蒙蒂霍尔游戏

Question

我是一名新程序员，目前正在学习 Java。我在这里有一个基于游戏 Let's make a deal & monty hall 问题 (https://www.youtube.com/watch?v=mhlc7peGlGg) 的游戏，我一直遇到此代码的逻辑问题。一切似乎都正常，除了我似乎无法让 user_door 切换到另一扇门并确定他们是否正确地成为赢家。如果有人可以帮助我理解我在这里做错了什么，我很乐意借此机会学习，谢谢！

import java.util.Random;
import java.util.Scanner;
public class GameShow {

public static void main(String[] args) {
    Scanner scan = new Scanner (System.in);
    Random generator = new Random();

    // Initialize Variables
    int user_door,
        open_door,
        other_door,
        prize_door;

    // Generate random value 1-3
    prize_door = generator.nextInt(3)+1;
    open_door = prize_door;

    while(open_door == prize_door){
        open_door = generator.nextInt(3)+1;
    }

    other_door = open_door;

    while (other_door == open_door || other_door == prize_door){
        other_door = generator.nextInt(3)+1;
    }

    // Begin Game
    System.out.println("*** Welcome to the game show! ***");  
    System.out.println("Select the door (1, 2, or 3): ");
    user_door = scan.nextInt();


    // User Validation
        if (user_door > 3 || user_door < 0) {
            System.out.println("Please select door 1, 2, or 3");
            user_door = scan.nextInt();
        } else if(user_door == 1 || user_door == 2 ||  user_door == 3) {

    //Continue Game
    System.out.println("\nIn a moment, I will show you where the prize is located,");
    System.out.println("but first I will show you what is behind one of the other doors");

    //Continue Dialogue
    System.out.println("\nBehind door number " + open_door + " are goats!");
    System.out.println("You selected door number " + user_door);
    System.out.println("\nWould you like to switch your door(y/n)? ");

    //User Input Yes or No
    char userReply = scan.next().charAt(0);

    //If statement with nested while statements for user input
        if (userReply == 'y'){
                user_door = other_door;
                } while(userReply != 'y' && userReply != 'n')
                {
                    //User Validation
                    System.out.println("Please enter either y/n");
                    userReply = scan.next().charAt(0);
                } 

    System.out.println("The prize is behind door number: " + prize_door); 

    //Check to see if user won or lost
        if(user_door == prize_door){
            System.out.println("Congratulations! You won the prize!");
            } else {
                    System.out.println("Sorry. You lost.");
                    }       

        }
    }
}

Answer 1

我同意 Srikanth 上面所说的。缺少的是您测试用户回复的部分。您应该切换“if”和“while”语句。这将有助于解决您的门开关问题。

之前

 if (userReply == 'y')
     {
     user_door = other_door;
     }  
 while(userReply != 'y' && userReply != 'n')
     {
     //User Validation
     System.out.println("Please enter either y/n");
     userReply = scan.next().charAt(0);
     } 

改成

 while(userReply != 'y' && userReply != 'n')
    {
        //User Validation
        System.out.println("Please enter either y/n");
        userReply = scan.next().charAt(0);
    }
 if (userReply == 'y')
    {
     user_door = other_door;
    } 

您执行此操作的方式仅检查第一次输入字符的时间。如果他们做错了并试图再做一次，那么它不会切换门，因为你已经通过了程序的那部分。

Answer 2

您永远不会正确分配另一扇门。您分配的另一扇门是一个随机数，而且您的实现也没有正确描述蒙蒂霍尔问题。

一个破坏程序的例子

假设奖品门 = 1，打开门 = 2，然后其他门 = 3 默认情况下。

假设用户选择了 2 号门。那么，当 2 号门打开时，实际上还有 2 个可能的其他门：1 号和 3 号。

Monty Hall 问题的一个关键部分是打开了一扇用户没有选择的门。

当你选择用户的门时，你需要确定other_door是什么。这段代码有点难以理解，所以我将回顾一下我在尝试找到您的问题时发现的一些问题。

奖门生成

// Generate random value 1-3
prize_door = generator.nextInt(3)+1; //assigns a value to the prize
open_door = prize_door;

while(open_door == prize_door){
    open_door = generator.nextInt(3)+1; //assigns a value to the door to open, not the prize
}

other_door = open_door;

while (other_door == open_door || other_door == prize_door){
    other_door = generator.nextInt(3)+1; //assigns the last value to this
}

此代码正确分配了prize_door，但其余部分还不需要完成。我们可以将其实现简化为。

prize_door = generator.nextInt(3) + 1; //we know what the prize is

这比 3 个变量更容易管理，并且对随机性的依赖性更小。我们只需要跟踪选择者的门，以及奖品后面的门。

打开用户的大门

user_door = -1; //instantiate to -1 here

// User Validation
while (user_door > 3 || user_door < 0) {
    System.out.println("Please select door 1, 2, or 3");
    user_door = scan.nextInt();
}

如果我在被告知不要输入 4 后尝试输入，您的用户验证将失败。您应该将它们锁定在循环中，直到它们选择一扇门。

这也消除了您对 if 的需求，因为一旦离开循环，数字将是 1、2 或 3

开门

此时，我们已经实例化了两个数字。

System.out.println("\nBehind door number " + open_door + " are goats!");
System.out.println("You selected door number " + user_door);

在最初的 Monty Hall 游戏中，您应该打开一扇选择者尚未选择的门。用户的门永远不会打开。在这种情况下，我们有两个输入，user_door 和 Prize_door 使用俗气的逻辑，我们可以通过将它们加在一起来确定打开哪扇门，或者你可以做一个 true if 来验证。我会介绍俗气的，因为它写起来更快而且仍然有意义。

if(prize_door + user_door == 4) { // 1 and 3
    open_door = 2;
}
else if(prize_door + user_door == 5) { // 2 and 3
    open_door = 1;
}
else { //1 and 2
    open_door = 3;
}

非俗气的选项是：

boolean oneIsTaken = false;
boolean twoIsTaken = false;

if(prize_door == 1 || user_door == 1) {
    oneIsTaken = true;
}
if(prize_door == 2 || user_door == 2) {
    twoIsTaken = true;
}

if(oneIsTaken && twoIsTaken) {
    open_door = 3;
}
else if(oneIsTaken) {
    open_door = 2;
}
else {
    open_door = 1;
}

获得可以切换到的门

使用打开的门和用户的原始门，我们可以选择切换的门，因为这两个不是你要切换的门。

if(open_door + user_door == 4) { // 1 and 3
    user_door = 2;
}
else if(open_door + user_door == 5) { // 2 and 3
    user_door = 1;
}
else { //1 and 2
    user_door = 3;
}

判断用户是否中奖

和以前一样：

    if(user_door == prize_door){
        System.out.println("Congratulations! You won the prize!");
        } else {
                System.out.println("Sorry. You lost.");
                }       

    }

这个没有问题。

Answer 3

我已经做了一些代码优化，重要的是我已经解决了它。您在代码中犯的错误将在下面解释。

import java.util.Random;
import java.util.Scanner;

public class table {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        Random generator = new Random();

        // Initialize Variables
        int user_door, open_door, other_door, prize_door;

        // Generate random value 1-3
        prize_door = generator.nextInt(3) + 1;
        other_door = prize_door;

        // Begin Game
        System.out.println("*** Welcome to the game show! ***");

        // User Validation
        do{
            System.out.println("Select the door (1, 2, or 3): ");
            user_door = scan.nextInt();
        }while(user_door > 3 || user_door < 0);
        do{
            open_door = generator.nextInt(3)+1;
        }while(open_door == prize_door || open_door == user_door);

        System.out.println("\nIn a moment, I will show you where the prize is located,");
        System.out.println("but first I will show you what is behind one of the other doors");
        System.out.println("\nBehind door number " + open_door+ " are goats!");
        System.out.println("You selected door number " + user_door);
        char userReply;
        do{
            System.out.println("\nWould you like to switch your door(y/n)? ");
            userReply = scan.next().charAt(0);
        }while(userReply!='y' && userReply!='n');       

        int user_duplicate = user_door;
        if (userReply == 'y') {
            do{
                user_door = generator.nextInt(3)+1;
            }while(user_door == open_door || user_door == user_duplicate);
        }

        System.out.println("The prize is behind door number: " + prize_door);

        if (user_door == prize_door) {
            System.out.println("Congratulations! You won the prize!");
        } else {
            System.out.println("Sorry. You lost.");
        }
    }
}

在获得user_door 之前，您应该只分配prize_door，获得user_door 之后，您必须分配open_door，如果您选择换门，还必须分配other_door。

Answer 4

让我们通过示例逐步了解您的代码的作用

import java.util.Random;
import java.util.Scanner;
public class GameShow {

public static void main(String[] args) {
Scanner scan = new Scanner (System.in);
Random generator = new Random();

// Initialize Variables
int user_door,
    open_door,
    other_door,
    prize_door;

// Generate random value 1-3
prize_door = generator.nextInt(3)+1; 

假设生成器将 2 分配给 price_door

open_door = prize_door;

while(open_door == prize_door){
    open_door = generator.nextInt(3)+1;
}

让我们假设在循环之后生成器将 1 分配给 open_door

other_door = open_door;

while (other_door == open_door || other_door == prize_door){
    other_door = generator.nextInt(3)+1;
}

让我们假设在循环之后生成器将 3 分配给 other_door

// Begin Game
System.out.println("*** Welcome to the game show! ***");  
System.out.println("Select the door (1, 2, or 3): ");

让我们回忆一下变量的值 open_door = 1other_door = 3prize_door = 2

user_door = scan.nextInt();
// User Validation
    if (user_door > 3 || user_door < 0) {
        System.out.println("Please select door 1, 2, or 3");
        user_door = scan.nextInt();
    } else if(user_door == 1 || user_door == 2 ||  user_door == 3) {

假设用户输入 1 user_door = 1

//Continue Game
System.out.println("\nIn a moment, I will show you where the prize is located,");
System.out.println("but first I will show you what is behind one of the other doors");

//Continue Dialogue
System.out.println("\nBehind door number " + open_door + " are goats!");

在这里，您正在打开与user_door = 1 相同的门 1 (open_door = 1)

根据游戏逻辑哪个是错误的

System.out.println("You selected door number " + user_door);
System.out.println("\nWould you like to switch your door(y/n)? ");

//User Input Yes or No
char userReply = scan.next().charAt(0);

//If statement with nested while statements for user input
    if (userReply == 'y'){
            user_door = other_door;
            } while(userReply != 'y' && userReply != 'n')
            {
                //User Validation
                System.out.println("Please enter either y/n");
                userReply = scan.next().charAt(0);
            } 

System.out.println("The prize is behind door number: " + prize_door); 

//Check to see if user won or lost
    if(user_door == prize_door){
        System.out.println("Congratulations! You won the prize!");
        } else {
                System.out.println("Sorry. You lost.");
                }       

    }
}
}

您需要根据用户的输入更改门的选择，因为作为主持人，您知道汽车/黄金在哪里以及程序中缺少的山羊在哪里

您应该在游戏开始前修复prize_door，您正在做的正确但open_door和other_door应该根据user_door来决定

Answer 5

很多答案，很多东西被指出是错误的或无效的（但不知何故，没有人会以这种方式随机化 3 个元素，而不是仅仅基于随机化器将它们改组一次），但没有人明确地解决了这个问题：“。 ..除了我似乎无法让 user_door 切换到另一扇门..”

三个门标记为prize_door、open_door 和other_door。 user_door 可以是其中任何一个。这是错误的，但无论如何。

if (userReply == 'y')
    user_door = other_door;

这里使用的语句将user_door更改为other_door（非奖品闭门）的具体值，而不考虑user_door的值是什么。它应该将其从 prize_door 更改为 other_door，反之亦然。

if (userReply == 'y')
{
    if (user_door == other_door)
        user_door = prize_door;
    else
        user_door = other_door;
}

除非user_door == open_door的情况，确实不应该存在。