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【博德之門3】?jī)?yōu)勢(shì)/重骰收益究竟多少?

2023-08-11 20:01 作者:半萌不萌  | 我要投稿

今天讓學(xué)生幫忙打工算了一下,有錯(cuò)誤會(huì)更正

太長(zhǎng)不看版:

1?

dn骰子的期望為%5Cfrac%7Bn%2B1%7D%7B2%7D%20 (地球人應(yīng)該都知道吧)

如d20的期望為10.5


2?

重骰1/2的專長(zhǎng)(如巨武器戰(zhàn)斗)帶給每個(gè)dn骰子的收益為%5Cfrac%7Bn-2%7D%7Bn%7D%20

如手持星界銀劍(2d6+3+1d6)蘸火(1d4)施放2環(huán)至圣斬(3d8)的收益為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

(聽說PHB里面只有武器骰才吃,但實(shí)測(cè)游戲里都吃,改了就只算武器骰,別來找我)


3

dn 優(yōu)勢(shì)骰/劣勢(shì)骰的收益/損失為%5Cfrac%7Bn%5E2-1%7D%7B6n%7D%20

如d20劣勢(shì)骰的期望會(huì)減少3.325,d8優(yōu)勢(shì)骰的期望會(huì)增加1.3125


4

已經(jīng)優(yōu)勢(shì)的情況下,重骰1/2的收益為(如兇蠻打手+巨武器戰(zhàn)斗)

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

如一個(gè)d6的收益為0.178,一個(gè)d8的收益為0.163


正篇

首先我們要知道如下期望公式:

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

以及兩個(gè)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

因此一個(gè)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)

重骰專長(zhǎng)生效時(shí),有%5Cfrac%7B2%7D%7Bn%7D%20的概率觸發(fā)重骰,得到原來的期望,有%5Cfrac%7Bn-2%7D%7Bn%7D%20的概率不發(fā)生重骰,這部分的期望為

%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

較原先(結(jié)論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


考慮n面骰的累計(jì)概率分布,投出至多x的概率為

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

在具有優(yōu)勢(shì)(雙骰取高)的情況下,很顯然想要令其結(jié)果至多為x,則兩骰皆至多為x??紤]到兩者獨(dú)立,優(yōu)勢(shì)骰不能超過y的概率為

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的概率為

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的概率為%5Cfrac%7B2%5Ctimes14-1%20%7D%7B20%5E2%7D%20%3D%5Cfrac%7B27%7D%7B400%7D%20

使用期望公式,將得到的結(jié)果乘以對(duì)應(yīng)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

減去無優(yōu)勢(shì)的期望,得到收益為

%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)


在同時(shí)具有優(yōu)勢(shì)和1,2重骰的情況下,期望為

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)

減去無優(yōu)勢(shì)期望%5Cfrac%7Bn%2B1%7D%7B2%7D%20,收益為

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

再減去結(jié)論3的收益量?%5Cfrac%7Bn%5E2-1%7D%7B6n%7D

結(jié)果為?

%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)


歡迎指正。

【博德之門3】?jī)?yōu)勢(shì)/重骰收益究竟多少?的評(píng)論 (共 條)

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