Monty Hall Problem

monty-hall.js

const doors = [0, 1, 2];
const tries = 100000;

let pickHasCar = 0;
let theOtherHasCar = 0;
for (let i = 0; i < tries; i++) {
  const car = doors[random(3)];
  const pick = doors[random(3)];

  // Open a door that doesn't have the car.
  const open =
    pick === car
      ? doors.filter(d => d !== pick)[random(2)]
      : doors.find(d => d !== pick && d !== car);
  const theOther = doors.find(d => d !== pick && d !== open);

  if (pick === car) {
    pickHasCar++;
  }
  if (theOther === car) {
    theOtherHasCar++;
  }
}
console.log({
  pickHasCar,
  theOtherHasCar,
  ratio: theOtherHasCar / pickHasCar
});

function random(upperBound) {
  return Math.floor(upperBound * Math.random());
}

// When the pick has the car (1/3), the other one doesn't have the car for sure.
// When the pick doesn't have the car (2/3), the other one has the car for sure.
// So, the other one has the car by 2/3 chance.