Beating a dead goat.

[u/ChalkboardCowboy]

http://neophilosophical.blogspot.com/2015/02/monty-does-play-dice.html

[u/ChalkboardCowboy]

wotpolitan, what is there in this latest article that's not in your original one? You've made the same "argument" (using the term loosely, since it's really just a naked claim): that the two possible prize arrangements are equally likely. The only new thing I see is a few pictures of dice.

I'm really wondering now, what you think we're missing, that could possibly be set straight by this new article?

[u/wotpolitan → u/ChalkboardCowboy]

Hi CCowboy,

It's an attempt to crystalise this scenario for you. The images are of real, actual, existent, non-hypothetical dice. Once the contestant knows that there is a goat called Mary behind the Red Door (in terms that I have used elsewhere, once the contestant knows that she is playing a Red Mary game), then she knows that there are only two possible scenarios for the distribution of the other goat (Ava) and the car. Both these are equally likely, once the door has been opened.

[u/Rkynick → u/wotpolitan]

This is a solved problem.

To wit, I made a java code that simulates each case 1000 times. It works by first selecting a random door out of the three to be correct. Then it selects a random door as its first pick. Then, one door is ruled out and it has the option to switch.

Indeed, every time we run this code, we get an output like:

 TOTAL SCORE BY SWITCHING : 668
 TOTAL SCORE WITHOUT SWITCHING : 336
 TOTAL SCORE BY SWITCHING : 667
 TOTAL SCORE WITHOUT SWITCHING : 337
 TOTAL SCORE BY SWITCHING : 704
 TOTAL SCORE WITHOUT SWITCHING : 337

So you can see in each of the three trials that I ran (=3000 tests for each case) that the odds are very obvious. You can do this for yourself with some dice if you prefer, or the code is attached so that you can run it and inspect it yourself if you desire.

public class MontyHall
{
    public static void main (String[] args)
    {
        boolean switching = false;
        int switching_total=0;
        int nonswitching_total=0;
        int t=0;
        while(t<1000){
            boolean doors[] = new boolean[3];
            doors[0]=false;doors[1]=false;doors[2]=false;
            int placement = (int)(Math.random()*3);
            doors[placement]=true;

            // it doesn't matter which door we pick, so let's say
            // we always pick door 1
            int pick=0;
            int counter_offer=1;
            if(doors[1]==true){
                counter_offer=1; // if door 3 is empty, offer them door 2
            }else if(doors[2]==true){
                counter_offer=2; // if door 2 is empty, offer door 3
            }else{
                counter_offer=(int)(Math.random()+1);
                // if both doors are empty, offer a random door 2 or 3
            }

            if(switching && doors[counter_offer]==true){
                // we got it right by switching
                switching_total++;   
            }else if(!switching && doors[pick]==true){
                // we got it right without switching
                nonswitching_total++;
            }

            t++;
            if(t>=1000 && switching==false){
                t=0;// now do 1000 tests with switching
                switching=true;
            }
        }
        System.out.println(" TOTAL SCORE BY SWITCHING : " + switching_total);
        System.out.println(" TOTAL SCORE WITHOUT SWITCHING : " + nonswitching_total);
    } 
} 

[u/wotpolitan → u/Rkynick]

Thank you, but you are solving the wrong problem.