D%20

如手持星界银剑(2d6+3+1d6)蘸火(1d4)施放2环至圣斩(3d8)的收益为2%5Ctimes%20%5Cfrac%7B4%7D%7B6%7D%20%2B1%5Ctimes%20%5Cfrac%7B2%7D%7B4%7D%20%2B3%5Ctimes%20%5Cfrac%7B6%7D%7B8%7D%20%3D%5Cfrac%7B49%7D%7B12%7D%3D4.08%20

(听说PHB里面只有武器骰才吃,但实测游戏里都吃,改了就只算武器骰,别来找我)


3

dn 优势骰/劣势骰的收益/损失为%5Cfrac%7Bn%5E2-1%7D%7B6n%7D%20

如d20劣势骰的期望会减少3.325,d8优势骰的期望会增加1.3125


4

已经优势的情况下,重骰1/2的收益为(如凶蛮打手+巨武器战斗)

%5Cfrac%7B4n%5E2-11n-1%7D%7B2n%5E3%7D%20

如一个d6的收益为0.178,一个d8的收益为0.163


正篇

首先我们要知道如下期望公式:

E%3D%5Csum_%7B%7Dx_%7Bi%7D%20p_%7Bi%7D%20

以及两个求和公式:

%5Csum_%7Bx%3D1%7D%5En%20x%20%3D%5Cfrac%7Bn(n%2B1)%7D%7B2%7D%20

%5Csum_%7Bx%3D1%7D%5En%20x%5E2%20%3D%5Cfrac%7Bn(n%2B1)(2n%2B1)%7D%7B6%7D%20

因此一个n面骰的期望就是:

%5Cfrac%7B1%7D%7Bn%7D%20%5B%E6%A6%82%E7%8E%87%5D(1%2B2%2B3%2B...n)%5B%E5%8F%96%E5%80%BC%5D%3D%5Cfrac%7B1%7D%7Bn%7D%5Cfrac%7Bn(n%2B1)%7D%7B2%7D%3D%5Cfrac%7Bn%2B1%7D%7B2%7D%20%20%20%20%20%20%20%20(%E7%BB%93%E8%AE%BA1)

重骰专长生效时,有%5Cfrac%7B2%7D%7Bn%7D%20的概率触发重骰,得到原来的期望,有%5Cfrac%7Bn-2%7D%7Bn%7D%20的概率不发生重骰,这部分的期望为

%5Cfrac%7B1%7D%7Bn%7D%20%5B%E6%A6%82%E7%8E%87%5D(3%2B...n)%5B%E5%8F%96%E5%80%BC%5D%3D%5Cfrac%7B1%7D%7Bn%7D%5Cfrac%7Bn(n%2B3)%7D%7B2%7D%3D%5Cfrac%7Bn%2B3%7D%7B2%7D

较原先(结论1)增加1,乘以概率可得

E%3D%5Cfrac%7B3%7D%7Bn%5E2%7D%5B%E9%87%8D%E9%AA%B0%E6%A6%82%E7%8E%87%5D%5Cfrac%7B(n%2B1)(4n-1)%7D%7B6n%7D%5B%E9%87%8D%E9%AA%B0%E7%BB%93%E6%9E%9C%E6%9C%9F%E6%9C%9B%5D%20%20%2B%5Csum_%7Bx%3D3%7D%5En%20%5Cfrac%7Bx(2x-1)%7D%7Bn%5E2%7D%20%5B%E6%9C%AA%E9%87%8D%E9%AA%B0%E9%83%A8%E5%88%86%E6%9C%9F%E6%9C%9B%5D


考虑n面骰的累计概率分布,投出至多x的概率为

P(X%5Cleq%20x)%20%3D%20%5Cfrac%7Bx%7D%7Bn%7D%20

 如d20有%5Cfrac%7B16%7D%7B20%7D%20概率投出小于等于16

在具有优势(双骰取高)的情况下,很显然想要令其结果至多为x,则两骰皆至多为x。考虑到两者独立,优势骰不能超过y的概率为

P(Y%5Cleq%20y)%3DP(X_1%5Cleq%20y)P(X_2%5Cleq%20y)%3D(P(X%5Cleq%20y))%5E2%3D%5Cfrac%7By%5E2%7D%7Bn%5E2%7D%20

所以投出y的概率为

P(Y%3D%20y)%3DP(Y%5Cleq%20y)-P(Y%5Cleq%20y-1)%3D%5Cfrac%7By%5E2-(y-1)%5E2%7D%7Bn%5E2%7D%20%3D%5Cfrac%7B2y-1%7D%7Bn%5E2%7D%20

如d20投出14的概率为%5Cfrac%7B2%5Ctimes14-1%20%7D%7B20%5E2%7D%20%3D%5Cfrac%7B27%7D%7B400%7D%20

使用期望公式,将得到的结果乘以对应的概率再求和,得

E%3D%5Cfrac%7B1%7D%7Bn%5E2%7D%20%5Csum_%7By%3D1%7D%5En%20y(2y-1)%3D%5Cfrac%7B1%7D%7Bn%5E2%7D(2%5Csum_%7By%3D1%7D%5En%20y%5E2-%5Csum_%7By%3D1%7D%5En%20y)

%3D%5Cfrac%7B1%7D%7Bn%5E2%7D(%5Cfrac%7Bn(n%2B1)(2n%2B1)%7D%7B3%7D%20-%5Cfrac%7Bn(n%2B1)%7D%7B2%7D)%3D%5Cfrac%7B(n%2B1)(4n-1)%7D%7B6n%7D

减去无优势的期望,得到收益为

%5Cfrac%7B(n%2B1)(4n-1)%7D%7B6n%7D-%5Cfrac%7Bn%2B1%7D%7B2%7D%20%20%3D%5Cfrac%7Bn%5E2-1%7D%7B6n%7D%20(%E7%BB%93%E8%AE%BA3)


在同时具有优势和1,2重骰的情况下,期望为

E%3D%5Cfrac%7B3%7D%7Bn%5E2%7D%5B%E9%87%8D%E9%AA%B0%E6%A6%82%E7%8E%87%5D%5Cfrac%7B(n%2B1)(4n-1)%7D%7B6n%7D%5B%E9%87%8D%E9%AA%B0%E7%BB%93%E6%9E%9C%E6%9C%9F%E6%9C%9B%5D%20%20%2B%5Csum_%7Bx%3D3%7D%5En%20%5Cfrac%7Bx(2x-1)%7D%7Bn%5E2%7D%20%5B%E6%9C%AA%E9%87%8D%E9%AA%B0%E9%83%A8%E5%88%86%E6%9C%9F%E6%9C%9B%5D

%3D%5Cfrac%7B3%7D%7Bn%5E2%7D%20%5Cfrac%7B(n%2B1)(4n-1)%7D%7B6n%7D%2B(%5Cfrac%7B(n%2B1)(4n-1)%7D%7B6n%7D-%5Cfrac%7B7%7D%7Bn%5E2%7D)

减去无优势期望%5Cfrac%7Bn%2B1%7D%7B2%7D%20,收益为

%5Cfrac%7Bn%5E4%2B11n%5E2-33n-3%7D%7B6n%5E3%7D%20

再减去结论3的收益量 %5Cfrac%7Bn%5E2-1%7D%7B6n%7D

结果为 

%5Cfrac%7Bn%5E4%2B11n%5E2-33n-3%7D%7B6n%5E3%7D%20-%5Cfrac%7Bn%5E2-1%7D%7B6n%7D%3D%5Cfrac%7B4n%5E2-11n-1%7D%7B2n%5E3%7D%20(%E7%BB%93%E8%AE%BA4)


欢迎指正。

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