Question

我正在尝试使用R中的简单蒙特卡洛采样程序来回答以下问题：一个骨灰盒包含10个球。两个红色，三个白色和五个黑色。一次抽出全部10张，无需更换。找出绘制的第一个和最后一个球都是黑色的可能性。

我尝试了两种方法，但都没有用。

更长的方法对我来说更直观：

balls <- c(1:10) #Consider 1-5 black, 6-8 white, and 9-10 red.

pick.ball <- function(balls){
sample(x = balls, 1, replace = FALSE)
}

experiment <- function(n){
picks = NULL
keep <- NULL
for(j in 1:n){
   for(i in 1:10){
   picks[i] <- pick.ball(balls = balls)
   }
keep[j] <- ifelse(picks[1] == any(1:5) & picks[10] == any(1:5), 1, 0)
 }
return(length(which(keep == 1))/n)
}

这是我的第二种更简单的方法，它显示了我对重复循环的理解不足。不用理会它-它会永远持续下去。但是，如果有人可以帮助我更好地理解原因，那将不胜感激！

balls <- c(1:10) #Consider 1-5 black, 6-8 white, and 9-10 red.

pick.ball <- function(balls, n){
  keep = NULL
  for(i in 1:n){
  picks <- sample(x = balls, 10, replace = FALSE)
  keep[i] <- ifelse(picks[1] == any(1:5) & picks[10] == any(1:5), 1, 0)
  repeat{
    picks
    if(length(keep) == n){
      break
      }
    }
  }
  return(which(keep == 1)/n)
}

Answer 1

这是我创建的循环。您可以根据需要将其包装在一个函数中。我没有使用数字编号，而是使用字母。

urn <- c(rep("B", 5), rep("W", 3), rep("R", 2))

# Set the number of times you want to run the loop

nloops <- 10000


# Create an empty data frame to hold the outcomes of the simulation

m <- structure(list(first = character(),
                    last = character(),
                    is_black = integer()),
               class = "data.frame")

现在运行循环


set.seed(456)
for (j in 1:nloops) {
  b <- sample(urn, 10, replace = FALSE)
  m[j, 1:2 ] <- b[c(1, 10)] 
  m[j, 3] <- ifelse(m[j, 1] == "B" & m[j, 2] == "B", 1, 0)
}

head(m)

  first last is_black
1     B    W        0
2     B    B        1
3     B    B        1
4     R    B        0
5     B    R        0
6     R    W        0

最后，答案：

# Proportion of cases where first and last ball drawn were black

sum(m[ , 3]) / nloops

# This was 0.22

蒙特卡洛R函数帮助查找概率（（问题引发的球）

1 个答案: