# Monty Hall problem

**Monty Hall problem**

You are encouraged to solve this task according to the task description, using any language you may know.

Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.

- Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice? (Krauss and Wang 2003:10)

Note that the player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.

Simulate at least a thousand games using three doors for each strategy __and show the results__ in such a way as to make it easy to compare the effects of each strategy.

## [edit] ActionScript

package {

import flash.display.Sprite;

public class MontyHall extends Sprite

{

public function MontyHall()

{

var iterations:int = 30000;

var switchWins:int = 0;

var stayWins:int = 0;

for (var i:int = 0; i < iterations; i++)

{

var doors:Array = [0, 0, 0];

doors[Math.floor(Math.random() * 3)] = 1;

var choice:int = Math.floor(Math.random() * 3);

var shown:int;

do

{

shown = Math.floor(Math.random() * 3);

} while (doors[shown] == 1 || shown == choice);

stayWins += doors[choice];

switchWins += doors[3 - choice - shown];

}

trace("Switching wins " + switchWins + " times. (" + (switchWins / iterations) * 100 + "%)");

trace("Staying wins " + stayWins + " times. (" + (stayWins / iterations) * 100 + "%)");

}

}

}

Output:

Switching wins 18788 times. (62.626666666666665%) Staying wins 11212 times. (37.37333333333333%)

## [edit] Ada

-- Monty Hall Game

with Ada.Text_Io; use Ada.Text_Io;

with Ada.Float_Text_Io; use Ada.Float_Text_Io;

with ada.Numerics.Discrete_Random;

procedure Monty_Stats is

Num_Iterations : Positive := 100000;

type Action_Type is (Stay, Switch);

type Prize_Type is (Goat, Pig, Car);

type Door_Index is range 1..3;

package Random_Prize is new Ada.Numerics.Discrete_Random(Door_Index);

use Random_Prize;

Seed : Generator;

Doors : array(Door_Index) of Prize_Type;

procedure Set_Prizes is

Prize_Index : Door_Index;

Booby_Prize : Prize_Type := Goat;

begin

Reset(Seed);

Prize_Index := Random(Seed);

Doors(Prize_Index) := Car;

for I in Doors'range loop

if I /= Prize_Index then

Doors(I) := Booby_Prize;

Booby_Prize := Prize_Type'Succ(Booby_Prize);

end if;

end loop;

end Set_Prizes;

function Play(Action : Action_Type) return Prize_Type is

Chosen : Door_Index := Random(Seed);

Monty : Door_Index;

begin

Set_Prizes;

for I in Doors'range loop

if I /= Chosen and Doors(I) /= Car then

Monty := I;

end if;

end loop;

if Action = Switch then

for I in Doors'range loop

if I /= Monty and I /= Chosen then

Chosen := I;

exit;

end if;

end loop;

end if;

return Doors(Chosen);

end Play;

Winners : Natural;

Pct : Float;

begin

Winners := 0;

for I in 1..Num_Iterations loop

if Play(Stay) = Car then

Winners := Winners + 1;

end if;

end loop;

Put("Stay : count" & Natural'Image(Winners) & " = ");

Pct := Float(Winners * 100) / Float(Num_Iterations);

Put(Item => Pct, Aft => 2, Exp => 0);

Put_Line("%");

Winners := 0;

for I in 1..Num_Iterations loop

if Play(Switch) = Car then

Winners := Winners + 1;

end if;

end loop;

Put("Switch : count" & Natural'Image(Winners) & " = ");

Pct := Float(Winners * 100) / Float(Num_Iterations);

Put(Item => Pct, Aft => 2, Exp => 0);

Put_Line("%");

end Monty_Stats;

Results

Stay : count 34308 = 34.31% Switch : count 65695 = 65.69%

## [edit] ALGOL 68

INT trials=100 000;

PROC brand = (INT n)INT: 1 + ENTIER (n * random);

PROC percent = (REAL x)STRING: fixed(100.0*x/trials,0,2)+"%";

main:

(

INT prize, choice, show, not shown, new choice;

INT stay winning:=0, change winning:=0, random winning:=0;

INT doors = 3;

[doors-1]INT other door;

TO trials DO

# put the prize somewhere #

prize := brand(doors);

# let the user choose a door #

choice := brand(doors);

# let us take a list of unchoosen doors #

INT k := LWB other door;

FOR j TO doors DO

IF j/=choice THEN other door[k] := j; k+:=1 FI

OD;

# Monty opens one... #

IF choice = prize THEN

# staying the user will win... Monty opens a random port#

show := other door[ brand(doors - 1) ];

not shown := other door[ (show+1) MOD (doors - 1 ) + 1]

ELSE # no random, Monty can open just one door... #

IF other door[1] = prize THEN

show := other door[2];

not shown := other door[1]

ELSE

show := other door[1];

not shown := other door[2]

FI

FI;

# the user randomly choose one of the two closed doors

(one is his/her previous choice, the second is the

one not shown ) #

other door[1] := choice;

other door[2] := not shown;

new choice := other door[ brand(doors - 1) ];

# now let us count if it takes it or not #

IF choice = prize THEN stay winning+:=1 FI;

IF not shown = prize THEN change winning+:=1 FI;

IF new choice = prize THEN random winning+:=1 FI

OD;

print(("Staying: ", percent(stay winning), new line ));

print(("Changing: ", percent(change winning), new line ));

print(("New random choice: ", percent(random winning), new line ))

)

Sample output:

Staying: 33.62% Changing: 66.38% New random choice: 50.17%

## [edit] APL

∇ Run runs;doors;i;chosen;cars;goats;swap;stay;ix;prices

[1] ⍝0: Monthy Hall problem

[2] ⍝1: http://rosettacode.org/wiki/Monty_Hall_problem

[3]

[4] (⎕IO ⎕ML)←0 1

[5] prices←0 0 1 ⍝ 0=Goat, 1=Car

[6]

[7] ix←⊃,/{3?3}¨⍳runs ⍝ random indexes of doors (placement of car)

[8] doors←(runs 3)⍴prices[ix] ⍝ matrix of doors

[9] stay←+⌿doors[;?3] ⍝ chose randomly one door - is it a car?

[10] swap←runs-stay ⍝ If not, then the other one is!

[11]

[12] ⎕←'Swap: ',(2⍕100×(swap÷runs)),'% it''s a car'

[13] ⎕←'Stay: ',(2⍕100×(stay÷runs)),'% it''s a car'

∇

Run 100000 Swap: 66.54% it's a car Stay: 33.46% it's a car

## [edit] AutoHotkey

#NoTrayIcon

#SingleInstance, OFF

#Persistent

SetBatchLines, -1

Iterations = 1000

Loop, %Iterations%

{

If Monty_Hall(1)

Correct_Change++

Else

Incorrect_Change++

If Monty_Hall(2)

Correct_Random++

Else

Incorrect_Random++

If Monty_Hall(3)

Correct_Stay++

Else

Incorrect_Stay++

}

Percent_Change := floor(Correct_Change / Iterations * 100)

Percent_Random := floor(Correct_Random / Iterations * 100)

Percent_Stay := floor(Correct_Stay / Iterations * 100)

MsgBox,, Monty Hall Problem, These are the results:`r`n`r`nWhen I changed my guess, I got %Correct_Change% of %Iterations% (that's %Incorrect_Change% incorrect). Thats %Percent_Change%`% correct.`r`nWhen I randomly changed my guess, I got %Correct_Random% of %Iterations% (that's %Incorrect_Random% incorrect). Thats %Percent_Random%`% correct.`r`nWhen I stayed with my first guess, I got %Correct_Stay% of %Iterations% (that's %Incorrect_Stay% incorrect). Thats %Percent_Stay%`% correct.

ExitApp

Monty_Hall(Mode) ;Mode is 1 for change, 2 for random, or 3 for stay

{

Random, prize, 1, 3

Random, guess, 1, 3

If (prize = guess && Mode != 3)

While show != 0 && show != guess

Random, show, 1, 3

Else

show := 6 - prize - guess

Random, change_guess, 0, 1

If (Mode = 1 || (change_guess && Mode = 2))

Return, (6 - show - guess) = prize

Else If (Mode = 3 || (!change_guess && Mode = 2))

Return, guess = prize

Else

Return

}

Sample output:

These are the results: When I changed my guess, I got 762 of 1000 (that's 238 incorrect). Thats 76% correct. When I randomly changed my guess, I got 572 of 1000 (that's 428 incorrect). Thats 57% correct. When I stayed with my first guess, I got 329 of 1000 (that's 671 incorrect). Thats 32% correct.

## [edit] AWK

#!/bin/gawk -f

# Monty Hall problem

BEGIN {

srand()

doors = 3

iterations = 10000

# Behind a door:

EMPTY = "empty"; PRIZE = "prize"

# Algorithm used

KEEP = "keep"; SWITCH="switch"; RAND="random";

#

}

function monty_hall( choice, algorithm ) {

# Set up doors

for ( i=0; i<doors; i++ ) {

door[i] = EMPTY

}

# One door with prize

door[int(rand()*doors)] = PRIZE

chosen = door[choice]

del door[choice]

#if you didn't choose the prize first time around then

# that will be the alternative

alternative = (chosen == PRIZE) ? EMPTY : PRIZE

if( algorithm == KEEP) {

return chosen

}

if( algorithm == SWITCH) {

return alternative

}

return rand() <0.5 ? chosen : alternative

}

function simulate(algo){

prizecount = 0

for(j=0; j< iterations; j++){

if( monty_hall( int(rand()*doors), algo) == PRIZE) {

prizecount ++

}

}

printf " Algorithm %7s: prize count = %i, = %6.2f%%\n", \

algo, prizecount,prizecount*100/iterations

}

BEGIN {

print "\nMonty Hall problem simulation:"

print doors, "doors,", iterations, "iterations.\n"

simulate(KEEP)

simulate(SWITCH)

simulate(RAND)

}

Sample output:

bash$ ./monty_hall.awk

Monty Hall problem simulation:

3 doors, 10000 iterations.

Algorithm keep: prize count = 3411, = 34.11%

Algorithm switch: prize count = 6655, = 66.55%

Algorithm random: prize count = 4991, = 49.91%

bash$

## [edit] BASIC

RANDOMIZE TIMER

DIM doors(3) '0 is a goat, 1 is a car

CLS

switchWins = 0

stayWins = 0

FOR plays = 0 TO 32767

winner = INT(RND * 3) + 1

doors(winner) = 1'put a winner in a random door

choice = INT(RND * 3) + 1'pick a door, any door

DO

shown = INT(RND * 3) + 1

'don't show the winner or the choice

LOOP WHILE doors(shown) = 1 OR shown = choice

stayWins = stayWins + doors(choice) 'if you won by staying, count it

switchWins = switchWins + doors(3 - choice - shown) 'could have switched to win

doors(winner) = 0 'clear the doors for the next test

NEXT plays

PRINT "Switching wins"; switchWins; "times."

PRINT "Staying wins"; stayWins; "times."

Output:

Switching wins 21805 times. Staying wins 10963 times.

## [edit] BBC BASIC

total% = 10000

FOR trial% = 1 TO total%

prize_door% = RND(3) : REM. The prize is behind this door

guess_door% = RND(3) : REM. The contestant guesses this door

IF prize_door% = guess_door% THEN

REM. The contestant guessed right, reveal either of the others

reveal_door% = RND(2)

IF prize_door% = 1 reveal_door% += 1

IF prize_door% = 2 AND reveal_door% = 2 reveal_door% = 3

ELSE

REM. The contestant guessed wrong, so reveal the non-prize door

reveal_door% = prize_door% EOR guess_door%

ENDIF

stick_door% = guess_door% : REM. The sticker doesn't change his mind

swap_door% = guess_door% EOR reveal_door% : REM. but the swapper does

IF stick_door% = prize_door% sticker% += 1

IF swap_door% = prize_door% swapper% += 1

NEXT trial%

PRINT "After a total of ";total%;" trials,"

PRINT "The 'sticker' won ";sticker%;" times (";INT(sticker%/total%*100);"%)"

PRINT "The 'swapper' won ";swapper%;" times (";INT(swapper%/total%*100);"%)"

Output:

After a total of 10000 trials, The 'sticker' won 3379 times (33%) The 'swapper' won 6621 times (66%)

## [edit] C

//Evidence of the Monty Hall solution.

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

#define GAMES 3000000

int main(void){

unsigned i, j, k, choice, winsbyswitch=0, door[3];

srand(time(NULL)); //initialize random seed.

for(i=0; i<GAMES; i++){

door[0] = (!(rand()%2)) ? 1: 0; //give door 1 either a car or a goat randomly.

if(door[0]) door[1]=door[2]=0; //if 1st door has car, give other doors goats.

else{ door[1] = (!(rand()%2)) ? 1: 0; door[2] = (!door[1]) ? 1: 0; } //else, give 2nd door car or goat, give 3rd door what's left.

choice = rand()%3; //choose a random door.

//if the next door has a goat, and the following door has a car, or vice versa, you'd win if you switch.

if(((!(door[((choice+1)%3)])) && (door[((choice+2)%3)])) || (!(door[((choice+2)%3)]) && (door[((choice+1)%3)]))) winsbyswitch++;

}

printf("\nAfter %u games, I won %u by switching. That is %f%%. ", GAMES, winsbyswitch, (float)winsbyswitch*100.0/(float)i);

}

Output of one run:

After 3000000 games, I won 1999747 by switching. That is 66.658233%.

## [edit] C#

using System;

class Program

{

static void Main(string[] args)

{

int switchWins = 0;

int stayWins = 0;

Random gen = new Random();

for(int plays = 0; plays < 1000000; plays++ )

{

int[] doors = {0,0,0};//0 is a goat, 1 is a car

var winner = gen.Next(3);

doors[winner] = 1; //put a winner in a random door

int choice = gen.Next(3); //pick a door, any door

int shown; //the shown door

do

{

shown = gen.Next(3);

}

while (doors[shown] == 1 || shown == choice); //don't show the winner or the 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];

}

Console.Out.WriteLine("Staying wins " + stayWins + " times.");

Console.Out.WriteLine("Switching wins " + switchWins + " times.");

}

}

Sample output:

Staying wins: 333830 Switching wins: 666170

## [edit] C++

#include <iostream>

#include <cstdlib>

#include <ctime>

int randint(int n)

{

return (1.0*n*std::rand())/(1.0+RAND_MAX);

}

int other(int doorA, int doorB)

{

int doorC;

if (doorA == doorB)

{

doorC = randint(2);

if (doorC >= doorA)

++doorC;

}

else

{

for (doorC = 0; doorC == doorA || doorC == doorB; ++doorC)

{

// empty

}

}

return doorC;

}

int check(int games, bool change)

{

int win_count = 0;

for (int game = 0; game < games; ++game)

{

int const winning_door = randint(3);

int const original_choice = randint(3);

int open_door = other(original_choice, winning_door);

int const selected_door = change?

other(open_door, original_choice)

: original_choice;

if (selected_door == winning_door)

++win_count;

}

return win_count;

}

int main()

{

std::srand(std::time(0));

int games = 10000;

int wins_stay = check(games, false);

int wins_change = check(games, true);

std::cout << "staying: " << 100.0*wins_stay/games << "%, changing: " << 100.0*wins_change/games << "%\n";

}

Sample output:

staying: 33.73%, changing: 66.9%

## [edit] Clojure

(ns monty-hall-problem

(:use [clojure.contrib.seq :only (shuffle)]))

(defn play-game [staying]

(let [doors (shuffle [:goat :goat :car])

choice (rand-int 3)

[a b] (filter #(not= choice %) (range 3))

alternative (if (= :goat (nth doors a)) b a)]

(= :car (nth doors (if staying choice alternative)))))

(defn simulate [staying times]

(let [wins (reduce (fn [counter _] (if (play-game staying) (inc counter) counter))

0

(range times))]

(str "wins " wins " times out of " times)))

monty-hall-problem> (println "staying:" (simulate true 1000))

staying: wins 337 times out of 1000

nil

monty-hall-problem> (println "switching:" (simulate false 1000))

switching: wins 638 times out of 1000

nil

## [edit] COBOL

IDENTIFICATION DIVISION.

PROGRAM-ID. monty-hall.

DATA DIVISION.

WORKING-STORAGE SECTION.

78 Num-Games VALUE 1000000.

*> These are needed so the values are passed to

*> get-rand-int correctly.

01 One PIC 9 VALUE 1.

01 Three PIC 9 VALUE 3.

01 doors-area.

03 doors PIC 9 OCCURS 3 TIMES.

01 choice PIC 9.

01 shown PIC 9.

01 winner PIC 9.

01 switch-wins PIC 9(7).

01 stay-wins PIC 9(7).

01 stay-wins-percent PIC Z9.99.

01 switch-wins-percent PIC Z9.99.

PROCEDURE DIVISION.

PERFORM Num-Games TIMES

MOVE 0 TO doors (winner)

CALL "get-rand-int" USING CONTENT One, Three,

REFERENCE winner

MOVE 1 TO doors (winner)

CALL "get-rand-int" USING CONTENT One, Three,

REFERENCE choice

PERFORM WITH TEST AFTER

UNTIL NOT(shown = winner OR choice)

CALL "get-rand-int" USING CONTENT One, Three,

REFERENCE shown

END-PERFORM

ADD doors (choice) TO stay-wins

ADD doors (6 - choice - shown) TO switch-wins

END-PERFORM

COMPUTE stay-wins-percent ROUNDED =

stay-wins / Num-Games * 100

COMPUTE switch-wins-percent ROUNDED =

switch-wins / Num-Games * 100

DISPLAY "Staying wins " stay-wins " times ("

stay-wins-percent "%)."

DISPLAY "Switching wins " switch-wins " times ("

switch-wins-percent "%)."

.

IDENTIFICATION DIVISION.

PROGRAM-ID. get-rand-int.

DATA DIVISION.

WORKING-STORAGE SECTION.

01 call-flag PIC X VALUE "Y".

88 first-call VALUE "Y", FALSE "N".

01 num-range PIC 9.

LINKAGE SECTION.

01 min-num PIC 9.

01 max-num PIC 9.

01 ret PIC 9.

PROCEDURE DIVISION USING min-num, max-num, ret.

*> Seed RANDOM once.

IF first-call

MOVE FUNCTION RANDOM(FUNCTION CURRENT-DATE (9:8))

TO num-range

SET first-call TO FALSE

END-IF

COMPUTE num-range = max-num - min-num + 1

COMPUTE ret =

FUNCTION MOD(FUNCTION RANDOM * 100000, num-range)

+ min-num

.

END PROGRAM get-rand-int.

END PROGRAM monty-hall.

- Output:

Staying wins 0333396 times (33.34%). Switching wins 0666604 times (66.66%).

## [edit] ColdFusion

<cfscript>

function runmontyhall(num_tests) {

// number of wins when player switches after original selection

switch_wins = 0;

// number of wins when players "sticks" with original selection

stick_wins = 0;

// run all the tests

for(i=1;i<=num_tests;i++) {

// unconditioned potential for selection of each door

doors = [0,0,0];

// winning door is randomly assigned...

winner = randrange(1,3);

// ...and actualized in the array of real doors

doors[winner] = 1;

// player chooses one of three doors

choice = randrange(1,3);

do {

// monty randomly reveals a door...

shown = randrange(1,3);

}

// ...but monty only reveals empty doors;

// he will not reveal the door that the player has choosen

// nor will he reveal the winning door

while(shown==choice || doors[shown]==1);

// when the door the player originally selected is the winner, the "stick" option gains a point

stick_wins += doors[choice];

// to calculate the number of times the player would have won with a "switch", subtract the

// "value" of the chosen, "stuck-to" door from 1, the possible number of wins if the player

// chose and stuck with the winning door (1), the player would not have won by switching, so

// the value is 1-1=0 if the player chose and stuck with a losing door (0), the player would

// have won by switching, so the value is 1-0=1

switch_wins += 1-doors[choice];

}

// finally, simply run the percentages for each outcome

stick_percentage = (stick_wins/num_tests)*100;

switch_percentage = (switch_wins/num_tests)*100;

writeoutput('Number of Tests: ' & num_tests);

writeoutput('<br />Stick Wins: ' & stick_wins & ' ['& stick_percentage &'%]');

writeoutput('<br />Switch Wins: ' & switch_wins & ' ['& switch_percentage &'%]');

}

runmontyhall(10000);

</cfscript>

Output:

Tests: 10,000 | Switching wins: 6655 [66.55%] | Sticking wins: 3345 [33.45%]

## [edit] Common Lisp

(defun make-round ()

(let ((array (make-array 3

:element-type 'bit

:initial-element 0)))

(setf (bit array (random 3)) 1)

array))

(defun show-goat (initial-choice array)

(loop for i = (random 3)

when (and (/= initial-choice i)

(zerop (bit array i)))

return i))

(defun won? (array i)

(= 1 (bit array i)))

CL-USER> (progn (loop repeat #1=(expt 10 6)

for round = (make-round)

for initial = (random 3)

for goat = (show-goat initial round)

for choice = (loop for i = (random 3)

when (and (/= i initial)

(/= i goat))

return i)

when (won? round (random 3))

sum 1 into result-stay

when (won? round choice)

sum 1 into result-switch

finally (progn (format t "Stay: ~S%~%" (float (/ result-stay

#1# 1/100)))

(format t "Switch: ~S%~%" (float (/ result-switch

#1# 1/100))))))

Stay: 33.2716%

Switch: 66.6593%

;Find out how often we win if we always switch

(defun rand-elt (s)

(elt s (random (length s))))

(defun monty ()

(let* ((doors '(0 1 2))

(prize (random 3));possible values: 0, 1, 2

(pick (random 3))

(opened (rand-elt (remove pick (remove prize doors))));monty opens a door which is not your pick and not the prize

(other (car (remove pick (remove opened doors))))) ;you decide to switch to the one other door that is not your pick and not opened

(= prize other))) ; did you switch to the prize?

(defun monty-trials (n)

(count t (loop for x from 1 to n collect (monty))))

## [edit] D

import std.stdio, std.random;

void main() {

int switchWins, stayWins;

while (switchWins + stayWins < 100_000) {

immutable carPos = uniform(0, 3); // Which door is car behind?

immutable pickPos = uniform(0, 3); // Contestant's initial pick.

int openPos; // Which door is opened by Monty Hall?

// Monty can't open the door you picked or the one with the car

// behind it.

do {

openPos = uniform(0, 3);

} while(openPos == pickPos || openPos == carPos);

int switchPos;

// Find position that's not currently picked by contestant and

// was not opened by Monty already.

for (; pickPos==switchPos || openPos==switchPos; switchPos++) {}

if (pickPos == carPos)

stayWins++;

else if (switchPos == carPos)

switchWins++;

else

assert(0); // Can't happen.

}

writefln("Switching/Staying wins: %d %d", switchWins, stayWins);

}

- Output:

Switching/Staying wins: 66609 33391

## [edit] Dart

The class Game attempts to hide the implementation as much as possible, the play() function does not use any specifics of the implementation.

int rand(int max) => (Math.random()*max).toInt();

class Game {

int _prize;

int _open;

int _chosen;

Game() {

_prize=rand(3);

_open=null;

_chosen=null;

}

void choose(int door) {

_chosen=door;

}

void reveal() {

if(_prize==_chosen) {

int toopen=rand(2);

if (toopen>=_prize)

toopen++;

_open=toopen;

} else {

for(int i=0;i<3;i++)

if(_prize!=i && _chosen!=i) {

_open=i;

break;

}

}

}

void change() {

for(int i=0;i<3;i++)

if(_chosen!=i && _open!=i) {

_chosen=i;

break;

}

}

bool hasWon() => _prize==_chosen;

String toString() {

String res="Prize is behind door $_prize";

if(_chosen!=null) res+=", player has chosen door $_chosen";

if(_open!=null) res+=", door $_open is open";

return res;

}

}

void play(int count, bool swap) {

int wins=0;

for(int i=0;i<count;i++) {

Game game=new Game();

game.choose(rand(3));

game.reveal();

if(swap)

game.change();

if(game.hasWon())

wins++;

}

String withWithout=swap?"with":"without";

double percent=(wins*100.0)/count;

print("playing $withWithout switching won $percent%");

}

test() {

for(int i=0;i<5;i++) {

Game g=new Game();

g.choose(i%3);

g.reveal();

print(g);

g.change();

print(g);

print("win==${g.hasWon()}");

}

}

main() {

play(10000,false);

play(10000,true);

}

playing without switching won 33.32% playing with switching won 67.63%

## [edit] Eiffel

note

description: "[

Monty Hall Problem as an Eiffel Solution

1. Set the stage: Randomly place car and two goats behind doors 1, 2 and 3.

2. Monty offers choice of doors --> Contestant will choose a random door or always one door.

2a. Door has Goat - door remains closed

2b. Door has Car - door remains closed

3. Monty offers cash --> Contestant takes or refuses cash.

3a. Takes cash: Contestant is Cash winner and door is revealed. Car Loser if car door revealed.

3b. Refuses cash: Leads to offer to switch doors.

4. Monty offers door switch --> Contestant chooses to stay or change.

5. Door reveal: Contestant refused cash and did or did not door switch. Either way: Reveal!

6. Winner and Loser based on door reveal of prize.

Car Winner: Chooses car door

Cash Winner: Chooses cash over any door

Goat Loser: Chooses goat door

Car Loser: Chooses cash over car door or switches from car door to goat door

]"

date: "$Date$"

revision: "$Revision$"

class

MH_APPLICATION

create

make

feature {NONE} -- Initialization

make

-- Initialize Current.

do

play_lets_make_a_deal

ensure

played_1000_games: game_count = times_to_play

end

feature {NONE} -- Implementation: Access

live_contestant: attached like contestant

-- Attached version of `contestant'

do

if attached contestant as al_contestant then

Result := al_contestant

else

create Result

check not_attached_contestant: False end

end

end

contestant: detachable TUPLE [first_door_choice, second_door_choice: like door_number_anchor; takes_cash, switches_door: BOOLEAN]

-- Contestant for Current.

active_stage_door (a_door: like door_anchor): attached like door_anchor

-- Attached version of `a_door'.

do

if attached a_door as al_door then

Result := al_door

else

create Result

check not_attached_door: False end

end

end

door_1, door_2, door_3: like door_anchor

-- Doors with prize names and flags for goat and open (revealed).

feature {NONE} -- Implementation: Status

game_count, car_win_count, cash_win_count, car_loss_count, goat_loss_count, goat_avoidance_count: like counter_anchor

switch_count, switch_win_count: like counter_anchor

no_switch_count, no_switch_win_count: like counter_anchor

-- Counts of games played, wins and losses based on car, cash or goat.

feature {NONE} -- Implementation: Basic Operations

prepare_stage

-- Prepare the stage in terms of what doors have what prizes.

do

inspect new_random_of (3)

when 1 then

door_1 := door_with_car

door_2 := door_with_goat

door_3 := door_with_goat

when 2 then

door_1 := door_with_goat

door_2 := door_with_car

door_3 := door_with_goat

when 3 then

door_1 := door_with_goat

door_2 := door_with_goat

door_3 := door_with_car

end

active_stage_door (door_1).number := 1

active_stage_door (door_2).number := 2

active_stage_door (door_3).number := 3

ensure

door_has_prize: not active_stage_door (door_1).is_goat or

not active_stage_door (door_2).is_goat or

not active_stage_door (door_3).is_goat

consistent_door_numbers: active_stage_door (door_1).number = 1 and

active_stage_door (door_2).number = 2 and

active_stage_door (door_3).number = 3

end

door_number_having_prize: like door_number_anchor

-- What door number has the car?

do

if not active_stage_door (door_1).is_goat then

Result := 1

elseif not active_stage_door (door_2).is_goat then

Result := 2

elseif not active_stage_door (door_3).is_goat then

Result := 3

else

check prize_not_set: False end

end

ensure

one_to_three: between_1_and_x_inclusive (3, Result)

end

door_with_car: attached like door_anchor

-- Create a door with a car.

do

create Result

Result.name := prize

ensure

not_empty: not Result.name.is_empty

name_is_prize: Result.name.same_string (prize)

end

door_with_goat: attached like door_anchor

-- Create a door with a goat

do

create Result

Result.name := gag_gift

Result.is_goat := True

ensure

not_empty: not Result.name.is_empty

name_is_prize: Result.name.same_string (gag_gift)

is_gag_gift: Result.is_goat

end

next_contestant: attached like live_contestant

-- The next contestant on Let's Make a Deal!

do

create Result

Result.first_door_choice := new_random_of (3)

Result.second_door_choice := choose_another_door (Result.first_door_choice)

Result.takes_cash := random_true_or_false

if not Result.takes_cash then

Result.switches_door := random_true_or_false

end

ensure

choices_one_to_three: Result.first_door_choice <= 3 and Result.second_door_choice <= 3

switch_door_implies_no_cash_taken: Result.switches_door implies not Result.takes_cash

end

choose_another_door (a_first_choice: like door_number_anchor): like door_number_anchor

-- Make a choice from the remaining doors

require

one_to_three: between_1_and_x_inclusive (3, a_first_choice)

do

Result := new_random_of (3)

from until Result /= a_first_choice

loop

Result := new_random_of (3)

end

ensure

first_choice_not_second: a_first_choice /= Result

result_one_to_three: between_1_and_x_inclusive (3, Result)

end

play_lets_make_a_deal

-- Play the game 1000 times

local

l_car_win, l_car_loss, l_cash_win, l_goat_loss, l_goat_avoided: BOOLEAN

do

from

game_count := 0

invariant

consistent_win_loss_counts: (game_count = (car_win_count + cash_win_count + goat_loss_count))

consistent_loss_avoidance_counts: (game_count = (car_loss_count + goat_avoidance_count))

until

game_count >= times_to_play

loop

prepare_stage

contestant := next_contestant

l_cash_win := (live_contestant.takes_cash)

l_car_win := (not l_cash_win and

(not live_contestant.switches_door and live_contestant.first_door_choice = door_number_having_prize) or

(live_contestant.switches_door and live_contestant.second_door_choice = door_number_having_prize))

l_car_loss := (not live_contestant.switches_door and live_contestant.first_door_choice /= door_number_having_prize) or

(live_contestant.switches_door and live_contestant.second_door_choice /= door_number_having_prize)

l_goat_loss := (not l_car_win and not l_cash_win)

l_goat_avoided := (not live_contestant.switches_door and live_contestant.first_door_choice = door_number_having_prize) or

(live_contestant.switches_door and live_contestant.second_door_choice = door_number_having_prize)

check consistent_goats: l_goat_loss implies not l_goat_avoided end

check consistent_car_win: l_car_win implies not l_car_loss and not l_cash_win and not l_goat_loss end

check consistent_cash_win: l_cash_win implies not l_car_win and not l_goat_loss end

check consistent_goat_avoidance: l_goat_avoided implies (l_car_win or l_cash_win) and not l_goat_loss end

check consistent_car_loss: l_car_loss implies l_cash_win or l_goat_loss end

if l_car_win then car_win_count := car_win_count + 1 end

if l_cash_win then cash_win_count := cash_win_count + 1 end

if l_goat_loss then goat_loss_count := goat_loss_count + 1 end

if l_car_loss then car_loss_count := car_loss_count + 1 end

if l_goat_avoided then goat_avoidance_count := goat_avoidance_count + 1 end

if live_contestant.switches_door then

switch_count := switch_count + 1

if l_car_win then

switch_win_count := switch_win_count + 1

end

else -- if not live_contestant.takes_cash and not live_contestant.switches_door then

no_switch_count := no_switch_count + 1

if l_car_win or l_cash_win then

no_switch_win_count := no_switch_win_count + 1

end

end

game_count := game_count + 1

end

print ("%NCar Wins:%T%T " + car_win_count.out +

"%NCash Wins:%T%T " + cash_win_count.out +

"%NGoat Losses:%T%T " + goat_loss_count.out +

"%N-----------------------------" +

"%NTotal Win/Loss:%T%T" + (car_win_count + cash_win_count + goat_loss_count).out +

"%N%N" +

"%NCar Losses:%T%T " + car_loss_count.out +

"%NGoats Avoided:%T%T " + goat_avoidance_count.out +

"%N-----------------------------" +

"%NTotal Loss/Avoid:%T" + (car_loss_count + goat_avoidance_count).out +

"%N-----------------------------" +

"%NStaying Count/Win:%T" + no_switch_count.out + "/" + no_switch_win_count.out + " = " + (no_switch_win_count / no_switch_count * 100).out + " %%" +

"%NSwitch Count/Win:%T" + switch_count.out + "/" + switch_win_count.out + " = " + (switch_win_count / switch_count * 100).out + " %%"

)

end

feature {NONE} -- Implementation: Random Numbers

last_random: like random_number_anchor

-- The last random number chosen.

random_true_or_false: BOOLEAN

-- A randome True or False

do

Result := new_random_of (2) = 2

end

new_random_of (a_number: like random_number_anchor): like door_number_anchor

-- A random number from 1 to `a_number'.

do

Result := (new_random \\ a_number + 1).as_natural_8

end

new_random: like random_number_anchor

-- Random integer

-- Each call returns another random number.

do

random_sequence.forth

Result := random_sequence.item

last_random := Result

ensure

old_random_not_new: old last_random /= last_random

end

random_sequence: RANDOM

-- Random sequence seeded from clock when called.

attribute

create Result.set_seed ((create {TIME}.make_now).milli_second)

end

feature {NONE} -- Implementation: Constants

times_to_play: NATURAL_16 = 1000

-- Times to play the game.

prize: STRING = "Car"

-- Name of the prize

gag_gift: STRING = "Goat"

-- Name of the gag gift

door_anchor: detachable TUPLE [number: like door_number_anchor; name: STRING; is_goat, is_open: BOOLEAN]

-- Type anchor for door tuples.

door_number_anchor: NATURAL_8

-- Type anchor for door numbers.

random_number_anchor: INTEGER

-- Type anchor for random numbers.

counter_anchor: NATURAL_16

-- Type anchor for counters.

feature {NONE} -- Implementation: Contract Support

between_1_and_x_inclusive (a_number, a_value: like door_number_anchor): BOOLEAN

-- Is `a_value' between 1 and `a_number'?

do

Result := (a_value > 0) and (a_value <= a_number)

end

end

- Output:

Car Wins: 177 Cash Wins: 486 Goat Losses: 337 ----------------------------- Total Win/Loss: 1000 Car Losses: 657 Goats Avoided: 343 ----------------------------- Total Loss/Avoid: 1000 ----------------------------- Staying Count/Win: 742/573 = 77.223719676549862 % Switch Count/Win: 258/90 = 34.883720930232556 %

## [edit] Emacs Lisp

(defun montyhall (keep)

(let

((prize (random 3))

(choice (random 3)))

(if keep (= prize choice)

(/= prize choice))))

(let ((cnt 0))

(dotimes (i 10000)

(and (montyhall t) (setq cnt (1+ cnt))))

(princ (format "Strategy keep: %.3f %%" (/ cnt 100.0))))

(let ((cnt 0))

(dotimes (i 10000)

(and (montyhall nil) (setq cnt (1+ cnt))))

(princ (format "Strategy switch: %.3f %%" (/ cnt 100.0))))

- Output:

Strategy keep: 34.410 % Strategy switch: 66.430 %

## [edit] Erlang

-module(monty_hall).

-export([main/0]).

main() ->

random:seed(now()),

{WinStay, WinSwitch} = experiment(100000, 0, 0),

io:format("Switching wins ~p times.\n", [WinSwitch]),

io:format("Staying wins ~p times.\n", [WinStay]).

experiment(0, WinStay, WinSwitch) ->

{WinStay, WinSwitch};

experiment(N, WinStay, WinSwitch) ->

Doors = setelement(random:uniform(3), {0,0,0}, 1),

SelectedDoor = random:uniform(3),

OpenDoor = open_door(Doors, SelectedDoor),

experiment(

N - 1,

WinStay + element(SelectedDoor, Doors),

WinSwitch + element(6 - (SelectedDoor + OpenDoor), Doors) ).

open_door(Doors,SelectedDoor) ->

OpenDoor = random:uniform(3),

case (element(OpenDoor, Doors) =:= 1) or (OpenDoor =:= SelectedDoor) of

true -> open_door(Doors, SelectedDoor);

false -> OpenDoor

end.

Sample Output:

Switching wins 66595 times. Staying wins 33405 times.

## [edit] Euphoria

integer switchWins, stayWins

switchWins = 0

stayWins = 0

integer winner, choice, shown

for plays = 1 to 10000 do

winner = rand(3)

choice = rand(3)

while 1 do

shown = rand(3)

if shown != winner and shown != choice then

exit

end if

end while

stayWins += choice = winner

switchWins += 6-choice-shown = winner

end for

printf(1, "Switching wins %d times\n", switchWins)

printf(1, "Staying wins %d times\n", stayWins)

Sample Output:

- Switching wins 6697 times

- Staying wins 3303 times

## [edit] F#

I don't bother with having Monty "pick" a door, since you only win if you initially pick a loser in the switch strategy and you only win if you initially pick a winner in the stay strategy so there doesn't seem to be much sense in playing around the background having Monty "pick" doors. Makes it pretty simple to see why it's always good to switch.

open System

let monty nSims =

let rnd = new Random()

let SwitchGame() =

let winner, pick = rnd.Next(0,3), rnd.Next(0,3)

if winner <> pick then 1 else 0

let StayGame() =

let winner, pick = rnd.Next(0,3), rnd.Next(0,3)

if winner = pick then 1 else 0

let Wins (f:unit -> int) = seq {for i in [1..nSims] -> f()} |> Seq.sum

printfn "Stay: %d wins out of %d - Switch: %d wins out of %d" (Wins StayGame) nSims (Wins SwitchGame) nSims

Sample Output:

Stay: 332874 wins out of 1000000 - Switch: 667369 wins out of 1000000

I had a very polite suggestion that I simulate Monty's "pick" so I'm putting in a version that does that. I compare the outcome with my original outcome and, unsurprisingly, show that this is essentially a noop that has no bearing on the output, but I (kind of) get where the request is coming from so here's that version...

let montySlower nSims =

let rnd = new Random()

let MontyPick winner pick =

if pick = winner then

[0..2] |> Seq.filter (fun i -> i <> pick) |> Seq.nth (rnd.Next(0,2))

else

3 - pick - winner

let SwitchGame() =

let winner, pick = rnd.Next(0,3), rnd.Next(0,3)

let monty = MontyPick winner pick

let pickFinal = 3 - monty - pick

// Show that Monty's pick has no effect...

if (winner <> pick) <> (pickFinal = winner) then

printfn "Monty's selection actually had an effect!"

if pickFinal = winner then 1 else 0

let StayGame() =

let winner, pick = rnd.Next(0,3), rnd.Next(0,3)

let monty = MontyPick winner pick

// This one's even more obvious than the above since pickFinal

// is precisely the same as pick

let pickFinal = pick

if (winner = pick) <> (winner = pickFinal) then

printfn "Monty's selection actually had an effect!"

if winner = pickFinal then 1 else 0

let Wins (f:unit -> int) = seq {for i in [1..nSims] -> f()} |> Seq.sum

printfn "Stay: %d wins out of %d - Switch: %d wins out of %d" (Wins StayGame) nSims (Wins SwitchGame) nSims

## [edit] Forth

include random.fs

variable stay-wins

variable switch-wins

: trial ( -- )

3 random 3 random ( prize choice )

= if 1 stay-wins +!

else 1 switch-wins +!

then ;

: trials ( n -- )

0 stay-wins ! 0 switch-wins !

dup 0 do trial loop

cr stay-wins @ . [char] / emit dup . ." staying wins"

cr switch-wins @ . [char] / emit . ." switching wins" ;

1000 trials

or in iForth:

0 value stay-wins

0 value switch-wins

: trial ( -- )

3 choose 3 choose ( -- prize choice )

= IF 1 +TO stay-wins exit ENDIF

1 +TO switch-wins ;

: trials ( n -- )

CLEAR stay-wins

CLEAR switch-wins

dup 0 ?DO trial LOOP

CR stay-wins DEC. ." / " dup DEC. ." staying wins,"

CR switch-wins DEC. ." / " DEC. ." switching wins." ;

With output:

FORTH> 100000000 trials 33336877 / 100000000 staying wins, 66663123 / 100000000 switching wins. ok

## [edit] Fortran

PROGRAM MONTYHALL

IMPLICIT NONE

INTEGER, PARAMETER :: trials = 10000

INTEGER :: i, choice, prize, remaining, show, staycount = 0, switchcount = 0

LOGICAL :: door(3)

REAL :: rnum

CALL RANDOM_SEED

DO i = 1, trials

door = .FALSE.

CALL RANDOM_NUMBER(rnum)

prize = INT(3*rnum) + 1

door(prize) = .TRUE. ! place car behind random door

CALL RANDOM_NUMBER(rnum)

choice = INT(3*rnum) + 1 ! choose a door

DO

CALL RANDOM_NUMBER(rnum)

show = INT(3*rnum) + 1

IF (show /= choice .AND. show /= prize) EXIT ! Reveal a goat

END DO

SELECT CASE(choice+show) ! Calculate remaining door index

CASE(3)

remaining = 3

CASE(4)

remaining = 2

CASE(5)

remaining = 1

END SELECT

IF (door(choice)) THEN ! You win by staying with your original choice

staycount = staycount + 1

ELSE IF (door(remaining)) THEN ! You win by switching to other door

switchcount = switchcount + 1

END IF

END DO

WRITE(*, "(A,F6.2,A)") "Chance of winning by not switching is", real(staycount)/trials*100, "%"

WRITE(*, "(A,F6.2,A)") "Chance of winning by switching is", real(switchcount)/trials*100, "%"

END PROGRAM MONTYHALL

Sample Output

Chance of winning by not switching is 32.82% Chance of winning by switching is 67.18%

## [edit] Go

package main

import (

"fmt"

"math/rand"

"time"

)

func main() {

games := 100000

r := rand.New(rand.NewSource(time.Now().UnixNano()))

var switcherWins, keeperWins, shown int

for i := 0; i < games; i++ {

doors := []int{0, 0, 0}

doors[r.Intn(3)] = 1 // Set which one has the car

choice := r.Intn(3) // Choose a door

for shown = r.Intn(3); shown == choice || doors[shown] == 1; shown = r.Intn(3) {}

switcherWins += doors[3 - choice - shown]

keeperWins += doors[choice]

}

floatGames := float32(games)

fmt.Printf("Switcher Wins: %d (%3.2f%%)\n",

switcherWins, (float32(switcherWins) / floatGames * 100))

fmt.Printf("Keeper Wins: %d (%3.2f%%)",

keeperWins, (float32(keeperWins) / floatGames * 100))

}

Output:

Switcher Wins: 66542 (66.54%) Keeper Wins: 33458 (33.46%)

## [edit] Haskell

import System.Random (StdGen, getStdGen, randomR)

trials :: Int

trials = 10000

data Door = Car | Goat deriving Eq

play :: Bool -> StdGen -> (Door, StdGen)

play switch g = (prize, new_g)

where (n, new_g) = randomR (0, 2) g

d1 = [Car, Goat, Goat] !! n

prize = case switch of

False -> d1

True -> case d1 of

Car -> Goat

Goat -> Car

cars :: Int -> Bool -> StdGen -> (Int, StdGen)

cars n switch g = f n (0, g)

where f 0 (cs, g) = (cs, g)

f n (cs, g) = f (n - 1) (cs + result, new_g)

where result = case prize of Car -> 1; Goat -> 0

(prize, new_g) = play switch g

main = do

g <- getStdGen

let (switch, g2) = cars trials True g

(stay, _) = cars trials False g2

putStrLn $ msg "switch" switch

putStrLn $ msg "stay" stay

where msg strat n = "The " ++ strat ++ " strategy succeeds " ++

percent n ++ "% of the time."

percent n = show $ round $

100 * (fromIntegral n) / (fromIntegral trials)

With a `State` monad, we can avoid having to explicitly pass around the `StdGen` so often. `play` and `cars` can be rewritten as follows:

import Control.Monad.State

play :: Bool -> State StdGen Door

play switch = do

i <- rand

let d1 = [Car, Goat, Goat] !! i

return $ case switch of

False -> d1

True -> case d1 of

Car -> Goat

Goat -> Car

where rand = do

g <- get

let (v, new_g) = randomR (0, 2) g

put new_g

return v

cars :: Int -> Bool -> StdGen -> (Int, StdGen)

cars n switch g = (numcars, new_g)

where numcars = length $ filter (== Car) prize_list

(prize_list, new_g) = runState (replicateM n (play switch)) g

Sample output (for either implementation):

The switch strategy succeeds 67% of the time.

The stay strategy succeeds 34% of the time.

## [edit] HicEst

REAL :: ndoors=3, doors(ndoors), plays=1E4

DLG(NameEdit = plays, DNum=1, Button='Go')

switchWins = 0

stayWins = 0

DO play = 1, plays

doors = 0 ! clear the doors

winner = 1 + INT(RAN(ndoors)) ! door that has the prize

doors(winner) = 1

guess = 1 + INT(RAN(doors)) ! player chooses his door

IF( guess == winner ) THEN ! Monty decides which door to open:

show = 1 + INT(RAN(2)) ! select 1st or 2nd goat-door

checked = 0

DO check = 1, ndoors

checked = checked + (doors(check) == 0)

IF(checked == show) open = check

ENDDO

ELSE

open = (1+2+3) - winner - guess

ENDIF

new_guess_if_switch = (1+2+3) - guess - open

stayWins = stayWins + doors(guess) ! count if guess was correct

switchWins = switchWins + doors(new_guess_if_switch)

ENDDO

WRITE(ClipBoard, Name) plays, switchWins, stayWins

END

! plays=1E3; switchWins=695; stayWins=305;

! plays=1E4; switchWins=6673; stayWins=3327;

! plays=1E5; switchWins=66811; stayWins=33189;

! plays=1E6; switchWins=667167; stayWins=332833;

## [edit] Icon and Unicon

procedure main(arglist)Sample Output:

rounds := integer(arglist[1]) | 10000

doors := '123'

strategy1 := strategy2 := 0

every 1 to rounds do {

goats := doors -- ( car := ?doors )

guess1 := ?doors

show := goats -- guess1

if guess1 == car then strategy1 +:= 1

else strategy2 +:= 1

}

write("Monty Hall simulation for ", rounds, " rounds.")

write("Strategy 1 'Staying' won ", real(strategy1) / rounds )

write("Strategy 2 'Switching' won ", real(strategy2) / rounds )

end

Monty Hall simulation for 10000 rounds. Strategy 1 'Staying' won 0.3266 Strategy 2 'Switching' won 0.6734

## [edit] Io

keepWins := 0Sample output:

switchWins := 0

doors := 3

times := 100000

pickDoor := method(excludeA, excludeB,

door := excludeA

while(door == excludeA or door == excludeB,

door = (Random value() * doors) floor

)

door

)

times repeat(

playerChoice := pickDoor()

carDoor := pickDoor()

shownDoor := pickDoor(carDoor, playerChoice)

switchDoor := pickDoor(playerChoice, shownDoor)

(playerChoice == carDoor) ifTrue(keepWins = keepWins + 1)

(switchDoor == carDoor) ifTrue(switchWins = switchWins + 1)

)

("Switching to the other door won #{switchWins} times.\n"\

.. "Keeping the same door won #{keepWins} times.\n"\

.. "Game played #{times} times with #{doors} doors.") interpolate println

Switching to the other door won 66935 times. Keeping the same door won 33065 times. Game played 100000 times with 3 doors.

## [edit] J

The core of this simulation is picking a random item from a set

pick=: {~ ?@#

And, of course, we will be picking one door from three doors

DOORS=:1 2 3

But note that the simulation code should work just as well with more doors.

Anyways the scenario where the contestant's switch or stay strategy makes a difference is where Monty has picked from the doors which are neither the user's door nor the car's door.

scenario=: ((pick@-.,])pick,pick) bind DOORS

(Here, I have decided that the result will be a list of three door numbers. The first number in that list is the number Monty picks, the second number represents the door the user picked, and the third number represents the door where the car is hidden.)

Once we have our simulation test results for the scenario, we need to test if staying would win. In other words we need to test if the user's first choice matches where the car was hidden:

stayWin=: =/@}.

In other words: drop the first element from the list representing our test results -- this leaves us with the user's choice and the door where the car was hidden -- and then insert the verb `=`

between those two values.

We also need to test if switching would win. In other words, we need to test if the user would pick the car from the doors other than the one Monty picked and the one the user originally picked:

switchWin=: pick@(DOORS -. }:) = {:

In other words, start with our list of all doors and then remove the door the monty picked and the door the user picked, and then pick one of the remaining doors at random (the pick at random part is only significant if there were originally more than 3 doors) and see if that matches the door where the car is.

Finally, we need to run the simulation a thousand times and count how many times each strategy wins:

+/ (stayWin,switchWin)@scenario"0 i.1000

320 680

Or, we could bundle this all up as a defined word. Here, the (optional) left argument "names" the doors and the right argument says how many simulations to run:

simulate=:3 :0

1 2 3 simulate y

:

pick=. {~ ?@#

scenario=. ((pick@-.,])pick,pick) bind x

stayWin=. =/@}.

switchWin=. pick@(x -. }:) = {:

r=.(stayWin,switchWin)@scenario"0 i.1000

labels=. ];.2 'limit stay switch '

smoutput labels,.":"0 y,+/r

)

Example use:

simulate 1000

limit 1000

stay 304

switch 696

Or, with more doors (and assuming this does not require new rules about how Monty behavior or how the player behaves):

1 2 3 4 simulate 1000

limit 1000

stay 233

switch 388

## [edit] Java

import java.util.Random;

public class Monty{

public static void main(String[] args){

int switchWins = 0;

int stayWins = 0;

Random gen = new Random();

for(int plays = 0;plays < 32768;plays++ ){

int[] doors = {0,0,0};//0 is a goat, 1 is a car

doors[gen.nextInt(3)] = 1;//put a winner in a random door

int choice = gen.nextInt(3); //pick 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("Switching wins " + switchWins + " times.");

System.out.println("Staying wins " + stayWins + " times.");

}

}

Output:

Switching wins 21924 times. Staying wins 10844 times.

## [edit] JavaScript

### [edit] Extensive Solution

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

function montyhall(tests, doors) {

'use strict';

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

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

};

}

- Output:

montyhall(1000, 3)

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

montyhall(1000, 4)

Object {stayWins: "253 25.3%", switchWins: "384 38.4%"}

montyhall(1000, 5)

Object {stayWins: "202 20.2%", switchWins: "265 26.5%"}

### [edit] Basic Solution

var totalGames = 10000,

selectDoor = function () {

return Math.floor(Math.random() * 3); // Choose a number from 0, 1 and 2.

},

games = (function () {

var i = 0, games = [];

for (; i < totalGames; ++i) {

games.push(selectDoor()); // Pick a door which will hide the prize.

}

return games;

}()),

play = function (switchDoor) {

var i = 0, j = games.length, winningDoor, randomGuess, totalTimesWon = 0;

for (; i < j; ++i) {

winningDoor = games[i];

randomGuess = selectDoor();

if ((randomGuess === winningDoor && !switchDoor) ||

(randomGuess !== winningDoor && switchDoor))

{

/*

* If I initially guessed the winning door and didn't switch,

* or if I initially guessed a losing door but then switched,

* I've won.

*

* The only time I lose is when I initially guess the winning door

* and then switch.

*/

totalTimesWon++;

}

}

return totalTimesWon;

};

/*

* Start the simulation

*/

console.log("Playing " + totalGames + " games");

console.log("Wins when not switching door", play(false));

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

- Output:

Playing 10000 games

Wins when not switching door 3326

Wins when switching door 6630

## [edit] Liberty BASIC

'adapted from BASIC solution

DIM doors(3) '0 is a goat, 1 is a car

total = 10000 'set desired number of iterations

switchWins = 0

stayWins = 0

FOR plays = 1 TO total

winner = INT(RND(1) * 3) + 1

doors(winner) = 1'put a winner in a random door

choice = INT(RND(1) * 3) + 1'pick a door, any door

DO

shown = INT(RND(1) * 3) + 1

'don't show the winner or the choice

LOOP WHILE doors(shown) = 1 OR shown = choice

if doors(choice) = 1 then

stayWins = stayWins + 1 'if you won by staying, count it

else

switchWins = switchWins + 1'could have switched to win

end if

doors(winner) = 0 'clear the doors for the next test

NEXT

PRINT "Result for ";total;" games."

PRINT "Switching wins "; switchWins; " times."

PRINT "Staying wins "; stayWins; " times."

Output:

Result for 10000 games. Switching wins 6634 times. Staying wins 3366 times.

## [edit] Lua

function playgame(player)

local car = math.random(3)

local pchoice = player.choice()

local function neither(a, b) --slow, but it works

local el = math.random(3)

return (el ~= a and el ~= b) and el or neither(a, b)

end

local el = neither(car, pchoice)

if(player.switch) then pchoice = neither(pchoice, el) end

player.wins = player.wins + (pchoice == car and 1 or 0)

end

for _, v in ipairs{true, false} do

player = {choice = function() return math.random(3) end,

wins = 0, switch = v}

for i = 1, 20000 do playgame(player) end

print(player.wins)

end

## [edit] Mathematica

montyHall[nGames_] :=

Module[{r, winningDoors, firstChoices, nStayWins, nSwitchWins, s},

r := RandomInteger[{1, 3}, nGames];

winningDoors = r;

firstChoices = r;

nStayWins = Count[Transpose[{winningDoors, firstChoices}], {d_, d_}];

nSwitchWins = nGames - nStayWins;

Grid[{{"Strategy", "Wins", "Win %"}, {"Stay", Row[{nStayWins, "/", nGames}], s=N[100 nStayWins/nGames]},

{"Switch", Row[{nSwitchWins, "/", nGames}], 100 - s}}, Frame -> All]]

- Usage

montyHall[100000]

## [edit] MATLAB

function montyHall(numDoors,numSimulations)

assert(numDoors > 2);

function num = randInt(n)

num = floor( n*rand()+1 );

end

%The first column will tallie wins, the second losses

switchedDoors = [0 0];

stayed = [0 0];

for i = (1:numSimulations)

availableDoors = (1:numDoors); %Preallocate the available doors

winningDoor = randInt(numDoors); %Define the winning door

playersOriginalChoice = randInt(numDoors); %The player picks his initial choice

availableDoors(playersOriginalChoice) = []; %Remove the players choice from the available doors

%Pick the door to open from the available doors

openDoor = availableDoors(randperm(numel(availableDoors))); %Sort the available doors randomly

openDoor(openDoor == winningDoor) = []; %Make sure Monty doesn't open the winning door

openDoor = openDoor(randInt(numel(openDoor))); %Choose a random door to open

availableDoors(availableDoors==openDoor) = []; %Remove the open door from the available doors

availableDoors(end+1) = playersOriginalChoice; %Put the player's original choice back into the pool of available doors

availableDoors = sort(availableDoors);

playersNewChoice = availableDoors(randInt(numel(availableDoors))); %Pick one of the available doors

if playersNewChoice == playersOriginalChoice

switch playersNewChoice == winningDoor

case true

stayed(1) = stayed(1) + 1;

case false

stayed(2) = stayed(2) + 1;

otherwise

error 'ERROR'

end

else

switch playersNewChoice == winningDoor

case true

switchedDoors(1) = switchedDoors(1) + 1;

case false

switchedDoors(2) = switchedDoors(2) + 1;

otherwise

error 'ERROR'

end

end

end

disp(sprintf('Switch win percentage: %f%%\nStay win percentage: %f%%\n', [switchedDoors(1)/sum(switchedDoors),stayed(1)/sum(stayed)] * 100));

end

Output:

>> montyHall(3,100000)

Switch win percentage: 66.705972%

Stay win percentage: 33.420062%

## [edit] MAXScript

fn montyHall choice switch =

(

doors = #(false, false, false)

doors[random 1 3] = true

chosen = doors[choice]

if switch then chosen = not chosen

chosen

)

fn iterate iterations switched =

(

wins = 0

for i in 1 to iterations do

(

if (montyHall (random 1 3) switched) then

(

wins += 1

)

)

wins * 100 / iterations as float

)

iterations = 10000

format ("Stay strategy:%\%\n") (iterate iterations false)

format ("Switch strategy:%\%\n") (iterate iterations true)

Output:

Stay strategy:33.77%

Switch strategy:66.84%

## [edit] NetRexx

/* NetRexx ************************************************************

* 30.08.2013 Walter Pachl translated from Java/REXX/PL/I

**********************************************************************/

options replace format comments java crossref savelog symbols nobinary

doors = create_doors

switchWins = 0

stayWins = 0

shown=0

Loop plays=1 To 1000000

doors=0

r=r3()

doors[r]=1

choice = r3()

loop Until shown<>choice & doors[shown]=0

shown = r3()

End

If doors[choice]=1 Then

stayWins=stayWins+1

Else

switchWins=switchWins+1

End

Say "Switching wins " switchWins " times."

Say "Staying wins " stayWins " times."

method create_doors static returns Rexx

doors = ''

doors[0] = 0

doors[1] = 0

doors[2] = 0

return doors

method r3 static

rand=random()

return rand.nextInt(3) + 1

Output

Switching wins 667335 times. Staying wins 332665 times.

## [edit] OCaml

let trials = 10000

type door = Car | Goat

let play switch =

let n = Random.int 3 in

let d1 = [|Car; Goat; Goat|].(n) in

if not switch then d1

else match d1 with

Car -> Goat

| Goat -> Car

let cars n switch =

let total = ref 0 in

for i = 1 to n do

let prize = play switch in

if prize = Car then

incr total

done;

!total

let () =

let switch = cars trials true

and stay = cars trials false in

let msg strat n =

Printf.printf "The %s strategy succeeds %f%% of the time.\n"

strat (100. *. (float n /. float trials)) in

msg "switch" switch;

msg "stay" stay

## [edit] PARI/GP

test(trials)={

my(stay=0,change=0);

for(i=1,trials,

my(prize=random(3),initial=random(3),opened);

while((opened=random(3))==prize | opened==initial,);

if(prize == initial, stay++, change++)

);

print("Wins when staying: "stay);

print("Wins when changing: "change);

[stay, change]

};

test(1e4)

Output:

Wins when staying: 3433 Wins when changing: 6567 %1 = [3433, 6567]

## [edit] Perl

#! /usr/bin/perl

use strict;

my $trials = 10000;

my $stay = 0;

my $switch = 0;

foreach (1 .. $trials)

{

my $prize = int(rand 3);

# let monty randomly choose a door where he puts the prize

my $chosen = int(rand 3);

# let us randomly choose a door...

my $show;

do { $show = int(rand 3) } while $show == $chosen || $show == $prize;

# ^ monty opens a door which is not the one with the

# prize, that he knows it is the one the player chosen

$stay++ if $prize == $chosen;

# ^ if player chose the correct door, player wins only if he stays

$switch++ if $prize == 3 - $chosen - $show;

# ^ if player switches, the door he picks is (3 - $chosen - $show),

# because 0+1+2=3, and he picks the only remaining door that is

# neither $chosen nor $show

}

print "Stay win ratio " . (100.0 * $stay/$trials) . "\n";

print "Switch win ratio " . (100.0 * $switch/$trials) . "\n";

## [edit] Perl 6

This implementation is parametric over the number of doors. Increasing the number of doors in play makes the superiority of the switch strategy even more obvious.

enum Prize <Car Goat>;

enum Strategy <Stay Switch>;

sub play (Strategy $strategy, Int :$doors = 3) returns Prize {

# Call the door with a car behind it door 0. Number the

# remaining doors starting from 1.

my Prize @doors = Car, Goat xx $doors - 1;

# The player chooses a door.

my Prize $initial_pick = @doors.splice(@doors.keys.pick,1)[0];

# Of the n doors remaining, the host chooses n - 1 that have

# goats behind them and opens them, removing them from play.

@doors.splice($_,1)

for pick @doors.elems - 1, grep { @doors[$_] == Goat }, keys @doors;

# If the player stays, they get their initial pick. Otherwise,

# they get whatever's behind the remaining door.

return $strategy === Stay ?? $initial_pick !! @doors[0];

}

constant TRIALS = 1000;

for 3, 10 -> $doors {

my %wins;

say "With $doors doors: ";

for Stay, 'Staying', Switch, 'Switching' -> $s, $name {

for ^TRIALS {

++%wins{$s} if play($s, doors => $doors) == Car;

}

say " $name wins ",

round(100*%wins{$s} / TRIALS),

'% of the time.'

}

}

- Output:

With 3 doors: Staying wins 31% of the time. Switching wins 68% of the time. With 10 doors: Staying wins 9% of the time. Switching wins 90% of the time.

## [edit] PHP

<?php

function montyhall($iterations){

$switch_win = 0;

$stay_win = 0;

foreach (range(1, $iterations) as $i){

$doors = array(0, 0, 0);

$doors[array_rand($doors)] = 1;

$choice = array_rand($doors);

do {

$shown = array_rand($doors);

} while($shown == $choice || $doors[$shown] == 1);

$stay_win += $doors[$choice];

$switch_win += $doors[3 - $choice - $shown];

}

$stay_percentages = ($stay_win/$iterations)*100;

$switch_percentages = ($switch_win/$iterations)*100;

echo "Iterations: {$iterations} - ";

echo "Stayed wins: {$stay_win} ({$stay_percentages}%) - ";

echo "Switched wins: {$switch_win} ({$switch_percentages}%)";

}

montyhall(10000);

?>

Output:

Iterations: 10000 - Stayed wins: 3331 (33.31%) - Switched wins: 6669 (66.69%)

## [edit] PicoLisp

(de montyHall (Keep)

(let (Prize (rand 1 3) Choice (rand 1 3))

(if Keep # Keeping the first choice?

(= Prize Choice) # Yes: Monty's choice doesn't matter

(<> Prize Choice) ) ) ) # Else: Win if your first choice was wrong

(prinl

"Strategy KEEP -> "

(let Cnt 0

(do 10000 (and (montyHall T) (inc 'Cnt)))

(format Cnt 2) )

" %" )

(prinl

"Strategy SWITCH -> "

(let Cnt 0

(do 10000 (and (montyHall NIL) (inc 'Cnt)))

(format Cnt 2) )

" %" )

Output:

Strategy KEEP -> 33.01 % Strategy SWITCH -> 67.73 %

## [edit] PL/I

*process source attributes xref;

ziegen: Proc Options(main);

/* REXX ***************************************************************

* 30.08.2013 Walter Pachl derived from Java

**********************************************************************/

Dcl (switchWins,stayWins) Bin Fixed(31) Init(0);

Dcl doors(3) Bin Fixed(31);

Dcl (plays,r,choice) Bin Fixed(31) Init(0);

Dcl c17 Char(17) Init((datetime()));

Dcl p9 Pic'(9)9' def(c17) pos(5);

i=random(p9);

Do plays=1 To 1000000;

doors=0;

r=r3();

doors(r)=1;

choice=r3();

Do Until(shown^=choice & doors(shown)=0);

shown=r3();

End;

If doors(choice)=1 Then

stayWins+=1;

Else

switchWins+=1;

End;

Put Edit("Switching wins ",switchWins," times.")(Skip,a,f(6),a);

Put Edit("Staying wins ",stayWins ," times.")(Skip,a,f(6),a);

r3: Procedure Returns(Bin Fixed(31));

/*********************************************************************

* Return a random integer: 1, 2, or 3

*********************************************************************/

Dcl r Bin Float(53);

Dcl res Bin Fixed(31);

r=random();

res=(r*3)+1;

Return(res);

End;

End;

Output:

Switching wins 665908 times. Staying wins 334092 times.

## [edit] Post Script

Use ghostscript or print this to a postscript printer

%!PS

/Courier % name the desired font

20 selectfont % choose the size in points and establish

% the font as the current one

% init random number generator

(%Calendar%) currentdevparams /Second get srand

1000000 % iteration count

0 0 % 0 wins on first selection 0 wins on switch

2 index % get iteration count

{

rand 3 mod % winning door

rand 3 mod % first choice

eq {

1 add

}

{

exch 1 add exch

} ifelse

} repeat

% compute percentages

2 index div 100 mul exch 2 index div 100 mul

% display result

70 600 moveto

(Switching the door: ) show

80 string cvs show (%) show

70 700 moveto

(Keeping the same: ) show

80 string cvs show (%) show

showpage % print all on the page

Sample output:

Keeping the same: 33.4163% Switching the door: 66.5837%

## [edit] PowerShell

#Declaring variables

$intIterations = 10000

$intKept = 0

$intSwitched = 0

#Creating a function

Function Play-MontyHall()

{

#Using a .NET object for randomization

$objRandom = New-Object -TypeName System.Random

#Generating the winning door number

$intWin = $objRandom.Next(1,4)

#Generating the chosen door

$intChoice = $objRandom.Next(1,4)

#Generating the excluded number

#Because there is no method to exclude a number from a range,

#I let it re-generate in case it equals the winning number or

#in case it equals the chosen door.

$intLose = $objRandom.Next(1,4)

While (($intLose -EQ $intWin) -OR ($intLose -EQ $intChoice))

{$intLose = $objRandom.Next(1,4)}

#Generating the 'other' door

#Same logic applies as for the chosen door: it cannot be equal

#to the winning door nor to the chosen door.

$intSwitch = $objRandom.Next(1,4)

While (($intSwitch -EQ $intLose) -OR ($intSwitch -EQ $intChoice))

{$intSwitch = $objRandom.Next(1,4)}

#Simple counters per win for both categories

#Because a child scope cannot change variables in the parent

#scope, the scope of the counters is expanded script-wide.

If ($intChoice -EQ $intWin)

{$script:intKept++}

If ($intSwitch -EQ $intWin)

{$script:intSwitched++}

}

#Looping the Monty Hall function for $intIterations times

While ($intIterationCount -LT $intIterations)

{

Play-MontyHall

$intIterationCount++

}

#Output

Write-Host "Results through $intIterations iterations:"

Write-Host "Keep : $intKept ($($intKept/$intIterations*100)%)"

Write-Host "Switch: $intSwitched ($($intSwitched/$intIterations*100)%)"

Write-Host ""

Output:

Results through 10000 iterations: Keep : 3336 (33.36%) Switch: 6664 (66.64%)

## [edit] PureBasic

Structure winsOutput:

stay.i

redecide.i

EndStructure

#goat = 0

#car = 1

Procedure MontyHall(*results.wins)

Dim Doors(2)

Doors(Random(2)) = #car

player = Random(2)

Select Doors(player)

Case #car

*results\redecide + #goat

*results\stay + #car

Case #goat

*results\redecide + #car

*results\stay + #goat

EndSelect

EndProcedure

OpenConsole()

#Tries = 1000000

Define results.wins

For i = 1 To #Tries

MontyHall(@results)

Next

PrintN("Trial runs for each option: " + Str(#Tries))

PrintN("Wins when redeciding: " + Str(results\redecide) + " (" + StrD(results\redecide / #Tries * 100, 2) + "% chance)")

PrintN("Wins when sticking: " + Str(results\stay) + " (" + StrD(results\stay / #Tries * 100, 2) + "% chance)")

Input()

Trial runs for each option: 1000000 Wins when redeciding: 666459 (66.65% chance) Wins when sticking: 333541 (33.35% chance)

## [edit] Python

'''

I could understand the explanation of the Monty Hall problem

but needed some more evidence

References:

http://www.bbc.co.uk/dna/h2g2/A1054306

http://en.wikipedia.org/wiki/Monty_Hall_problem especially:

http://en.wikipedia.org/wiki/Monty_Hall_problem#Increasing_the_number_of_doors

'''

from random import randrange

doors, iterations = 3,100000 # could try 100,1000

def monty_hall(choice, switch=False, doorCount=doors):

# Set up doors

door = [False]*doorCount

# One door with prize

door[randrange(doorCount)] = True

chosen = door[choice]

unpicked = door

del unpicked[choice]

# Out of those unpicked, the alternative is either:

# the prize door, or

# an empty door if the initial choice is actually the prize.

alternative = True in unpicked

if switch:

return alternative

else:

return chosen

print "\nMonty Hall problem simulation:"

print doors, "doors,", iterations, "iterations.\n"

print "Not switching allows you to win",

print sum(monty_hall(randrange(3), switch=False)

for x in range(iterations)),

print "out of", iterations, "times."

print "Switching allows you to win",

print sum(monty_hall(randrange(3), switch=True)

for x in range(iterations)),

print "out of", iterations, "times.\n"

Sample output:

Monty Hall problem simulation: 3 doors, 100000 iterations. Not switching allows you to win 33337 out of 100000 times. Switching allows you to win 66529 out of 100000 times.

### [edit] Python 3 version:

Another (simpler in my opinion), way to do this is below, also in python 3:

import random

#1 represents a car

#0 represent a goat

stay = 0 #amount won if stay in the same position

switch = 0 # amount won if you switch

for i in range(1000):

lst = [1,0,0] # one car and two goats

random.shuffle(lst) # shuffles the list randomly

ran = random.randrange(3) # gets a random number for the random guess

user = lst[ran] #storing the random guess

del(lst[ran]) # deleting the random guess

huh = 0

for i in lst: # getting a value 0 and deleting it

if i ==0:

del(lst[huh]) # deletes a goat when it finds it

break

huh+=1

if user ==1: # if the original choice is 1 then stay adds 1

stay+=1

if lst[0] == 1: # if the switched value is 1 then switch adds 1

switch+=1

print("Stay =",stay)

print("Switch = ",switch)

#Done by Sam Witton 09/04/2014

## [edit] R

# Since R is a vector based language that penalizes for loops, we will avoid

# for-loops, instead using "apply" statement variants (like "map" in other

# functional languages).

set.seed(19771025) # set the seed to set the same results as this code

N <- 10000 # trials

true_answers <- sample(1:3, N, replace=TRUE)

# We can assme that the contestant always choose door 1 without any loss of

# generality, by equivalence. That is, we can always relabel the doors

# to make the user-chosen door into door 1.

# Thus, the host opens door '2' unless door 2 has the prize, in which case

# the host opens door 3.

host_opens <- 2 + (true_answers == 2)

other_door <- 2 + (true_answers != 2)

## if always switch

summary( other_door == true_answers )

## if we never switch

summary( true_answers == 1)

## if we randomly switch

random_switch <- other_door

random_switch[runif(N) >= .5] <- 1

summary(random_switch == true_answers)

## To go with the exact parameters of the Rosetta challenge, complicating matters....

## Note that the player may initially choose any of the three doors (not just Door 1),

## that the host opens a different door revealing a goat (not necessarily Door 3), and

## that he gives the player a second choice between the two remaining unopened doors.

N <- 10000 #trials

true_answers <- sample(1:3, N, replace=TRUE)

user_choice <- sample(1:3, N, replace=TRUE)

## the host_choice is more complicated

host_chooser <- function(user_prize) {

# this could be cleaner

bad_choices <- unique(user_prize)

# in R, the x[-vector] form implies, choose the indices in x not in vector

choices <- c(1:3)[-bad_choices]

# if the first arg to sample is an int, it treats it as the number of choices

if (length(choices) == 1) { return(choices)}

else { return(sample(choices,1))}

}

host_choice <- apply( X=cbind(true_answers,user_choice), FUN=host_chooser,MARGIN=1)

not_door <- function(x){ return( (1:3)[-x]) } # we could also define this

# directly at the FUN argument following

other_door <- apply( X = cbind(user_choice,host_choice), FUN=not_door, MARGIN=1)

## if always switch

summary( other_door == true_answers )

## if we never switch

summary( true_answers == user_choice)

## if we randomly switch

random_switch <- user_choice

change <- runif(N) >= .5

random_switch[change] <- other_door[change]

summary(random_switch == true_answers)

Results: > ## if always switch > summary( other_door == true_answers ) Mode FALSE TRUE logical 3298 6702 > ## if we never switch > summary( true_answers == 1) Mode FALSE TRUE logical 6702 3298 > ## if we randomly switch > summary(random_switch == true_answers) Mode FALSE TRUE logical 5028 4972 > ## if always switch > summary( other_door == true_answers ) Mode FALSE TRUE logical 3295 6705 > ## if we never switch > summary( true_answers == user_choice) Mode FALSE TRUE logical 6705 3295 > ## if we randomly switch > summary(random_switch == true_answers) Mode FALSE TRUE logical 4986 5014

# As above, but generalized to K number of doors K = 4 # number of doors N = 1e4 # number of simulation trials chooser <- function(x) { i <- (1:K)[-x]; if (length(i)>1) sample(i,1) else i } p100 <- function(...) { cat("\nNumber of doors:", K, "\nSimulation yields % winning probability:", " (2nd choice after host reveal)\n"); print(c(...) * 100, digits=3) } prize_door <- sample(1:K, N, replace=TRUE) first_choice <- sample(1:K, N, replace=TRUE) host_opens <- apply(cbind(prize_door, first_choice), 1, chooser) second_choice <- apply(cbind(host_opens, first_choice), 1, chooser) p100("By first choice" = (Pr.first_win <- mean(first_choice == prize_door)), "By second choice" = (Pr.second_win <- mean(second_choice == prize_door)), " Change gain" = Pr.second_win / Pr.first_win - 1) #------- # # Sample output: Number of doors: 4 Simulation yields % winning probability: (2nd choice after host reveal) By first choice By second choice Change gain 24.7 36.5 48.0

## [edit] Racket

#lang racket

(define (get-last-door a b) ; assumes a != b

(vector-ref '#(- 2 1

2 - 0

1 0 -)

(+ a (* 3 b))))

(define (run-game strategy)

(define car-door (random 3))

(define first-choice (random 3))

(define revealed-goat

(if (= car-door first-choice)

(let ([r (random 2)]) (if (<= car-door r) (add1 r) r)) ; random

(get-last-door car-door first-choice))) ; reveal goat

(define final-choice (strategy first-choice revealed-goat))

(define win? (eq? final-choice car-door))

;; (printf "car: ~s\nfirst: ~s\nreveal: ~s\nfinal: ~s\n => ~s\n\n"

;; car-door first-choice revealed-goat final-choice

;; (if win? 'win 'lose))

win?)

(define (keep-choice first-choice revealed-goat)

first-choice)

(define (change-choice first-choice revealed-goat)

(get-last-door first-choice revealed-goat))

(define (test-strategy strategy)

(define N 10000000)

(define wins (for/sum ([i (in-range N)]) (if (run-game strategy) 1 0)))

(printf "~a: ~a%\n"

(object-name strategy)

(exact->inexact (/ wins N 1/100))))

(for-each test-strategy (list keep-choice change-choice))

Sample Output:

keep-choice: 33.33054% change-choice: 66.67613%

## [edit] REXX

### [edit] version 1

/* REXX ***************************************************************

* 30.08.2013 Walter Pachl derived from Java

**********************************************************************/

Call time 'R'

switchWins = 0;

stayWins = 0

Do plays = 1 To 1000000

doors.=0

r=r3()

doors.r=1

choice = r3()

Do Until shown<>choice & doors.shown=0

shown = r3()

End

If doors.choice=1 Then

stayWins=stayWins+1

Else

switchWins=switchWins+1

End

Say "Switching wins " switchWins " times."

Say "Staying wins " stayWins " times."

Say 'REXX:' time('E') 'seconds'

Call time 'R'

'ziegen'

Say 'PL/I:' time('E') 'seconds'

Say ' '

Call time 'R'

'java ziegen'

Say 'NetRexx:' time('E') 'seconds'

Exit

r3: Return random(2)+1

Output for 1000000 samples:

Switching wins 666442 times. Staying wins 333558 times. REXX: 4.321000 seconds Switching wins 665908 times. Staying wins 334092 times. PL/I: 0.328000 seconds Switching wins 667335 times. Staying wins 332665 times. NetRexx: 2.042000 seconds

### [edit] version 2

/*REXX program simulates a # of trials of the classic Monty Hall problem*/

parse arg t .; if t=='' then t=1000000 /*Not specified? Then use default*/

wins.=0 /*wins.0=stay; wins.1=switching.*/

/*door values: 0=goat 1=car */

do t /*perform this loop T times. */

door.=0 /*set all doors to zero. */

car=random(1,3); door.car=1 /*TV show hides a car randomly. */

?=random(1,3) ; _=door.? /*contestant picks a random door.*/

wins._=wins._+1 /*bump the type of win strategy. */

end /*DO t*/

say 'switching wins ' format(wins.0/t*100,,1)"% of the time."

say ' staying wins ' format(wins.1/t*100,,1)"% of the time."; say

say 'performed' t "times." /*stick a fork in it, we're done.*

**output** when using the default number of trials (one million):

switching wins 66.7% of the time. staying wins 33.3% of the time. performed 1000000 times.

## [edit] Ruby

n = 10_000 #number of times to play

stay = switch = 0 #sum of each strategy's wins

n.times do #play the game n times

#the doors reveal 2 goats and a car

doors = [ :goat, :goat, :car ].shuffle

#random guess

guess = rand(3)

#random door shown, but it is neither the guess nor the car

begin shown = rand(3) end while shown == guess || doors[shown] == :car

if doors[guess] == :car

#staying with the initial guess wins if the initial guess is the car

stay += 1

else

#switching guesses wins if the unshown door is the car

switch += 1

end

end

puts "Staying wins %.2f%% of the time." % (100.0 * stay / n)

puts "Switching wins %.2f%% of the time." % (100.0 * switch / n)

Sample Output:

Staying wins 33.84% of the time. Switching wins 66.16% of the time.

## [edit] Run BASIC

' adapted from BASIC solution

input "Number of tries;";tries ' gimme the number of iterations

FOR plays = 1 TO tries

winner = INT(RND(1) * 3) + 1

doors(winner) = 1 'put a winner in a random door

choice = INT(RND(1) * 3) + 1 'pick a door please

[DO] shown = INT(RND(1) * 3) + 1

' ------------------------------------------

' don't show the winner or the choice

if doors(shown) = 1 then goto [DO]

if shown = choice then goto [DO]

if doors(choice) = 1 then

stayWins = stayWins + 1 ' if you won by staying, count it

else

switchWins = switchWins + 1 ' could have switched to win

end if

doors(winner) = 0 'clear the doors for the next test

NEXT

PRINT " Result for ";tries;" games."

PRINT "Switching wins ";switchWins; " times."

PRINT " Staying wins ";stayWins; " times."

## [edit] Scala

import scala.util.Random

object MontyHallSimulation {

def main(args: Array[String]) {

val samples = if (args.size == 1 && (args(0) matches "\\d+")) args(0).toInt else 1000

val doors = Set(0, 1, 2)

var stayStrategyWins = 0

var switchStrategyWins = 0

1 to samples foreach { _ =>

val prizeDoor = Random shuffle doors head;

val choosenDoor = Random shuffle doors head;

val hostDoor = Random shuffle (doors - choosenDoor - prizeDoor) head;

val switchDoor = doors - choosenDoor - hostDoor head;

(choosenDoor, switchDoor) match {

case (`prizeDoor`, _) => stayStrategyWins += 1

case (_, `prizeDoor`) => switchStrategyWins += 1

}

}

def percent(n: Int) = n * 100 / samples

val report = """|%d simulations were ran.

|Staying won %d times (%d %%)

|Switching won %d times (%d %%)""".stripMargin

println(report

format (samples,

stayStrategyWins, percent(stayStrategyWins),

switchStrategyWins, percent(switchStrategyWins)))

}

}

Sample:

1000 simulations were ran. Staying won 333 times (33 %) Switching won 667 times (66 %)

## [edit] Scheme

(define (random-from-list list) (list-ref list (random (length list))))

(define (random-permutation list)

(if (null? list)

'()

(let* ((car (random-from-list list))

(cdr (random-permutation (remove car list))))

(cons car cdr))))

(define (random-configuration) (random-permutation '(goat goat car)))

(define (random-door) (random-from-list '(0 1 2)))

(define (trial strategy)

(define (door-with-goat-other-than door strategy)

(cond ((and (not (= 0 door)) (equal? (list-ref strategy 0) 'goat)) 0)

((and (not (= 1 door)) (equal? (list-ref strategy 1) 'goat)) 1)

((and (not (= 2 door)) (equal? (list-ref strategy 2) 'goat)) 2)))

(let* ((configuration (random-configuration))

(players-first-guess (strategy `(would-you-please-pick-a-door?)))

(door-to-show-player (door-with-goat-other-than players-first-guess

configuration))

(players-final-guess (strategy `(there-is-a-goat-at/would-you-like-to-move?

,players-first-guess

,door-to-show-player))))

(if (equal? (list-ref configuration players-final-guess) 'car)

'you-win!

'you-lost)))

(define (stay-strategy message)

(case (car message)

((would-you-please-pick-a-door?) (random-door))

((there-is-a-goat-at/would-you-like-to-move?)

(let ((first-choice (cadr message)))

first-choice))))

(define (switch-strategy message)

(case (car message)

((would-you-please-pick-a-door?) (random-door))

((there-is-a-goat-at/would-you-like-to-move?)

(let ((first-choice (cadr message))

(shown-goat (caddr message)))

(car (remove first-choice (remove shown-goat '(0 1 2))))))))

(define-syntax repeat

(syntax-rules ()

((repeat <n> <body> ...)

(let loop ((i <n>))

(if (zero? i)

'()

(cons ((lambda () <body> ...))

(loop (- i 1))))))))

(define (count element list)

(if (null? list)

0

(if (equal? element (car list))

(+ 1 (count element (cdr list)))

(count element (cdr list)))))

(define (prepare-result strategy results)

`(,strategy won with probability

,(exact->inexact (* 100 (/ (count 'you-win! results) (length results)))) %))

(define (compare-strategies times)

(append

(prepare-result 'stay-strategy (repeat times (trial stay-strategy)))

'(and)

(prepare-result 'switch-strategy (repeat times (trial switch-strategy)))))

;; > (compare-strategies 1000000)

;; (stay-strategy won with probability 33.3638 %

;; and switch-strategy won with probability 66.716 %)

## [edit] Seed7

$ include "seed7_05.s7i";

const proc: main is func

local

var integer: switchWins is 0;

var integer: stayWins is 0;

var integer: winner is 0;

var integer: choice is 0;

var integer: shown is 0;

var integer: plays is 0;

begin

for plays range 1 to 10000 do

winner := rand(1, 3);

choice := rand(1, 3);

repeat

shown := rand(1, 3)

until shown <> winner and shown <> choice;

stayWins +:= ord(choice = winner);

switchWins +:= ord(6 - choice - shown = winner);

end for;

writeln("Switching wins " <& switchWins <& " times");

writeln("Staying wins " <& stayWins <& " times");

end func;

Output:

Switching wins 6654 times Staying wins 3346 times

## [edit] Tcl

A simple way of dealing with this one, based on knowledge of the underlying probabilistic system, is to use code like this:

set stay 0; set change 0; set total 10000

for {set i 0} {$i<$total} {incr i} {

if {int(rand()*3) == int(rand()*3)} {

incr stay

} else {

incr change

}

}

puts "Estimate: $stay/$total wins for staying strategy"

puts "Estimate: $change/$total wins for changing strategy"

But that's not really the point of this challenge; it should add the concealing factors too so that we're simulating not just the solution to the game, but also the game itself. (Note that we are using Tcl's lists here to simulate sets.)

We include a third strategy that is proposed by some people (who haven't thought much about it) for this game: just picking at random between all the doors offered by Monty the second time round.

package require Tcl 8.5

# Utility: pick a random item from a list

proc pick list {

lindex $list [expr {int(rand()*[llength $list])}]

}

# Utility: remove an item from a list if it is there

proc remove {list item} {

set idx [lsearch -exact $list $item]

return [lreplace $list $idx $idx]

}

# Codify how Monty will present the new set of doors to choose between

proc MontyHallAction {doors car picked} {

set unpicked [remove $doors $picked]

if {$car in $unpicked} {

# Remove a random unpicked door without the car behind it

set carless [remove $unpicked $car]

return [list {*}[remove $carless [pick $carless]] $car]

# Expressed this way so Monty Hall isn't theoretically

# restricted to using 3 doors, though that could be written

# as just: return [list $car]

} else {

# Monty has a real choice now...

return [remove $unpicked [pick $unpicked]]

}

}

# The different strategies you might choose

proc Strategy:Stay {originalPick otherChoices} {

return $originalPick

}

proc Strategy:Change {originalPick otherChoices} {

return [pick $otherChoices]

}

proc Strategy:PickAnew {originalPick otherChoices} {

return [pick [list $originalPick {*}$otherChoices]]

}

# Codify one round of the game

proc MontyHallGameRound {doors strategy winCounter} {

upvar 1 $winCounter wins

set car [pick $doors]

set picked [pick $doors]

set newDoors [MontyHallAction $doors $car $picked]

set picked [$strategy $picked $newDoors]

# Check for win...

if {$car eq $picked} {

incr wins

}

}

# We're always using three doors

set threeDoors {a b c}

set stay 0; set change 0; set anew 0

set total 10000

# Simulate each of the different strategies

for {set i 0} {$i<$total} {incr i} {

MontyHallGameRound $threeDoors Strategy:Stay stay

MontyHallGameRound $threeDoors Strategy:Change change

MontyHallGameRound $threeDoors Strategy:PickAnew anew

}

# Print the results

puts "Estimate: $stay/$total wins for 'staying' strategy"

puts "Estimate: $change/$total wins for 'changing' strategy"

puts "Estimate: $anew/$total wins for 'picking anew' strategy"

This might then produce output like

Estimate: 3340/10000 wins for 'staying' strategy Estimate: 6733/10000 wins for 'changing' strategy Estimate: 4960/10000 wins for 'picking anew' strategy

Of course, this challenge could also be tackled by putting up a GUI and letting the user be the source of the randomness. But that's moving away from the letter of the challenge and takes a lot of effort anyway...

## [edit] UNIX Shell

#!/bin/bash

# Simulates the "monty hall" probability paradox and shows results.

# http://en.wikipedia.org/wiki/Monty_Hall_problem

# (should rewrite this in C for faster calculating of huge number of rounds)

# (Hacked up by Éric Tremblay, 07.dec.2010)

num_rounds=10 #default number of rounds

num_doors=3 # default number of doors

[ "$1" = "" ] || num_rounds=$[$1+0]

[ "$2" = "" ] || num_doors=$[$2+0]

nbase=1 # or 0 if we want to see door numbers zero-based

num_win=0; num_lose=0

echo "Playing $num_rounds times, with $num_doors doors."

[ "$num_doors" -lt 3 ] && {

echo "Hey, there has to be at least 3 doors!!"

exit 1

}

echo

function one_round() {

winning_door=$[$RANDOM % $num_doors ]

player_picks_door=$[$RANDOM % $num_doors ]

# Host leaves this door AND the player's first choice closed, opens all others

# (this WILL loop forever if there is only 1 door)

host_skips_door=$winning_door

while [ "$host_skips_door" = "$player_picks_door" ]; do

#echo -n "(Host looks at door $host_skips_door...) "

host_skips_door=$[$RANDOM % $num_doors]

done

# Output the result of this round

#echo "Round $[$nbase+current_round]: "

echo -n "Player chooses #$[$nbase+$player_picks_door]. "

[ "$num_doors" -ge 10 ] &&

# listing too many door numbers (10 or more) will just clutter the output

echo -n "Host opens all except #$[$nbase+$host_skips_door] and #$[$nbase+$player_picks_door]. " \

|| {

# less than 10 doors, we list them one by one instead of "all except ?? and ??"

echo -n "Host opens"

host_opens=0

while [ "$host_opens" -lt "$num_doors" ]; do

[ "$host_opens" != "$host_skips_door" ] && [ "$host_opens" != "$player_picks_door" ] && \

echo -n " #$[$nbase+$host_opens]"

host_opens=$[$host_opens+1]

done

echo -n " "

}

echo -n "(prize is behind #$[$nbase+$winning_door]) "

echo -n "Switch from $[$nbase+$player_picks_door] to $[$nbase+$host_skips_door]: "

[ "$winning_door" = "$host_skips_door" ] && {

echo "WIN."

num_win=$[num_win+1]

} || {

echo "LOSE."

num_lose=$[num_lose+1]

}

} # end of function one_round

# ok, let's go

current_round=0

while [ "$num_rounds" -gt "$current_round" ]; do

one_round

current_round=$[$current_round+1]

done

echo

echo "Wins (switch to remaining door): $num_win"

echo "Losses (first guess was correct): $num_lose"

exit 0

Output of a few runs:

$ ./monty_hall_problem.sh Playing 10 times, with 3 doors. Player chooses #2. Host opens #3 (prize is behind #1) Switch from 2 to 1: WIN. Player chooses #1. Host opens #3 (prize is behind #2) Switch from 1 to 2: WIN. Player chooses #2. Host opens #3 (prize is behind #2) Switch from 2 to 1: LOSE. Player chooses #1. Host opens #2 (prize is behind #1) Switch from 1 to 3: LOSE. Player chooses #2. Host opens #3 (prize is behind #1) Switch from 2 to 1: WIN. Player chooses #2. Host opens #1 (prize is behind #2) Switch from 2 to 3: LOSE. Player chooses #3. Host opens #1 (prize is behind #2) Switch from 3 to 2: WIN. Player chooses #2. Host opens #1 (prize is behind #3) Switch from 2 to 3: WIN. Player chooses #1. Host opens #3 (prize is behind #1) Switch from 1 to 2: LOSE. Player chooses #1. Host opens #2 (prize is behind #3) Switch from 1 to 3: WIN. Wins (switch to remaining door): 6 Losses (first guess was correct): 4 $ ./monty_hall_problem.sh 5 10 Playing 5 times, with 10 doors. Player chooses #1. Host opens all except #10 and #1. (prize is behind #10) Switch from 1 to 10: WIN. Player chooses #7. Host opens all except #8 and #7. (prize is behind #8) Switch from 7 to 8: WIN. Player chooses #6. Host opens all except #1 and #6. (prize is behind #1) Switch from 6 to 1: WIN. Player chooses #8. Host opens all except #3 and #8. (prize is behind #8) Switch from 8 to 3: LOSE. Player chooses #6. Host opens all except #5 and #6. (prize is behind #5) Switch from 6 to 5: WIN. Wins (switch to remaining door): 4 Losses (first guess was correct): 1 $ ./monty_hall_problem.sh 1000 Playing 1000 times, with 3 doors. Player chooses #2. Host opens #1 (prize is behind #2) Switch from 2 to 3: LOSE. Player chooses #3. Host opens #1 (prize is behind #2) Switch from 3 to 2: WIN. [ ... ] Player chooses #1. Host opens #3 (prize is behind #2) Switch from 1 to 2: WIN. Player chooses #3. Host opens #2 (prize is behind #1) Switch from 3 to 1: WIN. Wins (switch to remaining door): 655 Losses (first guess was correct): 345

## [edit] Ursala

This is the same algorithm as the Perl solution. Generate two lists of 10000 uniformly distributed samples from {1,2,3}, count each match as a win for the staying strategy, and count each non-match as a win for the switching strategy.

#import std

#import nat

#import flo

rounds = 10000

car_locations = arc{1,2,3}* iota rounds

initial_choices = arc{1,2,3}* iota rounds

staying_wins = length (filter ==) zip(car_locations,initial_choices)

switching_wins = length (filter ~=) zip(car_locations,initial_choices)

format = printf/'%0.2f'+ (times\100.+ div+ float~~)\rounds

#show+

main = ~&plrTS/<'stay: ','switch: '> format* <staying_wins,switching_wins>

Output will vary slightly for each run due to randomness.

stay: 33.95 switch: 66.05

## [edit] Vedit macro language

Vedit macro language does not have random number generator, so one is implemented in subroutine RANDOM (the algorithm was taken from ANSI C library).

#90 = Time_Tick // seed for random number generator

#91 = 3 // random numbers in range 0 to 2

#1 = 0 // wins for "always stay" strategy

#2 = 0 // wins for "always switch" strategy

for (#10 = 0; #10 < 10000; #10++) { // 10,000 iterations

Call("RANDOM")

#3 = Return_Value // #3 = winning door

Call("RANDOM")

#4 = Return_Value // #4 = players choice

do {

Call("RANDOM")

#5 = Return_Value // #5 = door to open

} while (#5 == #3 || #5 == #4)

if (#3 == #4) { // original choice was correct

#1++

}

if (#3 == 3 - #4 - #5) { // switched choice was correct

#2++

}

}

Ins_Text("Staying wins: ") Num_Ins(#1)

Ins_Text("Switching wins: ") Num_Ins(#2)

return

//--------------------------------------------------------------

// Generate random numbers in range 0 <= Return_Value < #91

// #90 = Seed (0 to 0x7fffffff)

// #91 = Scaling (0 to 0xffff)

:RANDOM:

#92 = 0x7fffffff / 48271

#93 = 0x7fffffff % 48271

#90 = (48271 * (#90 % #92) - #93 * (#90 / #92)) & 0x7fffffff

return ((#90 & 0xffff) * #91 / 0x10000)

Sample output:

Staying winns: 3354 Switching winns: 6646

## [edit] XPL0

def Games = 10000; \number of games simulated

int Game, Wins;

include c:\cxpl\codes;

proc Play(Switch); \Play one game

int Switch;

int Car, Player, Player0, Monty;

[Car:= Ran(3); \randomly place car behind a door

Player0:= Ran(3); \player randomly chooses a door

repeat Monty:= Ran(3); \Monty opens door revealing a goat

until Monty # Car and Monty # Player0;

if Switch then \player switches to remaining door

repeat Player:= Ran(3);

until Player # Player0 and Player # Monty

else Player:= Player0; \player sticks with original door

if Player = Car then Wins:= Wins+1;

];

[Format(2,1);

Text(0, "Not switching doors wins car in ");

Wins:= 0;

for Game:= 0 to Games-1 do Play(false);

RlOut(0, float(Wins)/float(Games)*100.0);

Text(0, "% of games.^M^J");

Text(0, "But switching doors wins car in ");

Wins:= 0;

for Game:= 0 to Games-1 do Play(true);

RlOut(0, float(Wins)/float(Games)*100.0);

Text(0, "% of games.^M^J");

]

Example output:

Not switching doors wins car in 33.7% of games. But switching doors wins car in 66.7% of games.

## [edit] zkl

const games=0d100_000;

reg switcherWins=0, keeperWins=0, shown=0;

do(games){

doors := L(0,0,0);

doors[(0).random(3)] = 1; // Set which one has the car

choice := (0).random(3); // Choose a door

while(1){ shown = (0).random(3);

if (not (shown == choice or doors[shown] == 1)) break; }

switcherWins += doors[3 - choice - shown];

keeperWins += doors[choice];

}

"Switcher Wins: %,d (%3.2f%%)".fmt(

switcherWins, switcherWins.toFloat() / games * 100).println();

"Keeper Wins: %,d (%3.2f%%)".fmt(

keeperWins, keeperWins.toFloat() / games * 100).println();

- Output:

Switcher Wins: 66,730 (66.73%) Keeper Wins: 33,270 (33.27%)

- Programming Tasks
- Discrete math
- Games
- Probability and statistics
- ActionScript
- Ada
- ALGOL 68
- APL
- AutoHotkey
- AWK
- BASIC
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- C
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