Question

please help in java

Monty Hall was a television game show host years ago. The contestant would have 3 doors to choose from. There was always a prize behind one door and goats behind the other two doors. The contestant would choose a door.

Monty Hall would then open a door following these rules: Not a door the contestant chose and not the winning door. Monty ALWAYS showed a goat. He'd then ask the contestant if they wanted to change doors. After the contestant made their final choice, Monty Hall would open the chosen door.

Program Requirements:

Your job here is to simulate this. Make it automatic. Randomly choose the winning door. Then randomly choose one of the 3 doors as the contestant's choice. Run this at least 100,000 times to calculate an accurate probability. You will be calculating for the probability of winning if the contestant switches doors when given the opportunity. Also compute the probability of winning if they don't switch.

There has a been a lot of discussion over the years about whether or not the contestant should switch doors. Here is a clip from the movie 21 (Links to an external site.)Links to an external site. that discusses this choice.

Some examples of the potential output from past students (with the answers hidden to protect the innocent).  

Example 1:

There is a xxx.xx% chance that you will win if you switch doors.
There is a yyyy.yy% chance that you will win if you do NOT switch doors.

Example 2:

Without switching:
xxxx / 100000 wins
xx.xxxx% chance to win

With switching:
yyyy / 100000 wins
yy.yyyy% chance to win

Example 3:

The probability of the contestant winning if they don't switch is : 0.xxxx
The probability of the contestant winning if they do switch is : 0.yyyy

again(y/n): y

The probability of the contestant winning if they don't switch is : 0.xxxx <different but similar values than the first run>
The probability of the contestant winning if they do switch is : 0.yyyy

Since we are calculating WINS, the two percentages should add up to 100% or 1

Additional Requirements:

• Do not hard code the answer. It must be calculated and averaged over a number of runs. If <when> I change the number of iterations of the loop, the average should change <minimally>.
• You can start testing with a small number of iterations but the program should ultimately loop 100,00 times. DON'T put any print statements inside the loop or you may kill my laptop...

Answer 1

thanks for the question, here is the MontyHall class that determines the probabilities of staying with the same door or switching the door

All important lines of code are commented so that you can understand whats going on

========================================================================================

import java.util.Random;
public class MontyHall{

    private static Random gen = new Random();

    public static void main(String[] args){
        int switchWins = 0;
        int stayWins = 0;

        for(int plays = 1;plays <= 100000;plays++ ){
            int[] doors = {0,0,0};//0 is a goat, 1 is a car
            doors[gen.nextInt(3)] = 1;//put a car in a random door
            int choice = gen.nextInt(3); //guess a door, any door
            int shown; //the shown door
            do{
                shown = gen.nextInt(3);
                //don't show the winner or the choice
            }while(doors[shown] == 1 || shown == choice);
            stayWins += doors[choice];//if you won by staying, count it
            //the switched (last remaining) door is (3 - choice - shown), because 0+1+2=3
            switchWins += doors[3 - choice - shown];
        }
        System.out.println("Without switching: ");
        System.out.println(stayWins+"/1000000 wins");
        System.out.println(String.format("%.4f",stayWins/10000.00)+"% chance to win");
        System.out.println("With switching: ");
        System.out.println(switchWins+"/1000000 wins");
        System.out.println(String.format("%.4f",switchWins/10000.00)+"% chance to win");


    }
}

=================================================================================

thanks !

let me know incase you have any questions or problem.

Answer 2

import java.io.*;
import java.util.Random;
class Monty
{
   public static void main(String args[])
   {
       int original,my_choice,first_show,user_change;
       long win = 0, change = 0, cases=100000;
       Random rand = new Random();
       for(long i=0;i<cases;i++)
       {
           original = rand.nextInt(3);   // Actual door
           my_choice = rand.nextInt(3);   // Contestent choose
           user_change = rand.nextInt(2);   // If user wants to change answer
           if(user_change == 1)
           {
               if(my_choice != original)   // User changed the answer and the answer is correct
               {
                   win++;
                   change++;
               }
           }
           else if(my_choice == original)       // user didn't changed the answer but the answer is correct
               win++;
       }
       System.out.println("\nWith switching the chances of wining "+((double)change)/((double)cases));
       System.out.println("\nWithout switching the chances of wining "+((double)(win-change))/((double)cases));
       System.out.println("\nChances of wining "+((double)win)/((double)cases));
   }
}

Set the file name as Monty.java

In Linux use javac Monty.java and java Monty.

Please let me know if anything else modification is required. If you like the answer please hit the thumbs up. Best of luck !!!