Monty Hall problem: Difference between revisions

===Extensive Solution===

This solution can test with n doors, the difference in probability of switching is shown to diminish as the number of doors increases.

<lang javascript>

function montyhall(tests, doors) {

tests = tests ? tests : 1000;

doors = doors ? doors : 3;

var prizeDoor, chosenDoor, shownDoor, switchDoor, chosenWins = 0, switchWins = 0;

// randomly pick a door excluding input doors

function pick(excludeA, excludeB) {

var door;

do {

door = Math.floor(Math.random() * doors);

} while (door === excludeA || door === excludeB);

return door;

}

// run tests

for (var i = 0; i < tests; i ++) {

// pick set of doors

prizeDoor = pick();

chosenDoor = pick();

shownDoor = pick(prizeDoor, chosenDoor);

switchDoor = pick(chosenDoor, shownDoor);

// test set for both choices

if (chosenDoor === prizeDoor) {

chosenWins ++;

} else if (switchDoor === prizeDoor) {

switchWins ++;

}

}

// results

return {

stayWins: chosenWins + ' / ' + tests + ' ' + (100 * chosenWins / tests) + '%',

switchWins: switchWins + ' / ' + tests + ' ' + (100 * switchWins / tests) + '%'

}

}

</lang>

Output:

<lang javascript>

montyhall(1000, 3)

Object {stayWins: "349 / 1000 34.9%", switchWins: "651 / 1000 65.1%"}

montyhall(1000, 4)

Object {stayWins: "247 / 1000 24.7%", switchWins: "351 / 1000 35.1%"}

montyhall(1000, 5)

Object {stayWins: "205 / 1000 20.5%", switchWins: "268 / 1000 26.8%"}

</lang>

===Basic Solution===

<lang javascript>

var totalGames = 10000,

console.log("Wins when switching door", play(true));

</lang>

Output:

<lang javascript>

Playing 10000 games

Wins when not switching door 3326

Wins when switching door 6630

</lang>