根据我的理解,您可以使用此代码解决问题。这适用于从每个堆中单独给定1张牌的情况,或者在替换后进行的后续给定抽取。如果您对多次抽取的概率或在没有替换的情况下进行的后续抽取感兴趣,则不起作用。它不是一种基于重复抽样的计算方法。
所有可能的抽取组合,即从每个堆中是否为King:
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))
每个组合的综合概率:
pile.possibilities$Combo.prob <- pile.possibilities$H.prob *
pile.possibilities$S.prob *
pile.possibilities$C.prob *
pile.possibilities$D.prob
您将肯定拥有其中一种组合。
> sum(Pile.combo.prob)
[1] 1
筛选您感兴趣的组合:
pile.possibilities$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
> 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
rep
是错误的,请查看rep(1:4, 1:4)
以了解如何修正;相反,请尝试rep(cars,4)
。其次,你正在测试单张牌,而不是四张一组的牌。第三,你正在进行有替换的抽样,这意味着可能会得到四张黑桃A。 - r2evanstable(replicate(1000, any(sample(rep(cards, 4), size=4) == 'K')))
。 - r2evanslength(rep(cards,4))
会给你什么结果?table(rep(cards,4))
呢?第一个不是应该得到52
,第二个是得到4
的计数吗?如果是这样的话,请再次向我解释问题,我可能漏掉了什么。(如果不是这样的话,那么我还是漏掉了什么。) - r2evans