Monty Hall problem solved (this is stupid)

monty.js

monty = function()
    {
    /*
    Assumptions:
       1. The host will always give you a second choice.
       2. The host will never open the door with the prize
       because that would end the mystery.
    Hypothesis:
       Most people are right and this is math foolishness. This is not
       the prisoner's dilemma. This is liar's paradox which is solved
       by asking the liar what the truth-teller would say and doing
       the opposite. 
    Parameters:
       The monty function takes a number, either 2 or 3 and
       returns  [ 0,1,2 ] or [ 0,1 ]. You
       always choose door 2 and the host always opens door 0
       or 1. Randomly chosing a different door doesn't change
       the outcome over time.
    */

   monty_ = function(x) { return Math.floor(Math.random(1) * x); }

        win = 0;
        lie = 0;
    for( x = 0; x < 1000000; x++) {
       door = monty_(3);
       lie = monty_(2);
       if(door == 2) {
           if(lie)win++;  }  
        else
        if(door == 1) 
           if(lie)win++;
           
    }
        console.log(  win / 10000 );

        /* By running monty_ repeatedly with "3", we obviously get
        a 33% success rate across the million executions. We
        would get the same with a trillion or a zillion 
        zillion. Making the number larger is pointless. Running
        monty_ with "2" a million, billion, or trillion times would also
        give us a 50% success rate. No need to do this. So the 
        only thing we can do is run monty(2) against the 
        results of monty(3) and see if the win rate changes. With
        this version, we still get the same odds. */
}