Question

我正在尝试编写代码来获取某个场景的概率。共有52张卡分为4个套装。从每堆中随机抽取1张卡片制作4张卡片组合，然后将这些卡片放回到它们的堆中。你如何计算组合只有1王的概率？我尝试了以下但我认为我做错了

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，您可以使用此代码解决问题。这适用于从每一堆中单张一张牌，或者在替换后随后给出的抽牌。如果您对多次抽奖的概率感兴趣，或者在没有替换的情况下进行后续抽奖，则不起作用。这不是基于重复抽样的计算方法。

所有可能的抽取组合，即每个堆中的国王或不是国王：

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))

每个组合的概率组合：

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

确定你将拥有其中一种组合。

> 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

R和概率

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