java - Monty hall 模拟未按预期运行

Question

    import java.util.Scanner;
    import static java.lang.System.*;
    import static java.lang.Math.*;
    import java.util.Random; 


    public class Montyhall 
   {
    public static void main(String[] args) 
    {
        Scanner keyboard = new Scanner(System.in);
        System.out.print("Enter number of attempts:");
        int attempts = keyboard.nextInt();/*Decide how many times you want this to
        run. The larger the number, the more realistic an answer you will get.*/

        int curtains[] = new int[3]; //like the game show, one of these will be a winner

        Random rand = new Random(); /*sets the methods up with the same random seed, which
        otherwise changes by the milisecond and could influence the response*/

        withoutswitch(curtains, attempts, rand);
        withswitch(curtains, attempts, rand);



    }

    public static void withoutswitch(int curtains[], int attempts, Random rand)
    {

        int nsCorrect=0;

        for(int x=0; x < attempts; x++)
        {
            int winner = rand.nextInt(3);//Sets the winner to 1, leaving the other two at 0.
            curtains[winner] = 1;        //It then checks to see if they are the same, 
            int guess = rand.nextInt(3); //a 1/3 chance, roughly.

            if(curtains[guess]==1){
                nsCorrect++;
            }
            curtains[0]=0;
            curtains[1]=0;
            curtains[2]=0;
        //the player never changes their decision, so it does not matter if a door is opened.
        }
        System.out.println("Number of successes with no switch: " + nsCorrect);


    }

    public static void withswitch(int curtains[], int attempts, Random rand)
    {

        int ysCorrect=0;
        int goat = 0;

        for(int x=0; x < attempts; x++)
        {
            int winner = rand.nextInt(3);
            curtains[winner] = 1;
            int guess = rand.nextInt(3);
            goat = rand.nextInt(3);//one of the doors is opened
            while(goat == winner || goat == guess)//the opened door is randomized until
                goat = rand.nextInt(3); //it isn't the guess or the winner.


            int guess2 = rand.nextInt(3);
            while(guess2 == guess || guess2 == goat)
                guess2 = rand.nextInt(3);//the second guess goes through a similar process

            if(curtains[guess2]==1){

                ysCorrect++;
                }
            curtains[0]=0;
            curtains[1]=0;
            curtains[2]=0;
        }
        System.out.println("Number of successes with a switch: " + ysCorrect);
        }   

}

抱歉，它有点乱，我正试图在中断近一年后重新开始编码。第一部分按预期运行，返回大约 1/3 的成功机会。然而，第二个应该给我 2/3 的机会，但我仍然得到与没有开关时大致相同的正确数量。我浏览了该站点，大部分发现了我不熟悉的 Java 之外的内容。 This一个非常相似，但似乎没有人真正帮助解决主要问题。

我该怎么做才能使赔率更真实？关于清理代码的建议也将不胜感激。

编辑:代码现在可以运行了，我现在只是想把它精简一下。

Answer 1

在你的方法withoutswitch你需要改变

if(guess==1)
  nsCorrect++;

到

if (curtains[guess] == 1)
  nsCorrect++;

与 withswitch 方法相同。每次运行 for 循环后，您需要将 curtains 重置为 0。否则之前的 1 将在那里，并且在几次运行后 curtains 将只包含 1。

private static void resetCurtains(int[] curtains) {
  for (int i = 0; i < curtains.length; i++) {
    curtains[i] = 0;
  }
}

每次在 for 循环中运行后调用该方法。

此外，我建议使用 {}，即使语句是 1-liner:

if (curtrains[guess] == 1) {
  nsCorrect++;
}