R 和概率

Question

我正在尝试编写代码来获取特定场景的概率。有 52 张牌，分为 4 个花色堆。从每堆中随机抽取 1 张牌以形成 4 张牌组合，然后将牌放回各自的牌堆中。你如何算出该组合只有 1 个 King 的概率？ 我已经尝试过以下方法，但我认为我做错了什么

cards <- c(2:10,'J','Q', 'K','A') 
v <- sample(rep(cards,1:13),1000,replace=T)
cat('The probability of getting a King is approximately:',sum(v=='K')/length(v),'\n')

Answer 1

据我了解您的问题，您可以使用此代码来解决。这适用于从每堆中抽取 1 张牌的单次给定抽取，或替换后的后续给定抽取。如果您对多次抽奖或后续抽奖（无替换）的概率感兴趣，则此方法不起作用。这不是基于重复采样的计算方式。

所有可能的抽奖组合，即每堆中的国王或非国王：

Hearts <- rep(c((rep("k",1)),(rep("n",1))),8)
Spades <- rep(c((rep("k",2)),(rep("n",2))),4)
Clubs <- rep(c((rep("k",4)),(rep("n",4))),2)
Diamonds <- rep(c((rep("k",8)),(rep("n",8))),1)

pile.possibilities <- data.frame(Hearts,Spades,Clubs,Diamonds)

并绘制每堆的概率：

pile.possibilities$H.prob <- ifelse (pile.possibilities$Hearts == "k", (1/13), (12/13))
pile.possibilities$S.prob <- ifelse (pile.possibilities$Spades == "k", (1/13), (12/13))
pile.possibilities$C.prob <- ifelse (pile.possibilities$Clubs == "k", (1/13), (12/13))
pile.possibilities$D.prob <- ifelse (pile.possibilities$Diamonds == "k", (1/13), (12/13))

每个组合的组合概率：

pile.possibilities$Combo.prob <- pile.possibilities$H.prob *  
                                 pile.possibilities$S.prob *   
                                 pile.possibilities$C.prob *   
                                 pile.possibilities$D.prob

您肯定会拥有这些组合之一。

> sum(Pile.combo.prob)
[1] 1

过滤您感兴趣的组合：

pile.possibilities$one.king.combo <- paste(pile.possibilities$Hearts,pile.possibilities$Spades,pile.possibilities$Clubs,pile.possibilities$Diamonds,sep = "")
pile.possibilities$one.king.combo <- sapply(strsplit(pile.possibilities$one.king, NULL), function(x) paste(sort(x), collapse = ''))

one.king.probability<- sum(subset(pile.possibilities, one.king.combo == "knnn")$Combo.prob)
one.king.probability
[1] 0.2420083

#Final data frame used
> pile.possibilities
   Hearts Spades Clubs Diamonds     H.prob     S.prob     C.prob     D.prob Combo.prob one.king.combo
1       k      k     k        k 0.07692308 0.07692308 0.07692308 0.07692308 3.501278e-05           kkkk
2       n      k     k        k 0.92307692 0.07692308 0.07692308 0.07692308 4.201534e-04           kkkn
3       k      n     k        k 0.07692308 0.92307692 0.07692308 0.07692308 4.201534e-04           kkkn
4       n      n     k        k 0.92307692 0.92307692 0.07692308 0.07692308 5.041840e-03           kknn
5       k      k     n        k 0.07692308 0.07692308 0.92307692 0.07692308 4.201534e-04           kkkn
6       n      k     n        k 0.92307692 0.07692308 0.92307692 0.07692308 5.041840e-03           kknn
7       k      n     n        k 0.07692308 0.92307692 0.92307692 0.07692308 5.041840e-03           kknn
8       n      n     n        k 0.92307692 0.92307692 0.92307692 0.07692308 6.050208e-02           knnn
9       k      k     k        n 0.07692308 0.07692308 0.07692308 0.92307692 4.201534e-04           kkkn
10      n      k     k        n 0.92307692 0.07692308 0.07692308 0.92307692 5.041840e-03           kknn
11      k      n     k        n 0.07692308 0.92307692 0.07692308 0.92307692 5.041840e-03           kknn
12      n      n     k        n 0.92307692 0.92307692 0.07692308 0.92307692 6.050208e-02           knnn
13      k      k     n        n 0.07692308 0.07692308 0.92307692 0.92307692 5.041840e-03           kknn
14      n      k     n        n 0.92307692 0.07692308 0.92307692 0.92307692 6.050208e-02           knnn
15      k      n     n        n 0.07692308 0.92307692 0.92307692 0.92307692 6.050208e-02           knnn
16      n      n     n        n 0.92307692 0.92307692 0.92307692 0.92307692 7.260250e-01           nnnn

Answer 2

不要使用

sample

来计算精确概率。

使用二项式公式可以轻松计算 4 张牌组合中只有 1 张 K 的概率，因为该实验满足二项式实验的所有标准。

P(x) = choose(n, x) * p^x * (1 - p)^(n-x)

哪里， n = 试验次数， x = n 次试验中特定结果发生的次数， p = 成功的概率。

在 R 中：

n <- 4
x <- 1
p <- 1/13

choose(n, x) * p^x * (1 - p)^(n-x)
# [1] 0.2420083

https://en.wikipedia.org/wiki/Binomial_distribution