关于三门问题的代码实现

Posted 2020-09-29

首先附三门问题题目和连接https://baike.baidu.com/item/%E4%B8%89%E9%97%A8%E9%97%AE%E9%A2%98/1242689?fr=aladdin

其次根据题目需求，代码实现，这里用的是java

 1 public static void main(String[] args) {
 2         Random r = new Random();
 3         
 4         int no=0;
 5         int yes = 0;
 6         int doors = 3;
 7         int playCount =1000000;
 8         for(int i=0;i<playCount;i++){
 9             //定义三个门
10             ArrayList<Integer> nums = new ArrayList<>();
11             for(int j=1;j<=doors;j++){
12                 nums.add(j);
13             }
14             
15             //rightNum门后是汽车
16             int rightNum = r.nextInt(doors)+1;
17             
18             //参赛者第一次选择的是firstNum门
19             int firstNum = r.nextInt(doors)+1;
20             
21             //主持人去掉一扇后面是山羊的门
22             int cutNum = r.nextInt(doors)+1;
23             while(cutNum == rightNum || cutNum ==firstNum){
24                 cutNum = r.nextInt(doors)+1;
25             }
26             
27             //去掉主持人选中的门
28             for (int j = 0; j < nums.size(); j++) {
29                 if(nums.get(j)==cutNum){
30                     nums.remove(j);
31                 }
32             }
33             
34             //参赛者第二次选择一扇门
35             int secondNum = r.nextInt(doors)+1;
36             while(secondNum == cutNum || secondNum ==firstNum){
37                 secondNum = r.nextInt(doors)+1;
38             }
39             
40             //参赛者第二次选择，如果不换门选中汽车，则no加一
41             if(firstNum == rightNum){
42                 no++;
43             }
44             
45             
46             //参赛者第二次选择，如果换门选中汽车，则yes加一
47             if(secondNum==rightNum){
48                 yes++;
49             }
50             
51         }
52         //循环实验，输出结果
53         System.out.println("no:"+no+"\tyes:"+yes+"\ttotal:"+(no+yes)+"\tplayCount:"+playCount);
54         
55     }

 

结果是换门有更大的概率获得汽车，但门的数量越多，换与不换的概率越相近