作业:使用R模拟抛硬币直到连续的掷硬币 [英] Homework: Simulating coin tosses until consecutive heads using R

问题描述

我是R的新手，因此在这里提出了疑问(未能找到非常有用的模拟详细教程.)

I am new to R hence asking here (haven't been able to find very helpful tutorials for simulation that are detailed.)

问题陈述是这样

1. 模拟抛硬币20次并记录正面数 &最长的元首.
2. 模拟抛硬币并记录所需的翻动次数，直到依次(连续)出现2、3、4个头(负二项式?)
3. 用不同的种子运行100次以查找记录的项目的分布.

1. Simulate a coin toss for 20 times and record the number of heads & longest run of heads.
2. Simulate a coin toss and record the number of flips necessary until 2,3,4 heads occur in sequence (consecutively) (negative binomial?)
3. Make 100 runs with different seeds to find the distribution of items recorded.

如何用编程语言R解决这个问题?

How does one go about solving this in the programming language R ?

对于1，我做了以下事情:

For 1, I did the following:

n=20                #no of trials
y=NULL              #initializing a vector of NULL values   
for (i in 1:n) {
    x=runif(1)      #random uniform
    if (x<0.5) {    #if condition for assigning heads / tails
        y[i]='H'
    } else {
       y[i]='T'
    }
}
y                   #print the vector storing the heads and tails.

对于2，我理解这是负二项式的情况，因为它说直到依次出现2、3、4个头".但是我不确定如何继续用R编写代码或提出逻辑.

For 2, I understand that this is a case of negative binomial because it says "until 2,3,4 heads occur in sequence" . But I am not sure how to proceed to write the code in R or come up with the logic.

对于3，我猜想必须计算100次运行中的磁头数量，即磁头的游程长度.但是，设置不同种子意味着什么?我不确定.

For 3, I am guessing have to calculate the number of heads got in 100 runs, run lengths of heads. But what does it imply to set different seeds? I am unsure of that.

注意，这是一个作业问题，是的(我无法对其进行标记).我不需要代码，只需一些指针/更正/说明/建议即可.

note This is a homework problem, yes (I am not able to tag it so). I do not need the code, just some pointers/ corrections/clarifications/suggestions would help.

编辑为了模拟2，我在MATLAB上尝试了以下代码-

edit For simulating 2, I tried out the following code on MATLAB -

head_count = 0;
flip_count = 0;

while (head_count < 3)                  #counting for 3 multiple heads
    flip = rand(1);
    flip_count = flip_count+1;
    if (flip > 0.5)
        head_count = head_count + 1;
    else 
         head_count = 0;       #invalidate head counts if not successive.
    end
end
disp(flip_count)

请让我知道此逻辑是否正确.我可以使用R重新创建它.

Please let me know if this logic is correct. And I can recreate it using R.

推荐答案

由于这是家庭作业，所以我想给您一些有关#2和#3的指针.

As this is homework, I would like to give you some pointers for #2 and #3.

对于#2，进行一个循环，继续进行抛硬币并计算连续的头数.在每次掷球时，将掷球数增加1，而在达到要求的杆头数时，只需返回掷球数即可.

For #2, make a loop which keeps doing coin tosses and count the number of heads in a row. At every toss increase the count of tosses by 1 and when reaching the number of heads requested, just return the count of tosses.

为了让您入门，这将使nbTosses tossesL

To get you started, this will do nbTosses tossesL

coin <- c('h','t')

ComputeNbTosses <- function(targetTosses) {
  nbtosses = 0;
  while (nbtosses < targetTosses) {
    nbtosses = nbtosses + 1
  }
  return(nbtosses);
} 

您只需要对其进行修改，以接受参数targetHeads并仅在到达tagetHeads时退出while循环.

You just need to amend it to take in parameter targetHeads and exit the while loop only when you have reached the tagetHeads.

对于#3，将#2放入函数中，然后使用set.seed(integer)设置种子.计算机上的随机数并不是真正的随机数.大多数随机生成器都允许您设置种子.如果使用相同的种子，则将获得两次相同的伪随机数序列，这可以进行调试.

For #3, put #2 in a function and then use set.seed(integer), to set the seed. Random numbers on a computer are not really random. Most of random generators allow you to set a seed. If you use the same seed you will get twice the same sequence of pseudo random numbers, this allows for debugging.

我将#2从递归更改为普通循环，为他编写代码会更容易.

I changed #2 from a recursion to a normal loop, it will be easier to code for him.

编辑(提示): 这将计算您连续有多少个头，现在，当达到头的目标值时，您必须使循环停止:

EDIT (Tip): This will count how many heads in a row you have, now you have to make the loop stop when you reach the target value of heads:

coin <- c('h','t')

ComputeNbTosses <- function(targetTosses) {
  nbTosses = 0;
  nbHeadsInRow = 0;
  allTosses = c();
  while (nbTosses < targetTosses) {
    toss = sample(coin,1,T)
    allTosses = c(allTosses, toss);
    if (toss == 'h') {
      nbHeadsInRow = nbHeadsInRow + 1;
    } else {
       nbHeadsInRow = 0;
    }
    nbTosses = nbTosses + 1
  }
  ret = list();
  ret$nbTosses = nbTosses;
  ret$nbHeadsInRowAtEnd = nbHeadsInRow;
  ret$allTosses = allTosses;
  return(ret);
} 

ComputeNbTosses(5);
ComputeNbTosses(8);

这将打印您到达nbTosses时的行头数.这不是您想要的.您想要的是:

This will print how many heads in row you had when it reached nbTosses. This is not what you want. What you want is:

• 将参数从targetTosses更改为targetHeadsInRow
• 更改while中的条件，以使其在nbHeadsInRow达到targetHeadsInRow时停止.
• 您需要退货的是nbTosses(达到您的状况所需要的抛掷次数)

• change the argument from targetTosses to targetHeadsInRow
• change the condition in the while so that it stops when nbHeadsInRow reaches targetHeadsInRow.
• what you need to return is nbTosses (the number of tosses to reach your condition)

让我知道是否有不清楚的地方.

Let me know if anything is not clear.

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