Monty Hall problem
You are encouraged to solve this task according to the task description, using any language you may know.
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.
 Rules of the game
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?"
 The question
Is it to your advantage to change your choice?
 Note
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.
 Task
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.
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.
 References

 Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 322 DOI: 10.1037/00963445.132.1.3
 A YouTube video: Monty Hall Problem  Numberphile.
11l
V stay = 0
V sw = 0
L 1000
V lst = [1, 0, 0]
random:shuffle(&lst)
V ran = random:(3)
V user = lst[ran]
lst.pop(ran)
V huh = 0
L(i) lst
I i == 0
lst.pop(huh)
L.break
huh++
I user == 1
stay++
I lst[0] == 1
sw++
print(‘Stay = ’stay)
print(‘Switch = ’sw)
8086 Assembly
time: equ 2Ch ; MSDOS syscall to get current time
puts: equ 9 ; MSDOS syscall to print a string
cpu 8086
bits 16
org 100h
section .text
;;; Initialize the RNG with the current time
mov ah,time
int 21h
mov di,cx ; RNG state is kept in DI and BP
mov bp,dx
mov dx,sw ; While switching doors,
mov bl,1
call simrsl ; run simulations,
mov dx,nsw ; While not switching doors,
xor bl,bl ; run simulations.
;;; Print string in DX, run 65536 simulations (according to BL),
;;; then print the amount of cars won.
simrsl: mov ah,puts ; Print the string
int 21h
xor cx,cx ; Run 65536 simulations
call simul
mov ax,si ; Print amount of cars
mov bx,number ; String pointer
mov cx,10 ; Divisor
.dgt: xor dx,dx ; Divide AX by ten
div cx
add dl,'0' ; Add ASCII '0' to the remainder
dec bx ; Move string pointer backwards
mov [bx],dl ; Store digit in string
test ax,ax ; If quotient not zero,
jnz .dgt ; calculate next digit.
mov dx,bx ; Print string starting at first digit
mov ah,puts
int 21h
ret
;;; Run CX simulations.
;;; If BL = 0, don't switch doors, otherwise, always switch
simul: xor si,si ; SI is the amount of cars won
.loop: call door ; Behind which door is the car?
xchg dl,al ; DL = car door
call door ; Which door does the contestant choose?
xchg ah,al ; AH = contestant door
.monty: call door ; Which door does Monty open?
cmp al,dl ; It can't be the door with the car,
je .monty
cmp al,ah ; or the door the contestant picked.
je .monty
test bl,bl ; Will the contestant switch doors?
jz .nosw
xor ah,al ; If so, he switches
.nosw: cmp ah,dl ; Did he get the car?
jne .next
inc si ; If so, add a car
.next: loop .loop
ret
;;; Generate a pseudorandom byte in AL using "X ABC" method
;;; Use it to select a door (1,2,3).
door: xchg bx,bp ; Load RNG state into byteaddressable
xchg cx,di ; registers.
.loop: inc bl ; X++
xor bh,ch ; A ^= C
xor bh,bl ; A ^= X
add cl,bh ; B += A
mov al,cl ; C' = B
shr al,1 ; C' >>= 1
add al,ch ; C' += C
xor al,bh ; C' ^= A
mov ch,al ; C = C'
and al,3 ; ...but we only want the last two bits,
jz .loop ; and if it was 0, get a new random number.
xchg bx,bp ; Restore the registers
xchg cx,di
ret
section .data
sw: db 'When switching doors: $'
nsw: db 'When not switching doors: $'
db '*****'
number: db 13,10,'$'
 Output:
When switching doors: 42841 When not switching doors: 22395
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%)
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%
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;
[doors1]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%
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←runsstay ⍝ 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
Arturo
stay: 0
swit: 0
loop 1..1000 'i [
lst: shuffle new [1 0 0]
rand: random 0 2
user: lst\[rand]
remove 'lst rand
huh: 0
loop lst 'i [
if zero? i [
remove 'lst huh
break
]
huh: huh + 1
]
if user=1 > stay: stay+1
if and? [0 < size lst] [1 = first lst] > swit: swit+1
]
print ["Stay:" stay]
print ["Switch:" swit]
 Output:
Stay: 297 Switch: 549
AutoHotkey
#SingleInstance, Force
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 := round(Correct_Change / Iterations * 100)
Percent_Stay := round(Correct_Stay / Iterations * 100)
Percent_Random := round(Correct_Random / 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). That's %Percent_Change%`% correct.`r`n`r`nWhen I randomly changed my guess, I got %Correct_Random% of %Iterations% (that's %Incorrect_Random% incorrect). That's %Percent_Random%`% correct.`r`n`r`nWhen I stayed with my first guess, I got %Correct_Stay% of %Iterations% (that's %Incorrect_Stay% incorrect). That's %Percent_Stay%`% correct.
ExitApp
Monty_Hall(Mode) ;Mode is 1 for change, 2 for random, or 3 for stay
{
Random, guess, 1, 3
Random, actual, 1, 3
Random, rand, 1, 2
show := guess = actual ? guess = 3 ? guess  rand : guess = 1 ? guess+rand : guess + 2*rand  3 : 6  guess  actual
Mode := Mode = 2 ? 2*rand  1: Mode
Return, Mode = 1 ? 6  guess  show = actual : guess = actual
}
Sample output:
These are the results: When I changed my guess, I got 659 of 1000 (that's 341 incorrect). That's 66% correct. When I randomly changed my guess, I got 505 of 1000 (that's 495 incorrect). That's 51% correct. When I stayed with my first guess, I got 329 of 1000 (that's 671 incorrect). That's 32% correct.
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$
BASIC
ANSI BASIC
100 PROGRAM MontyHallProblem
110 DEF NGames = 10000
120 RANDOMIZE
130 LET NWins = 0
140 FOR Game = 0 TO NGames
150 IF IsGameWon(0) <> 0 THEN LET NWins = NWins + 1
160 NEXT Game
170 PRINT "NOT switching doors wins car in ";
180 PRINT USING "##.#": NWins / NGames * 100;
190 PRINT "% of games."
200 LET NWins = 0
210 FOR Game = 0 TO NGames
220 IF IsGameWon(1) <> 0 THEN LET NWins = NWins + 1
230 NEXT Game
240 PRINT "But switching doors wins car in ";
250 PRINT USING "##.#": NWins / NGames * 100;
260 PRINT "% of games."
270 END
280 REM ***
290 EXTERNAL FUNCTION IsGameWon(Sw)
300 REM Play one game.
310 REM Switching if and only if Sw <> 0.
320 REM Returns 1 if the game is won, 0 otherwise.
330 LET Car = INT(RND * 3) ! Randomly place car behind a door.
340 LET Player0 = INT(RND * 3) ! Player randomly chooses a door.
350 DO
360 LET Monty = INT(RND * 3) ! Monty opens door revealing a goat.
370 LOOP UNTIL (Monty <> Car) AND (Monty <> Player0)
380 IF Sw <> 0 THEN ! Player switches TO remaining door.
390 DO
400 LET Player = INT(RND * 3)
410 LOOP UNTIL (Player <> Player0) AND (Player <> Monty)
420 ELSE
430 LET Player = Player0 ! Player sticks with original door.
440 END IF
450 IF Player = Car THEN
460 LET IsGameWon = 1
470 ELSE
480 LET IsGameWon = 0
490 END IF
500 END FUNCTION
 Output:
(example)
NOT switching doors wins car in 32.3% of games. But switching doors wins car in 67.3% of games.
BASIC256
numTiradas = 1000000
permanece = 0
cambia = 0
for i = 1 to numTiradas
pta_coche = int(rand * 3) + 1
pta_elegida = int(rand * 3) + 1
if pta_coche <> pta_elegida then
pta_montys = 6  pta_coche  pta_elegida
else
do
pta_montys = int(Rand * 3) + 1
until pta_montys <> pta_coche
end if
# manteenr elección
if pta_coche = pta_elegida then permanece += 1
# cambiar elección
if pta_coche = 6  pta_montys  pta_elegida then cambia +=1
next i
print "Si mantiene su elección, tiene un "; permanece / numTiradas * 100 ;"% de probabilidades de ganar."
print "Si cambia, tiene un "; cambia / numTiradas * 100; "% de probabilidades de ganar."
end
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 nonprize 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%)
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 += 6choiceshown = winner
end for
printf(1, "Switching wins %d times.\n", switchWins)
printf(1, "Staying wins %d times.\n", stayWins)
 Output:
(sample)
Switching wins 6697 times. Staying wins 3303 times.
FreeBASIC
' version 19012019
' compile with: fbc s console
Const As Integer max = 1000000
Randomize Timer
Dim As UInteger i, car_door, chosen_door, montys_door, stay, switch
For i = 1 To max
car_door = Fix(Rnd * 3) + 1
chosen_door = Fix(Rnd * 3) + 1
If car_door <> chosen_door Then
montys_door = 6  car_door  chosen_door
Else
Do
montys_door = Fix(Rnd * 3) + 1
Loop Until montys_door <> car_door
End If
'Print car_door,chosen_door,montys_door
' stay
If car_door = chosen_door Then stay += 1
' switch
If car_door = 6  montys_door  chosen_door Then switch +=1
Next
Print Using "If you stick to your choice, you have a ##.## percent" _
+ " chance to win"; stay / max * 100
Print Using "If you switched, you have a ##.## percent chance to win"; _
switch / max * 100
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
 Output:
If you stick to your choice, you have a 33.32 percent chance to win If you switched, you have a 66.68 percent chance to win
FutureBasic
// Monty Hall problem
// May 2024  Rich Love
local fn IsGameWon(Sw as int) as bool
REM Play one game.
REM Switching if and only if Sw <> 0.
REM Returns 1 if the game is won, 0 otherwise.
short Car = INT(RND(3)) //Randomly place car behind a door.
short Player0 = INT(RND(3)) // Player randomly chooses a door.
short Monty
short IsGameWon
short Player
DO
Monty = INT(RND(3)) // Monty opens door revealing a goat.
UNTIL (Monty <> Car) AND (Monty <> Player0)
380
IF Sw <> 0 // Player switches TO remaining door.
DO
Player = INT(RND(3))
UNTIL (Player <> Player0) AND (Player <> Monty)
Monty = INT(RND(3))
ELSE
Player = Player0 // Player sticks with original door.
END IF
IF Player = Car
IsGameWon = 1
ELSE
IsGameWon = 0
END IF
END fn = IsgameWon
_NGames = 10000
float NWins = 0
short Game
FOR Game = 0 TO _NGames
IF fn IsGameWon(0) <> 0 THEN NWins = NWins + 1
NEXT Game
PRINT "NOT switching doors wins car in ";
PRINT USING "##.##" ; NWins / 100 ;
PRINT " % of games."
NWins = 0
FOR Game = 0 TO _NGames
IF fn IsGameWon(1) <> 0 THEN NWins = NWins + 1
NEXT Game
PRINT "But switching doors wins car in ";
PRINT USING "##.##" ; NWins / _NGames * 100;
PRINT " % of games."
handleevents
 Output:
NOT switching doors wins car in 33.48% of games. But switching doors wins car in 67.15% of games.
ISBASIC
100 PROGRAM "MontyH.bas"
110 RANDOMIZE
120 LET NUMGAMES=1000
130 LET CHANGING,NOTCHANGING=0
140 FOR I=0 TO NUMGAMES1
150 LET PRIZEDOOR=RND(3)+1:LET CHOSENDOOR=RND(3)+1
160 IF CHOSENDOOR=PRIZEDOOR THEN
170 LET NOTCHANGING=NOTCHANGING+1
180 ELSE
190 LET CHANGING=CHANGING+1
200 END IF
210 NEXT
220 PRINT "Num of games:";NUMGAMES
230 PRINT "Wins not changing doors:";NOTCHANGING,NOTCHANGING/NUMGAMES*100;"% of total."
240 PRINT "Wins changing doors:",CHANGING,CHANGING/NUMGAMES*100;"% of total."
Liberty BASIC
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:
(example)
Result for 10000 games. Switching wins 6634 times. Staying wins 3366 times.
Minimal BASIC
10 REM Monty Hall problem
20 LET N = 10000
30 RANDOMIZE
40 LET W = 0
50 FOR G = 0 TO N
60 LET S = 0
70 GOSUB 230
80 IF V = 0 THEN 100
90 LET W = W+1
100 NEXT G
110 PRINT "NOT switching doors wins car in";
120 PRINT W/N*100; "per cent of games."
130 LET W = 0
140 FOR G = 0 TO N
150 LET S = 1
160 GOSUB 230
170 IF V = 0 THEN 190
180 LET W = W+1
190 NEXT G
200 PRINT "But switching doors wins car in";
210 PRINT W/N*100; "per cent of games."
220 END
230 REM ** Is game won?
240 REM Play one game.
250 REM Switching if and only if S <> 0.
260 REM Randomly place car behind a door.
270 LET C = INT(RND*3)
280 REM Player randomly chooses a door.
290 LET P0 = INT(RND*3)
300 REM Monty opens door revealing a goat.
310 LET M = INT(RND*3)
320 IF M = C THEN 310
330 IF M = P0 THEN 310
340 IF S = 0 THEN 410
350 REM Player switches to remaining door.
360 LET P = INT(RND*3)
370 IF P = P0 THEN 360
380 IF P = M THEN 360
390 GOTO 430
400 REM Player sticks with original door.
410 LET P = P0
420 REM Victory?
430 IF P <> C THEN 460
440 LET V = 1
450 RETURN
460 LET V = 0
470 RETURN
PureBasic
Structure wins
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()
Output:
Trial runs for each option: 1000000 Wins when redeciding: 666459 (66.65% chance) Wins when sticking: 333541 (33.35% chance)
QuickBASIC
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.
Run BASIC
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."
Sinclair ZX81 BASIC
Works with 1k of RAM.
This program could certainly be made more efficient. What is really going on, after all, is
if initial guess = car then sticker wins else switcher wins;
but I take it that the point is to demonstrate the outcome to people who may not see that that's what is going on. I have therefore written the program in a deliberately naïve style, not assuming anything.
10 PRINT " WINS IF YOU"
20 PRINT "STICK","SWITCH"
30 LET STICK=0
40 LET SWITCH=0
50 FOR I=1 TO 1000
60 LET CAR=INT (RND*3)
70 LET GUESS=INT (RND*3)
80 LET SHOW=INT (RND*3)
90 IF SHOW=GUESS OR SHOW=CAR THEN GOTO 80
100 LET NEWGUESS=INT (RND*3)
110 IF NEWGUESS=GUESS OR NEWGUESS=SHOW THEN GOTO 100
120 IF GUESS=CAR THEN LET STICK=STICK+1
130 IF NEWGUESS=CAR THEN LET SWITCH=SWITCH+1
140 NEXT I
150 PRINT AT 2,0;STICK,SWITCH
 Output:
WINS IF YOU STICK SWITCH 341 659
True BASIC
OPTION BASE 0
DIM puertas(3)
LET numTiradas = 1000000
FOR i = 0 TO numTiradas
LET pta_coche = INT(RND * 3) + 1
LET puertas(pta_coche) = 1
LET pta_elegida = INT(RND * 3) + 1
DO
LET pta_montys = INT(RND * 3) + 1
LOOP WHILE puertas(pta_montys) = 1 OR pta_montys = pta_elegida
IF puertas(pta_elegida) = 1 THEN
LET cambia = cambia + 1
ELSE
LET permanece = permanece + 1
END IF
LET puertas(pta_coche) = 0
NEXT i
PRINT "Cambiar gana el"; permanece / numTiradas * 100; "% de las veces."
PRINT "Mantenerse gana el"; cambia / numTiradas * 100; "% de las veces."
END
Yabasic
numTiradas = 1000000
for i = 1 to numTiradas
pta_coche = int(ran(3)) + 1
pta_elegida = int(ran(3)) + 1
if pta_coche <> pta_elegida then
pta_montys = 6  pta_coche  pta_elegida
else
repeat
pta_montys = int(ran(3)) + 1
until pta_montys <> pta_coche
end if
// manteenr elección
if pta_coche = pta_elegida then permanece = permanece + 1 : fi
// cambiar elección
if pta_coche = 6  pta_montys  pta_elegida then cambia = cambia + 1 : fi
next i
print "Si mantiene su eleccion, tiene un ", permanece / numTiradas * 100, "% de probabilidades de ganar."
print "Si cambia, tiene un ", cambia / numTiradas * 100, "% de probabilidades de ganar."
end
C
//Evidence of the Monty Hall solution of marquinho1986 in C [github.com/marquinho1986]
#include <stdlib.h>
#include <stdio.h>
#include <stdbool.h>
#include <time.h>
#include <math.h>
#define NumSim 1000000000 // one billion of simulations! using the Law of large numbers concept [https://en.wikipedia.org/wiki/Law_of_large_numbers]
void main() {
unsigned long int i,stay=0;
int ChosenDoor,WinningDoor;
bool door[3]={0,0,0};
srand(time(NULL)); //initialize random seed.
for(i=0;i<=NumSim;i++){
WinningDoor=rand() % 3; // choosing winning door.
ChosenDoor=rand() % 3; // selected door.
if(door[WinningDoor]=true,door[ChosenDoor])stay++;
door[WinningDoor]=false;
}
printf("\nAfter %lu games, I won %u by staying. That is %f%%. and I won by switching %lu That is %f%%",NumSim, stay, (float)stay*100.0/(float)i,abs(NumSimstay),100(float)stay*100.0/(float)i);
}
Output of one run:
After 1000000000 games, I won 333332381 by staying. That is 33.333238%. and I won by switching 666667619 That is 66.666762%
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
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%
Chapel
Version 1 : using task parallelism.
use Random;
param doors: int = 3;
config const games: int = 1000;
config const maxTasks = 32;
var numTasks = 1;
while( games / numTasks > 1000000 && numTasks < maxTasks ) do numTasks += 1;
const tasks = 1..#numTasks;
const games_per_task = games / numTasks ;
const remaining_games = games % numTasks ;
var wins_by_stay: [tasks] int;
coforall task in tasks {
var rand = new RandomStream();
for game in 1..#games_per_task {
var player_door = (rand.getNext() * 1000): int % doors ;
var winning_door = (rand.getNext() * 1000): int % doors ;
if player_door == winning_door then
wins_by_stay[ task ] += 1;
}
if task == tasks.last then {
for game in 1..#remaining_games {
var player_door = (rand.getNext() * 1000): int % doors ;
var winning_door = (rand.getNext() * 1000): int % doors ;
if player_door == winning_door then
wins_by_stay[ task ] += 1;
}
}
}
var total_by_stay = + reduce wins_by_stay;
var total_by_switch = games  total_by_stay;
var percent_by_stay = ((total_by_stay: real) / games) * 100;
var percent_by_switch = ((total_by_switch: real) / games) * 100;
writeln( "Wins by staying: ", total_by_stay, " or ", percent_by_stay, "%" );
writeln( "Wins by switching: ", total_by_switch, " or ", percent_by_switch, "%" );
if ( total_by_stay > total_by_switch ){
writeln( "Staying is the superior method." );
} else if( total_by_stay < total_by_switch ){
writeln( "Switching is the superior method." );
} else {
writeln( "Both methods are equal." );
}
Sample output:
Wins by staying: 354 or 35.4% Wins by switching: 646 or 64.6% Switching is the superior method.
Version 2 : using data parallelism.
use Random;
config const numGames = 100_000_000;
var switch, stick: uint;
// have a separate RNG for each task; add together the results at the end
forall i in 1..numGames
with (var rand = new RandomStream(uint, parSafe = false), + reduce stick)
{
var chosen_door = rand.getNext() % 3;
var winner_door = rand.getNext() % 3;
if chosen_door == winner_door then
stick += 1;
}
// if you lost by sticking it means you would have won by switching
switch = numGames  stick;
writeln("Over ", numGames, " games:\n  switching wins ",
100.0*switch / numGames, "% of the time and\n  sticking wins ",
100.0*stick / numGames, "% of the time");
Sample output:
Over 1000000 games:  switching wins 66.6937% of the time and  sticking wins 33.3063% of the time
Clojure
(ns montyhallproblem
(:use [clojure.contrib.seq :only (shuffle)]))
(defn playgame [staying]
(let [doors (shuffle [:goat :goat :car])
choice (randint 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 (playgame staying) (inc counter) counter))
0
(range times))]
(str "wins " wins " times out of " times)))
montyhallproblem> (println "staying:" (simulate true 1000))
staying: wins 337 times out of 1000
nil
montyhallproblem> (println "switching:" (simulate false 1000))
switching: wins 638 times out of 1000
nil
COBOL
IDENTIFICATION DIVISION.
PROGRAMID. montyhall.
DATA DIVISION.
WORKINGSTORAGE SECTION.
78 NumGames VALUE 1000000.
*> These are needed so the values are passed to
*> getrandint correctly.
01 One PIC 9 VALUE 1.
01 Three PIC 9 VALUE 3.
01 doorsarea.
03 doors PIC 9 OCCURS 3 TIMES.
01 choice PIC 9.
01 shown PIC 9.
01 winner PIC 9.
01 switchwins PIC 9(7).
01 staywins PIC 9(7).
01 staywinspercent PIC Z9.99.
01 switchwinspercent PIC Z9.99.
PROCEDURE DIVISION.
PERFORM NumGames TIMES
MOVE 0 TO doors (winner)
CALL "getrandint" USING CONTENT One, Three,
REFERENCE winner
MOVE 1 TO doors (winner)
CALL "getrandint" USING CONTENT One, Three,
REFERENCE choice
PERFORM WITH TEST AFTER
UNTIL NOT(shown = winner OR choice)
CALL "getrandint" USING CONTENT One, Three,
REFERENCE shown
ENDPERFORM
ADD doors (choice) TO staywins
ADD doors (6  choice  shown) TO switchwins
ENDPERFORM
COMPUTE staywinspercent ROUNDED =
staywins / NumGames * 100
COMPUTE switchwinspercent ROUNDED =
switchwins / NumGames * 100
DISPLAY "Staying wins " staywins " times ("
staywinspercent "%)."
DISPLAY "Switching wins " switchwins " times ("
switchwinspercent "%)."
.
IDENTIFICATION DIVISION.
PROGRAMID. getrandint.
DATA DIVISION.
WORKINGSTORAGE SECTION.
01 callflag PIC X VALUE "Y".
88 firstcall VALUE "Y", FALSE "N".
01 numrange PIC 9.
LINKAGE SECTION.
01 minnum PIC 9.
01 maxnum PIC 9.
01 ret PIC 9.
PROCEDURE DIVISION USING minnum, maxnum, ret.
*> Seed RANDOM once.
IF firstcall
MOVE FUNCTION RANDOM(FUNCTION CURRENTDATE (9:8))
TO numrange
SET firstcall TO FALSE
ENDIF
COMPUTE numrange = maxnum  minnum + 1
COMPUTE ret =
FUNCTION MOD(FUNCTION RANDOM * 100000, numrange)
+ minnum
.
END PROGRAM getrandint.
END PROGRAM montyhall.
 Output:
Staying wins 0333396 times (33.34%). Switching wins 0666604 times (66.66%).
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, "stuckto" 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 11=0 if the player chose and stuck with a losing door (0), the player would
// have won by switching, so the value is 10=1
switch_wins += 1doors[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%]
Common Lisp
(defun makeround ()
(let ((array (makearray 3
:elementtype 'bit
:initialelement 0)))
(setf (bit array (random 3)) 1)
array))
(defun showgoat (initialchoice array)
(loop for i = (random 3)
when (and (/= initialchoice i)
(zerop (bit array i)))
return i))
(defun won? (array i)
(= 1 (bit array i)))
CLUSER> (progn (loop repeat #1=(expt 10 6)
for round = (makeround)
for initial = (random 3)
for goat = (showgoat 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 resultstay
when (won? round choice)
sum 1 into resultswitch
finally (progn (format t "Stay: ~S%~%" (float (/ resultstay
#1# 1/100)))
(format t "Switch: ~S%~%" (float (/ resultswitch
#1# 1/100))))))
Stay: 33.2716%
Switch: 66.6593%
;Find out how often we win if we always switch
(defun randelt (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 (randelt (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 montytrials (n)
(count t (loop for x from 1 to n collect (monty))))
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
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%
Delphi
program MontyHall;
{$APPTYPE CONSOLE}
{$R *.res}
uses
System.SysUtils;
const
numGames = 1000000; // Number of games to run
var
switchWins, stayWins, plays: Int64;
doors: array[0..2] of Integer;
i, winner, choice, shown: Integer;
begin
switchWins := 0;
stayWins := 0;
for plays := 1 to numGames do
begin
//0 is a goat, 1 is a car
for i := 0 to 2 do
doors[i] := 0;
//put a winner in a random door
winner := Random(3);
doors[winner] := 1;
//pick a door, any door
choice := Random(3);
//don't show the winner or the choice
repeat
shown := Random(3);
until (doors[shown] <> 1) and (shown <> choice);
//if you won by staying, count it
stayWins := stayWins + doors[choice];
//the switched (last remaining) door is (3  choice  shown), because 0+1+2=3
switchWins := switchWins + doors[3  choice  shown];
end;
WriteLn('Staying wins ' + IntToStr(stayWins) + ' times.');
WriteLn('Switching wins ' + IntToStr(switchWins) + ' times.');
end.
 Output:
Staying wins 333253 times. Switching wins 666747 times.
Dyalect
var switchWins = 0
var stayWins = 0
for plays in 0..1000000 {
var doors = [0 ,0, 0]
var winner = rnd(max: 3)
doors[winner] = 1
var choice = rnd(max: 3)
var shown = rnd(max: 3)
while doors[shown] == 1  shown == choice {
shown = rnd(max: 3)
}
stayWins += doors[choice]
switchWins += doors[3  choice  shown]
}
print("Staying wins \(stayWins) times.")
print("Switching wins \(switchWins) times.")
 Output:
Staying wins 286889 times. Switching wins 713112 times.
EasyLang
max = 1000000
for i = 1 to max
car_door = randint 3
chosen_door = randint 3
if car_door <> chosen_door
montys_door = 6  car_door  chosen_door
else
repeat
montys_door = randint 3
until montys_door <> car_door
.
.
if car_door = chosen_door
stay += 1
.
if car_door = 6  montys_door  chosen_door
switch += 1
.
.
print "If you stick to your choice, you have a " & stay / max * 100 & " percent chance to win"
print "If you switched, you have a " & switch / max * 100 & " percent chance to win"
 Output:
If you stick to your choice, you have a 33.36 percent chance to win If you switched, you have a 66.64 percent chance to win
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 %
Elixir
defmodule MontyHall do
def simulate(n) do
{stay, switch} = simulate(n, 0, 0)
:io.format "Staying wins ~w times (~.3f%)~n", [stay, 100 * stay / n]
:io.format "Switching wins ~w times (~.3f%)~n", [switch, 100 * switch / n]
end
defp simulate(0, stay, switch), do: {stay, switch}
defp simulate(n, stay, switch) do
doors = Enum.shuffle([:goat, :goat, :car])
guess = :rand.uniform(3)  1
[choice] = [0,1,2]  [guess, shown(doors, guess)]
if Enum.at(doors, choice) == :car, do: simulate(n1, stay, switch+1),
else: simulate(n1, stay+1, switch)
end
defp shown(doors, guess) do
[i, j] = Enum.shuffle([0,1,2]  [guess])
if Enum.at(doors, i) == :goat, do: i, else: j
end
end
MontyHall.simulate(10000)
 Output:
Staying wins 3397 times (33.970%) Switching wins 6603 times (66.030%)
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))))
(message "Strategy keep: %.3f%%" (/ cnt 100.0)))
(let ((cnt 0))
(dotimes (i 10000)
(and (montyhall nil) (setq cnt (1+ cnt))))
(message "Strategy switch: %.3f%%" (/ cnt 100.0)))
 Output:
Strategy keep: 34.410% Strategy switch: 66.430%
EMal
type Prize
enum do int GOAT, CAR end
type Door
model
int id
Prize prize
new by int =id, Prize =prize do end
fun asText = text by block do return "(id:" + me.id + ", prize:" + me.prize.value + ")" end
end
type Player
model
Door choice
fun choose = void by List doors
me.choice = doors[random(3)]
end
end
type Monty
model
fun setPrize = void by List doors, Prize prize
doors[random(3)].prize = prize
end
end
type MontyHallProblem
int ITERATIONS = 1000000
Map counter = text%int[ "keep" => 0, "switch" => 0 ]
writeLine("Simulating " + ITERATIONS + " games:")
for int i = 0; i < ITERATIONS; i++
if i % 100000 == 0 do write(".") end
^three numbered doors with no cars for now^
List doors = Door[Door(1, Prize.GOAT), Door(2, Prize.GOAT), Door(3, Prize.GOAT)]
Monty monty = Monty() # set up Monty
monty.setPrize(doors, Prize.CAR) # Monty randomly sets the car behind one door
Player player = Player() # set up the player
player.choose(doors) # the player makes a choice
^here Monty opens a door with a goat;
behind the ones that are still closed there is a car and a goat,
so that the player *always* wins by keeping or switching.
^
counter[when(player.choice.prize == Prize.CAR, "keep", "switch")]++
end
writeLine()
writeLine(counter)
 Output:
Simulating 1000000 games: .......... [keep:332376,switch:667624]
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.
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
Forth
version 1
include random.fs
variable staywins
variable switchwins
: trial (  )
3 random 3 random ( prize choice )
= if 1 staywins +!
else 1 switchwins +!
then ;
: trials ( n  )
0 staywins ! 0 switchwins !
dup 0 do trial loop
cr staywins @ . [char] / emit dup . ." staying wins"
cr switchwins @ . [char] / emit . ." switching wins" ;
1000 trials
or in iForth:
0 value staywins
0 value switchwins
: trial (  )
3 choose 3 choose (  prize choice )
= IF 1 +TO staywins exit ENDIF
1 +TO switchwins ;
: trials ( n  )
CLEAR staywins
CLEAR switchwins
dup 0 ?DO trial LOOP
CR staywins DEC. ." / " dup DEC. ." staying wins,"
CR switchwins DEC. ." / " DEC. ." switching wins." ;
With output:
FORTH> 100000000 trials 33336877 / 100000000 staying wins, 66663123 / 100000000 switching wins. ok
version 2
While Forthers are known (and regarded) for always simplifying the problem, I think version 1 is missing the point here. The optimization can only be done if one already understands the game. For what it's worth, here is a simulation that takes all the turns of the game.
require random.fs
here seed !
1000000 constant rounds
variable wins
variable car
variable firstPick
variable revealed
defer applyStrategy
: isCar ( u  f ) car @ = ;
: remaining ( u u  u ) 3 swap  swap  ;
: setup 3 random car ! ;
: choose 3 random firstPick ! ;
: otherGoat (  u ) car @ firstPick @ remaining ;
: randomGoat (  u ) car @ 1+ 2 random + 3 mod ;
: reveal firstPick @ isCar IF randomGoat ELSE otherGoat THEN revealed ! ;
: keep (  u ) firstPick @ ;
: switch (  u ) firstPick @ revealed @ remaining ;
: open ( u  f ) isCar ;
: play (  f ) setup choose reveal applyStrategy open ;
: record ( f ) 1 and wins +! ;
: run 0 wins ! rounds 0 ?DO play record LOOP ;
: result wins @ 0 d>f rounds 0 d>f f/ 100e f* ;
: .result result f. '%' emit ;
' keep IS applyStrategy run ." Keep door => " .result cr
' switch IS applyStrategy run ." Switch door => " .result cr
bye
 Output:
Keep door => 33.2922 % Switch door => 66.7207 %
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%
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website.
In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.
Solution
The following program makes a given number of simulations. On each, three options are evaluated:
 If the player keeps his/her selection
 If the player randomly chooses between hs/her selection and the other (closed) door.
 If the player switches his/her selection
Finally, it shows the number of wins for each case.
It can be seen that:
 If the player keeps his/her selection, he/she wins around 1/3 of times
 If the player randomly chooses between his/her selection and the other (closed) door, he/she wins around 1/2 of times
 If the player switches his/her selection, he/she wins around 2/3 of times
The following variation shows the evolution of the probabilities for each case:
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%)
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.
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 goatdoor
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;
Icon and Unicon
Sample Output:
Monty Hall simulation for 10000 rounds. Strategy 1 'Staying' won 0.3266 Strategy 2 'Switching' won 0.6734
Io
keepWins := 0
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
Sample output:
Switching to the other door won 66935 times. Keeping the same door won 33065 times. Game played 100000 times with 3 doors.
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.y
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 behaves or how the player behaves):
1 2 3 4 simulate 1000
limit 1000
stay 233
switch 388
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.
JavaScript
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%"}
In the above code/problem version with n doors, only one "losing" door is opened/shown by the show host before the possibility of switch. There is a generalization to the problem in which the show host progressively opens losing doors one by one until two remains. In this case, the win probability of switching increases as the number of door increases. This has been discussed in a [1] 2009 article.
Slight modification of the script above for modularity inside of HTML.
<html>
<body>
<input id="userInputMH" value="1000">
<input id="userInputDoor" value="3">
<br>
<button onclick="montyhall()">Calculate</button>
<p id="firstPickWins"></p>
<p id="switchPickWins"></p>
</body>
</html>
<script>
function montyhall() {
var tests = document.getElementById("userInputMH").value;
var doors = document.getElementById("userInputDoor").value;
var prizeDoor, chosenDoor, shownDoor, switchDoor, chosenWins = 0,switchWins = 0;
function pick(excludeA, excludeB) {
var door;
do {
door = Math.floor(Math.random() * doors);
} while (door === excludeA  door === excludeB);
return door;
}
for (var i = 0; i < tests; i++) {
prizeDoor = pick();
chosenDoor = pick();
shownDoor = pick(prizeDoor, chosenDoor);
switchDoor = pick(chosenDoor, shownDoor);
if (chosenDoor === prizeDoor) {
chosenWins++;
} else if (switchDoor === prizeDoor) {
switchWins++;
}
}
document.getElementById("firstPickWins").innerHTML = 'First Door Wins: ' + chosenWins + '  ' + (100 * chosenWins / tests) + '%';
document.getElementById("switchPickWins").innerHTML = 'Switched Door Wins: ' + switchWins + '  ' + (100 * switchWins / tests) + '%';
}
</script>
Output:
(1000, 3)
First Door Wins: 346  34.6%
Switching Door Wins: 654  65.4%
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.
*
* I lose when I initially guess the winning door and then switch,
* or initially guess a losing door and don't 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
jq
Works with gojq, the Go implementation of jq
Neither the C nor the Go implementations of jq currently provides a PRN generator, so this entry uses /dev/urandom as an external source of entropy as follows:
cat /dev/urandom  tr cd '09'  fold w 1  jq nrf montyhall.jq
where montyhall.jq contains one of the following jq programs.
Basic solution
This solution is based on the observation:
def rand:
input as $r
 if $r < . then $r else rand end;
def logical_montyHall:
. as $games
 {switchWins: 0, stayWins: 0}
 reduce range (0; $games) as $_ (.;
(3rand) as $car # put car in a random door
 (3rand) as $choice # choose a door at random
 if $choice == $car then .stayWins += 1
else .switchWins += 1
end )
 "Simulating \($games) games:",
"Staying wins \(.stayWins) times",
"Switching wins \(.switchWins) times\n" ;
1e3, 1e6  logical_montyHall
Simulation
def rand:
input as $r
 if $r < . then $r else rand end;
def montyHall:
. as $games
 [range(0;3)  0 ] as $doors0
 {switchWins: 0, stayWins: 0}
 reduce range (0; $games) as $_ (.;
($doors0  .[3rand] = 1) as $doors # put car in a random door
 (3rand) as $choice # choose a door at random
 .stop = false
 until (.stop;
.shown = (3rand) # the shown door
 if ($doors[.shown] != 1 and .shown != $choice)
then .stop=true
else .
end)
 .stayWins += $doors[$choice]
 .switchWins += $doors[3  $choice  .shown]
)
 "Simulating \($games) games:",
"Staying wins \(.stayWins) times",
"Switching wins \(.switchWins) times\n" ;
1e3, 1e6  montyHall
 Output:
Simulating 1000 games: Staying wins 325 times Switching wins 675 times Simulating 1000000 games: Staying wins 333253 times Switching wins 666747 times
Julia
To make things interesting, I decided to generalize the problem to ncur doors and ncar cars. To allow the MC to always show a goat behind a door after the contestant chooses, .
I was was of two minds on the type of simulation to provide, so I wrote two different simulators. The literal simulator mimics the mechanics of the game as literally as possible, shuffling the arrangement of cars behind doors and randomizes all choices. This avoids any feel of cheating but results in rather complex code. The clean simulator implements the game more elegantly but it might look like cheating.
The Literal Simulation Function
using Printf
function play_mh_literal{T<:Integer}(ncur::T=3, ncar::T=1)
ncar < ncur  throw(DomainError())
curtains = shuffle(collect(1:ncur))
cars = curtains[1:ncar]
goats = curtains[(ncar+1):end]
pick = rand(1:ncur)
isstickwin = pick in cars
deleteat!(curtains, findin(curtains, pick))
if !isstickwin
deleteat!(goats, findin(goats, pick))
end
if length(goats) > 0 # reveal goat
deleteat!(curtains, findin(curtains, shuffle(goats)[1]))
else # no goats, so reveal car
deleteat!(curtains, rand(1:(ncur1)))
end
pick = shuffle(curtains)[1]
isswitchwin = pick in cars
return (isstickwin, isswitchwin)
end
The Clean Simulation Function
function play_mh_clean{T<:Integer}(ncur::T=3, ncar::T=1)
ncar < ncur  throw(DomainError())
pick = rand(1:ncur)
isstickwin = pick <= ncar
pick = rand(1:(ncur2))
if isstickwin # remove initially picked car from consideration
pick += 1
end
isswitchwin = pick <= ncar
return (isstickwin, isswitchwin)
end
Supporting Functions
function mh_results{T<:Integer}(ncur::T, ncar::T,
nruns::T, play_mh::Function)
stickwins = 0
switchwins = 0
for i in 1:nruns
(isstickwin, isswitchwin) = play_mh(ncur, ncar)
if isstickwin
stickwins += 1
end
if isswitchwin
switchwins += 1
end
end
return (stickwins/nruns, switchwins/nruns)
end
function mh_analytic{T<:Integer}(ncur::T, ncar::T)
stickodds = ncar/ncur
switchodds = (ncar  stickodds)/(ncur2)
return (stickodds, switchodds)
end
function show_odds{T<:Real}(a::T, b::T)
@sprintf " %.1f %.1f %.2f" 100.0*a 100*b 1.0*b/a
end
function show_simulation{T<:Integer}(ncur::T, ncar::T, nruns::T)
println()
print("Simulating a ", ncur, " door, ", ncar, " car ")
println("Monty Hall problem with ", nruns, " runs.\n")
println(" Solution Stick Switch Improvement")
(a, b) = mh_results(ncur, ncar, nruns, play_mh_literal)
println(@sprintf("%10s: ", "Literal"), show_odds(a, b))
(a, b) = mh_results(ncur, ncar, nruns, play_mh_clean)
println(@sprintf("%10s: ", "Clean"), show_odds(a, b))
(a, b) = mh_analytic(ncur, ncar)
println(@sprintf("%10s: ", "Analytic"), show_odds(a, b))
println()
return nothing
end
Main
for i in 3:5, j in 1:(i2)
show_simulation(i, j, 10^5)
end
This code shows, for a variety of configurations, the results for 3 solutions: literal simulation, clean simulation, analytic. Stick is the percentage of times that the player wins a car by sticking to an initial choice. Switch is the winning percentage the comes with switching one's selection following the goat reveal. Improvement is the ratio of switch to stick.
 Output:
Simulating a 3 door, 1 car Monty Hall problem with 100000 runs. Solution Stick Switch Improvement Literal: 33.2 66.8 2.02 Clean: 33.4 66.6 2.00 Analytic: 33.3 66.7 2.00 Simulating a 4 door, 1 car Monty Hall problem with 100000 runs. Solution Stick Switch Improvement Literal: 25.1 37.5 1.49 Clean: 24.7 37.6 1.52 Analytic: 25.0 37.5 1.50 Simulating a 4 door, 2 car Monty Hall problem with 100000 runs. Solution Stick Switch Improvement Literal: 49.7 75.3 1.51 Clean: 49.9 74.9 1.50 Analytic: 50.0 75.0 1.50 Simulating a 5 door, 1 car Monty Hall problem with 100000 runs. Solution Stick Switch Improvement Literal: 20.2 26.5 1.31 Clean: 20.0 26.8 1.34 Analytic: 20.0 26.7 1.33 Simulating a 5 door, 2 car Monty Hall problem with 100000 runs. Solution Stick Switch Improvement Literal: 40.0 53.5 1.34 Clean: 40.4 53.4 1.32 Analytic: 40.0 53.3 1.33 Simulating a 5 door, 3 car Monty Hall problem with 100000 runs. Solution Stick Switch Improvement Literal: 60.3 79.9 1.33 Clean: 59.9 80.1 1.34 Analytic: 60.0 80.0 1.33
Literal versus Clean
The clean simulation runs significantly faster and uses less memory.
julia> @time mh_results(3, 1, 10^5, play_mh_literal) elapsed time: 0.346965522 seconds (183790752 bytes allocated, 27.56% gc time) (0.33216,0.66784) julia> @time mh_results(3, 1, 10^5, play_mh_clean) elapsed time: 0.046481738 seconds (9600160 bytes allocated) (0.33241,0.66759)
K
montyhall:{t:,/ 1_ x {`prng@`t[];ch:1?3;pr:1?3;sw:1?2;$[sw;a:ch=pr;a:~(ch=pr)];a}\0N;("KEEP %";(+/t)%x;"SWAP %";(+/~t)%x)}
montyhall 100000
Kotlin
// version 1.1.2
import java.util.Random
fun montyHall(games: Int) {
var switchWins = 0
var stayWins = 0
val rnd = Random()
(1..games).forEach {
val doors = IntArray(3) // all zero (goats) by default
doors[rnd.nextInt(3)] = 1 // put car in a random door
val choice = rnd.nextInt(3) // choose a door at random
var shown: Int
do {
shown = rnd.nextInt(3) // the shown door
}
while (doors[shown] == 1  shown == choice)
stayWins += doors[choice]
switchWins += doors[3  choice  shown]
}
println("Simulating $games games:")
println("Staying wins $stayWins times")
println("Switching wins $switchWins times")
}
fun main(args: Array<String>) {
montyHall(1_000_000)
}
Sample output:
 Output:
Simulating 1000000 games: Staying wins 333670 times Switching wins 666330 times
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
Lua/Torch
function montyHall(n)
local car = torch.LongTensor(n):random(3)  door with car
local choice = torch.LongTensor(n):random(3)  player's choice
local opens = torch.LongTensor(n):random(2):csub(1):mul(2):csub(1)  1 or +1
local iscarchoice = choice:eq(car)
local nocarchoice = 1iscarchoice
opens[iscarchoice] = (((opens + choice  1) % 3):abs() + 1)[iscarchoice]
opens[nocarchoice] = (6  car  choice)[nocarchoice]
local change = torch.LongTensor(n):bernoulli()  0: stay, 1: change
local win = iscarchoice:long():cmul(1change) + nocarchoice:long():cmul(change)
return car, choice, opens, change, win
end
function montyStats(n)
local car, pchoice, opens, change, win = montyHall(n)
local change_and_win = change [ win:byte()]:sum()/ change :sum()*100
local no_change_and_win = (1change)[ win:byte()]:sum()/(1change):sum()*100
local change_and_win_not = change [1win:byte()]:sum()/ change :sum()*100
local no_change_and_win_not = (1change)[1win:byte()]:sum()/(1change):sum()*100
print(string.format(" %9s %9s" , "no change", "change" ))
print(string.format("win %8.4f%% %8.4f%%", no_change_and_win , change_and_win ))
print(string.format("win not %8.4f%% %8.4f%%", no_change_and_win_not, change_and_win_not))
end
montyStats(1e7)
Output for 10 million samples:
no change change win 33.3008% 66.6487% win not 66.6992% 33.3513%
M2000 Interpreter
Module CheckIt {
Enum Strat {Stay, Random, Switch}
total=10000
Print $("0.00")
player_win_stay=0
player_win_switch=0
player_win_random=0
For i=1 to total {
Dim doors(1 to 3)=False
doors(Random(1,3))=True
guess=Random(1,3)
Inventory other
for k=1 to 3 {
If k <> guess Then Append other, k
}
If doors(guess) Then {
Mont_Hall_show=other(Random(0,1)!)
} Else {
If doors(other(0!)) Then {
Mont_Hall_show=other(1!)
} Else Mont_Hall_show=other(0!)
Delete Other, Mont_Hall_show
}
Strategy=Each(Strat)
While Strategy {
Select Case Eval(strategy)
Case Random
{
If Random(1,2)=1 Then {
If doors(guess) Then player_win_Random++
} else If doors(other(0!)) Then player_win_Random++
}
Case Switch
If doors(other(0!)) Then player_win_switch++
Else
If doors(guess) Then player_win_stay++
End Select
}
}
Print "Stay: ";player_win_stay/total*100;"%"
Print "Random: ";player_win_Random/total*100;"%"
Print "Switch: ";player_win_switch/total*100;"%"
}
CheckIt
 Output:
Stay: 33.39% Random: 51.00% Switch: 66.61%
Mathematica/Wolfram Language
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]
MATLAB
wins = ceil(3*rand(1e8,1))  ceil(3*rand(1e8,1))
mprintf('chance to win for staying: %1.6f %%\nchance to win for changing: %1.6f %%', 100*length(wins(wins==0))/length(wins), 100*length(wins(wins<>0))/length(wins))
OUTPUT:
chance to win for staying: 33.334694 %
chance to win for changing: 66.665306 %
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%
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%
Modula2
MODULE MontyHallProblem;
FROM STextIO IMPORT
WriteLn, WriteString;
FROM RandomNumbers IMPORT
Randomize, Rnd;
FROM SRealIO IMPORT
WriteFixed;
CONST
NGames = 10000; (* number of games simulated *)
VAR
NWins, Game: CARDINAL;
PROCEDURE IsGameWon(Sw: BOOLEAN): BOOLEAN;
(* Play one game. *)
VAR
Car, Player, Player0, Monty: CARDINAL;
BEGIN
Car := Rnd(3); (* Randomly place car behind a door. *)
Player0 := Rnd(3); (* Player randomly chooses a door. *)
REPEAT
Monty := Rnd(3); (* Monty opens door revealing a goat. *)
UNTIL (Monty <> Car) AND (Monty <> Player0);
IF Sw THEN
(* Player switches to remaining door. *)
REPEAT
Player := Rnd(3);
UNTIL (Player <> Player0) AND (Player <> Monty)
ELSE
Player := Player0 (* Player sticks with original door. *)
END;
RETURN (Player = Car);
END IsGameWon;
BEGIN
Randomize(0);
NWins := 0;
FOR Game := 0 TO NGames DO
IF IsGameWon(FALSE) THEN
NWins := NWins + 1
END
END;
WriteString('NOT switching doors wins car in ');
WriteFixed(FLOAT(NWins) / FLOAT(NGames) * 100.0, 1, 4);
WriteString('% of games.');
WriteLn;
NWins := 0;
FOR Game := 0 TO NGames DO
IF IsGameWon(TRUE) THEN
NWins := NWins + 1
END
END;
WriteString('But switching doors wins car in ');
WriteFixed(FLOAT(NWins) / FLOAT(NGames) * 100.0, 1, 4);
WriteString('% of games.');
WriteLn;
END MontyHallProblem.
 Output:
(example)
NOT switching doors wins car in 33.4% of games. But switching doors wins car in 66.5% of games.
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.
Nim
import random
randomize()
proc shuffle[T](x: var seq[T]) =
for i in countdown(x.high, 0):
let j = rand(i)
swap(x[i], x[j])
# 1 represents a car
# 0 represent a goat
var
stay = 0 # amount won if stay in the same position
switch = 0 # amount won if you switch
for i in 1..1000:
var lst = @[1,0,0] # one car and two goats
shuffle(lst) # shuffles the list randomly
let ran = rand(2 ) # gets a random number for the random guess
let user = lst[ran] # storing the random guess
del lst, ran # deleting the random guess
var 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
inc huh
if user == 1: # if the original choice is 1 then stay adds 1
inc stay
if lst[0] == 1: # if the switched value is 1 then switch adds 1
inc switch
echo "Stay = ",stay
echo "Switch = ",switch
Output:
Stay = 337 Switch = 663
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
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]
Pascal
program MontyHall;
uses
sysutils;
const
NumGames = 1000;
{Randomly pick a door(a number between 0 and 2}
function PickDoor(): Integer;
begin
Exit(Trunc(Random * 3));
end;
var
i: Integer;
PrizeDoor: Integer;
ChosenDoor: Integer;
WinsChangingDoors: Integer = 0;
WinsNotChangingDoors: Integer = 0;
begin
Randomize;
for i := 0 to NumGames  1 do
begin
//randomly picks the prize door
PrizeDoor := PickDoor;
//randomly chooses a door
ChosenDoor := PickDoor;
//if the strategy is not changing doors the only way to win is if the chosen
//door is the one with the prize
if ChosenDoor = PrizeDoor then
Inc(WinsNotChangingDoors);
//if the strategy is changing doors the only way to win is if we choose one
//of the two doors that hasn't the prize, because when we change we change to the prize door.
//The opened door doesn't have a prize
if ChosenDoor <> PrizeDoor then
Inc(WinsChangingDoors);
end;
Writeln('Num of games:' + IntToStr(NumGames));
Writeln('Wins not changing doors:' + IntToStr(WinsNotChangingDoors) + ', ' +
FloatToStr((WinsNotChangingDoors / NumGames) * 100) + '% of total.');
Writeln('Wins changing doors:' + IntToStr(WinsChangingDoors) + ', ' +
FloatToStr((WinsChangingDoors / NumGames) * 100) + '% of total.');
end.
Output:
Num of games:1000 Wins not changing doors:359, 35,9% of total. Wins changing doors:641, 64,1% of total.
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";
Phix
Modified copy of Euphoria
with javascript_semantics integer swapWins = 0, stayWins = 0, winner, choice, reveal, other atom t0 = time() for game=1 to 1_000_000 do winner = rand(3) choice = rand(3) while 1 do reveal = rand(3) if reveal!=winner and reveal!=choice then exit end if end while stayWins += (choice==winner) other = 6choicereveal  (as 1+2+3=6, and reveal!=choice) swapWins += (other==winner) end for printf(1, "Stay: %,d\nSwap: %,d\nTime: %s\n",{stayWins,swapWins,elapsed(time()t0)})
 Output:
Stay: 333,292 Swap: 666,708 Time: 0.2s
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%)
Picat
go =>
_ = random2(), % different seed
member(Rounds,[1000,10_000,100_000,1_000_000,10_000_000]),
println(rounds=Rounds),
SwitchWins = 0,
StayWins = 0,
NumDoors = 3,
foreach(_ in 1..Rounds)
Winner = choice(NumDoors),
Choice = choice(NumDoors),
% Shown is not needed for the simulation
% Shown = pick([Door : Door in 1..NumDoors, Door != Winner, Door != Choice]),
if Choice == Winner then
StayWins := StayWins + 1
else
SwitchWins := SwitchWins + 1
end
end,
printf("Switch win ratio %0.5f%%\n", 100.0 * SwitchWins/Rounds),
printf("Stay win ratio %0.5f%%\n", 100.0 * StayWins/Rounds),
nl,
fail,
nl.
% pick a number from 1..N
choice(N) = random(1,N).
pick(L) = L[random(1,L.len)].
 Output:
rounds = 1000 Switch win ratio 68.80000% Stay win ratio 31.20000% rounds = 10000 Switch win ratio 67.25000% Stay win ratio 32.75000% rounds = 100000 Switch win ratio 66.69700% Stay win ratio 33.30300% rounds = 1000000 Switch win ratio 66.65520% Stay win ratio 33.34480% rounds = 10000000 Switch win ratio 66.66641% Stay win ratio 33.33359%
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 %
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.
PostScript
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%
PowerShell
#Declaring variables
$intIterations = 10000
$intKept = 0
$intSwitched = 0
#Creating a function
Function PlayMontyHall()
{
#Using a .NET object for randomization
$objRandom = NewObject 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 regenerate 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 = $objRandom.Next(1,4)}
#Simple counters per win for both categories
#Because a child scope cannot change variables in the parent
#scope, the scope of the counters is expanded scriptwide.
If ($intChoice EQ $intWin)
{$script:intKept++}
If ($intSwitch EQ $intWin)
{$script:intSwitched++}
}
#Looping the Monty Hall function for $intIterations times
While ($intIterationCount LT $intIterations)
{
PlayMontyHall
$intIterationCount++
}
#Output
WriteHost "Results through $intIterations iterations:"
WriteHost "Keep : $intKept ($($intKept/$intIterations*100)%)"
WriteHost "Switch: $intSwitched ($($intSwitched/$intIterations*100)%)"
WriteHost ""
Output:
Results through 10000 iterations: Keep : 3336 (33.36%) Switch: 6664 (66.64%)
Prolog
: initialization(main).
% Simulate a play.
play(Switch, Won) :
% Random prize door
random(1, 4, P),
% Random contestant door
random(1, 4, C),
% Random reveal door, not prize or contestant door
repeat,
random(1, 4, R),
R \= P,
R \= C,
!,
% Final door
(
Switch, between(1, 3, F), F \= C, F \= R, !;
\+ Switch, F = C
),
% Check result.
(F = P > Won = true ; Won = false).
% Count wins.
win_count(0, _, Total, Total).
win_count(I, Switch, A, Total) :
I > 0,
I1 is I  1,
play(Switch, Won),
(Won, A1 is A + 1;
\+ Won, A1 is A),
win_count(I1, Switch, A1, Total).
main :
randomize,
win_count(1000, true, 0, SwitchTotal),
format('Switching wins ~d out of 1000.\n', [SwitchTotal]),
win_count(1000, false, 0, StayTotal),
format('Staying wins ~d out of 1000.\n', [StayTotal]).
 Output:
Switching wins 667 out of 1000. Staying wins 332 out of 1000.
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.
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
Quackery
[ $ "bigrat.qky" loadfile ] now!
[ 0 ( number of cars when not changing choice )
0 ( number of cars when changing choice )
rot times
[ 3 random ( door with goat )
3 random ( contestant's choice )
= ( If the two numbers are equal then the contestant
wins a car if they change their mind, and they win
a goat if they don't change their mind. The wins
are reversed if the numbers are not equal. )
if dip 1+ ] ( increment the relevant count )
say "Strategy A is that the contestant changes their mind." cr
say "Strategy B is that the contestant does not their mind." cr
say "Approximate ratio of car wins with strategy A over strategy B: "
swap 100 round
vulgar$ echo$ cr ] is trials ( n > )
 Output:
Running the simulation in the Quackery shell.
/O> 1000 trials ... Strategy A is that the contestant changes their mind. Strategy B is that the contestant does not their mind. Approximate ratio of car wins with strategy A over strategy B: 15/8 Stack empty. /O> 1000000 trials ... Strategy A is that the contestant changes their mind. Strategy B is that the contestant does not their mind. Approximate ratio of car wins with strategy A over strategy B: 2/1 Stack empty.
From this we can conclude that strategy B is preferable, as goats are the GOAT!
R
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 userchosen 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
Racket
#lang racket
(define (getlastdoor a b) ; assumes a != b
(vectorref '#( 2 1
2  0
1 0 )
(+ a (* 3 b))))
(define (rungame strategy)
(define cardoor (random 3))
(define firstchoice (random 3))
(define revealedgoat
(if (= cardoor firstchoice)
(let ([r (random 2)]) (if (<= cardoor r) (add1 r) r)) ; random
(getlastdoor cardoor firstchoice))) ; reveal goat
(define finalchoice (strategy firstchoice revealedgoat))
(define win? (eq? finalchoice cardoor))
;; (printf "car: ~s\nfirst: ~s\nreveal: ~s\nfinal: ~s\n => ~s\n\n"
;; cardoor firstchoice revealedgoat finalchoice
;; (if win? 'win 'lose))
win?)
(define (keepchoice firstchoice revealedgoat)
firstchoice)
(define (changechoice firstchoice revealedgoat)
(getlastdoor firstchoice revealedgoat))
(define (teststrategy strategy)
(define N 10000000)
(define wins (for/sum ([i (inrange N)]) (if (rungame strategy) 1 0)))
(printf "~a: ~a%\n"
(objectname strategy)
(exact>inexact (/ wins N 1/100))))
(foreach teststrategy (list keepchoice changechoice))
Sample Output:
keepchoice: 33.33054% changechoice: 66.67613%
Raku
(formerly 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 = flat Car, Goat xx $doors  1;
# The player chooses a door.
my Prize $initial_pick = @doors.splice(@doors.keys.pick,1)[0];
# Of the n doors remaining, the host chooses n  1 that have
# goats behind them and opens them, removing them from play.
while @doors > 1 {
@doors.splice($_,1)
when Goat
given @doors.keys.pick;
}
# 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 = 10_000;
for 3, 10 > $doors {
my atomicint @wins[2];
say "With $doors doors: ";
for Stay, 'Staying', Switch, 'Switching' > $s, $name {
(^TRIALS).race.map: {
@wins[$s]⚛++ if play($s, doors => $doors) == Car;
}
say " $name wins ",
round(100*@wins[$s] / TRIALS),
'% of the time.'
}
}
 Output:
With 3 doors: Staying wins 34% of the time. Switching wins 66% of the time. With 10 doors: Staying wins 10% of the time. Switching wins 90% of the time.
REXX
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
From the Rosetta Code:Village Pump/Run times on examples?
As per Michael Mol about showing timings for program execution times:
 Discouraging timing comparisons between different languages.
 Allowing detailed timings, if someone wants to, in the talk pages.
 But generally  like now, leaving them out.
version 2
/*REXX program simulates any number of trials of the classic TV show Monty Hall problem.*/
parse arg # seed . /*obtain the optional args from the CL.*/
if #==''  #=="," then #= 1000000 /*Not specified? Then 1 million trials*/
if datatype(seed, 'W') then call random ,, seed /*Specified? Use as a seed for RANDOM.*/
wins.= 0 /*wins.0 ≡ stay, wins.1 ≡ switching.*/
do #; door. = 0 /*initialize all doors to a value of 0.*/
car= random(1, 3); door.car= 1 /*the TV show hides a car randomly. */
?= random(1, 3); _= door.? /*the contestant picks a (random) door.*/
wins._ = wins._ + 1 /*bump count of type of win strategy.*/
end /*#*/ /* [↑] perform the loop # times. */
/* [↑] door values: 0≡goat 1≡car */
say 'switching wins ' format(wins.0 / # * 100, , 1)"% of the time."
say ' staying wins ' format(wins.1 / # * 100, , 1)"% of the time." ; say
say 'performed ' # " times with 3 doors." /*stick a fork in it, we're all done. */
 output when using the default inputs:
switching wins 66.7% of the time. staying wins 33.3% of the time. performed 1000000 times with 3 doors.
Ring
total = 10000
swapper = 0
sticker = 0
revealdoor = 0
for trial = 1 to total
prizedoor = random(3) + 1
guessdoor = random(3) + 1
if prizedoor = guessdoor
revealdoor = random(2) + 1
if prizedoor = 1 revealdoor += 1 ok
if (prizedoor = 2 and revealdoor = 2) revealdoor = 3 ok
else
revealdoor = (prizedoor ^ guessdoor)
ok
stickdoor = guessdoor
swapdoor = (guessdoor ^ revealdoor)
if stickdoor = prizedoor sticker += 1 ok
if swapdoor = prizedoor swapper += 1 ok
next
see "after a total of " + total + " trials," + nl
see "the 'sticker' won " + sticker + " times (" + floor(sticker/total*100) + "%)" + nl
see "the 'swapper' won " + swapper + " times (" + floor(swapper/total*100) + "%)" + nl
Output:
after a total of 10000 trials, the 'sticker' won 2461 times (24%) the 'swapper' won 7539 times (75%)
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.
Rust
extern crate rand;
use rand::Rng;
use rand::seq::SliceRandom;
#[derive(Clone, Copy, PartialEq)]
enum Prize {Goat , Car}
const GAMES: usize = 3_000_000;
fn main() {
let mut switch_wins = 0;
let mut rng = rand::thread_rng();
for _ in 0..GAMES {
let mut doors = [Prize::Goat; 3];
*doors.choose_mut(&mut rng).unwrap() = Prize::Car;
// You only lose by switching if you pick the car the first time
if doors.choose(&mut rng).unwrap() != &Prize::Car {
switch_wins += 1;
}
}
println!("I played the game {total} times and won {wins} times ({percent}%).",
total = GAMES,
wins = switch_wins,
percent = switch_wins as f64 / GAMES as f64 * 100.0
);
}
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 %)
Scheme
(define (randomfromlist list) (listref list (random (length list))))
(define (randompermutation list)
(if (null? list)
'()
(let* ((car (randomfromlist list))
(cdr (randompermutation (remove car list))))
(cons car cdr))))
(define (randomconfiguration) (randompermutation '(goat goat car)))
(define (randomdoor) (randomfromlist '(0 1 2)))
(define (trial strategy)
(define (doorwithgoatotherthan door strategy)
(cond ((and (not (= 0 door)) (equal? (listref strategy 0) 'goat)) 0)
((and (not (= 1 door)) (equal? (listref strategy 1) 'goat)) 1)
((and (not (= 2 door)) (equal? (listref strategy 2) 'goat)) 2)))
(let* ((configuration (randomconfiguration))
(playersfirstguess (strategy `(wouldyoupleasepickadoor?)))
(doortoshowplayer (doorwithgoatotherthan playersfirstguess
configuration))
(playersfinalguess (strategy `(thereisagoatat/wouldyouliketomove?
,playersfirstguess
,doortoshowplayer))))
(if (equal? (listref configuration playersfinalguess) 'car)
'youwin!
'youlost)))
(define (staystrategy message)
(case (car message)
((wouldyoupleasepickadoor?) (randomdoor))
((thereisagoatat/wouldyouliketomove?)
(let ((firstchoice (cadr message)))
firstchoice))))
(define (switchstrategy message)
(case (car message)
((wouldyoupleasepickadoor?) (randomdoor))
((thereisagoatat/wouldyouliketomove?)
(let ((firstchoice (cadr message))
(showngoat (caddr message)))
(car (remove firstchoice (remove showngoat '(0 1 2))))))))
(definesyntax repeat
(syntaxrules ()
((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 (prepareresult strategy results)
`(,strategy won with probability
,(exact>inexact (* 100 (/ (count 'youwin! results) (length results)))) %))
(define (comparestrategies times)
(append
(prepareresult 'staystrategy (repeat times (trial staystrategy)))
'(and)
(prepareresult 'switchstrategy (repeat times (trial switchstrategy)))))
;; > (comparestrategies 1000000)
;; (staystrategy won with probability 33.3638 %
;; and switchstrategy won with probability 66.716 %)
Scilab
// How it works:
// MontyHall() is a function with argument switch:
// it will be called 100000 times with switch=%T,
// and another 100000 times with switch=%F
function win=MontyHall(switch) //If switch==%T the player will switch
doors=zeros(1,3) //All goats
car=grand(1,1,'uin',1,3)
a(car)=1 //Place a car somewher
pick=grand(1,1,'uin',1,3) //The player picks...
if pick==car then //If the player picks right...
if switch==%T then //...and switches he will be wrong
win=%F
else //...but if he doesn't, he will be right
win=%T
end
else //If the player picks a goat...
if switch==%T then //...and switches: the other door with the goat shall be
win=%T // opened: the player will switch to the car and win
else //...but if he doesn't, he will remain by his goat
win=%F
end
end
endfunction
wins_switch=0
wins_stay=0
games=100000
for i=1:games
if MontyHall(%T)==%T then
wins_switch=wins_switch+1
end
if MontyHall(%F)==%T then
wins_stay=wins_stay+1
end
end
disp("Switching, one wins"+ascii(10)+string(wins_switch)+" games out of "+string(games))
disp("Staying, one wins"+ascii(10)+string(wins_stay)+" games out of "+string(games))
Output:
Switching, one wins 66649 games out of 100000 Staying, one wins 33403 games out of 100000
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
SETL
program monty_hall;
setrandom(0);
n_simulations := 100000;
print('Chance to win:');
print('When switching doors:', win_chance(true, n_simulations) * 100, '%');
print('When not switching doors:', win_chance(false, n_simulations) * 100, '%');
proc win_chance(switch, n_simulations);
wins := 0;
loop for i in [1..n_simulations] do
wins +:= if simulate(switch) then 1 else 0 end;
end loop;
return wins / n_simulations;
end proc;
proc simulate(switch);
doors := {1, 2, 3};
car := random doors;
goats := doors less car;
choice := random doors;
opened := random (goats less choice);
if switch then
choice := arb (doors less choice less opened);
end if;
return choice = car;
end proc;
end program;
 Output:
Chance to win: When switching doors: 66.584 % When not switching doors: 33.093 %
Sidef
var n = 1000 # number of times to play
var switchWins = (var stayWins = 0) # sum of each strategy's wins
n.times { # play the game n times
var prize = pick(^3)
var chosen = pick(^3)
var show;
do {
show = pick(^3)
} while (show ~~ [chosen, prize])
given(chosen) {
when (prize) { stayWins += 1 }
when ([3  show  prize]) { switchWins += 1 }
default { die "~ error ~" }
}
}
say ("Staying wins %.2f%% of the time." % (100.0 * stayWins / n))
say ("Switching wins %.2f%% of the time." % (100.0 * switchWins / n))
 Output:
Staying wins 31.20% of the time. Switching wins 68.80% of the time.
SPAD
montyHall(n) ==
wd:=[1+random(3) for j in 1..n]
fc:=[1+random(3) for j in 1..n]
st:=reduce(_+,[1 for j in 1..n  wd.j=fc.j])
p:=(st/n)::DoubleFloat
FORMAT(nil,"stay: ~A, switch: ~A",p,1p)$Lisp
Domain:Integer
 Output:
(1) > montyHall(1000) Compiling function montyHall with type PositiveInteger > SExpression (1) stay: 0.319, switch: 0.681 Type: SExpression (2) > montyHall(10000) (2) stay: 0.3286, switch: 0.6714 Type: SExpression (3) > montyHall(100000) (3) stay: 0.33526, switch: 0.66474 Type: SExpression
SparForte
As a structured script.
#!/usr/local/bin/spar
pragma annotate( summary, "monty" )
@( description, "Run random simulations of the Monty Hall game. Show the" )
@( description, "effects of a strategy of the contestant always keeping" )
@( description, "his first guess so it can be contrasted with the" )
@( description, "strategy of the contestant always switching his guess." )
@( description, "Simulate at least a thousand games using three doors" )
@( description, "for each strategy and show the results in such a way as" )
@( description, "to make it easy to compare the effects of each strategy." )
@( see_also, "http://rosettacode.org/wiki/Monty_Hall_problem" )
@( author, "Ken O. Burtch" );
pragma license( unrestricted );
pragma restriction( no_external_commands );
procedure monty is
num_iterations : constant positive := 100_000;
type action_type is (stay, switch);
type prize_type is (goat, pig, car);
doors : array(1..3) of prize_type;
type door_index is new positive;
 place the prizes behind random doors
procedure set_prizes is
begin
doors( 1 ) := goat;
doors( 2 ) := pig;
doors( 3 ) := car;
arrays.shuffle( doors );
end set_prizes;
 determine if the prize was chosen based on strategy
function play( action : action_type ) return prize_type is
chosen : door_index := door_index( numerics.rnd(3) );
monty : door_index;
begin
set_prizes;
for i in arrays.first(doors)..arrays.last(doors) loop
if i /= chosen and doors(i) /= car then
monty := i;
end if;
end loop;
if action = switch then
for i in arrays.first(doors)..arrays.last(doors) loop
if i /= monty and i /= chosen then
chosen := i;
exit;
end if;
end loop;
end if;
return doors( chosen );
end play;
winners : natural;  times won
pct : float;  percentage won
begin
winners := 0;
for i in 1..num_iterations loop
if play( stay ) = car then
winners := @+1;
end if;
end loop;
pct := float( winners * 100 ) / float( num_iterations );
put( "Stay: count" ) @ ( winners ) @ ( " = " ) @ ( pct ) @ ( "%" );
new_line;
winners := 0;
for i in 1..num_iterations loop
if play( switch ) = car then
winners := @+1;
end if;
end loop;
pct := float( winners * 100 ) / float( num_iterations );
put( "Switch: count" ) @ ( winners ) @ ( " = " ) @ ( pct ) @ ( "%" );
new_line;
end monty;
Standard ML
Works with SML/NJ or with MLton using the SML/NJ Util library.
val pidint = Word64.toInt(Posix.Process.pidToWord(Posix.ProcEnv.getpid()));
val rand = Random.rand(LargeInt.toInt(Time.toSeconds(Time.now())), pidint);
fun stick_win 0 wins = wins
 stick_win trial wins =
let
val winner_door = (Random.randNat rand) mod 3;
val chosen_door = (Random.randNat rand) mod 3;
in
if winner_door = chosen_door then
stick_win (trial1) (wins+1)
else
stick_win (trial1) wins
end
val trials = 1000000;
val sticks = stick_win trials 0;
val stick_winrate = 100.0 * Real.fromInt(sticks) / Real.fromInt(trials);
(* if you lost by sticking you would have won by swapping *)
val swap_winrate = 100.0  stick_winrate;
(print ("sticking = " ^ Real.toString(stick_winrate) ^ "% win rate\n");
print ("swapping = " ^ Real.toString(swap_winrate) ^ "% win rate\n"));
Output
sticking = 33.3449% win rate swapping = 66.6551% win rate
Stata
clear
set obs 1000000
gen car=runiformint(1,3)
gen choice1=runiformint(1,3)
gen succ1=car==choice1
gen shown=cond(succ1,runiformint(1,2),6carchoice1)
replace shown=shown+1 if succ1 & (car==1  car==shown)
gen choice2=6shownchoice1
gen succ2=car==choice2
tabstat succ1 succ2, s(mean)
Output
stats  succ1 succ2 + mean  .333632 .666368 
Swift
import Foundation
func montyHall(doors: Int = 3, guess: Int, switch: Bool) > Bool {
guard doors > 2, guess > 0, guess <= doors else { fatalError() }
let winningDoor = Int.random(in: 1...doors)
return winningDoor == guess ? !`switch` : `switch`
}
var switchResults = [Bool]()
for _ in 0..<1_000 {
let guess = Int.random(in: 1...3)
let wasRight = montyHall(guess: guess, switch: true)
switchResults.append(wasRight)
}
let switchWins = switchResults.filter({ $0 }).count
print("Switching would've won \((Double(switchWins) / Double(switchResults.count)) * 100)% of games")
print("Not switching would've won \(((Double(switchResults.count  switchWins)) / Double(switchResults.count)) * 100)% of games")
 Output:
Switching would've won 66.8% of games Not switching would've won 33.2% of games
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...
Transact SQL
TSQL for general case:
 BEGIN 
create table MONTY_HALL(
NOE int,
CAR int,
ALTERNATIVE int,
ORIGIN int,
[KEEP] int,
[CHANGE] int,
[RANDOM] int
)
 INIT
truncate table MONTY_HALL
declare @N int , @i int  No of Experiments and their counter
declare @rooms int ,  number of rooms
@origin int,  original choice
@car int ,  room with car
@alternative int  alternative room
select @rooms = 3, @N = 100000 , @i = 0
 EXPERIMENTS LOOP
while @i < @N begin
select @car = FLOOR(rand()*@rooms)+1 , @origin = FLOOR(rand()*@rooms)+1
select @alternative = FLOOR(rand()*(@rooms1))+1
select @alternative = case when @alternative < @origin then @alternative else @alternative + 1 end
select @alternative = case when @origin = @car then @alternative else @car end
insert MONTY_HALL
select @i,@car,@alternative,@origin,@origin,@alternative,case when rand() < 5e1 then @origin else @alternative end
select @i = @i + 1
end
 RESULTS
select avg (case when [KEEP] = CAR then 1e0 else 0e0 end )*1e2 as [% OF WINS FOR KEEP],
avg (case when [CHANGE] = CAR then 1e0 else 0e0 end )*1e2 as [% OF WINS FOR CHANGE],
avg (case when [RANDOM] = CAR then 1e0 else 0e0 end )*1e2 as [% OF WINS FOR RANDOM]
from MONTY_HALL
 END 
% OF WINS FOR KEEP % OF WINS FOR CHANGE % OF WINS FOR RANDOM    33.607 66.393 49.938
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 zerobased
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
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 nonmatch 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
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
V (Vlang)
import rand
fn main() {
games := 1_000_000
mut doors := [3]int{}
mut switch_wins, mut stay_wins, mut shown, mut guess := 0, 0, 0, 0
for _ in 1..games + 1 {
doors[rand.int_in_range(0, 3) or {exit(1)}] = 1 // Set which one has the car
guess = rand.int_in_range(0, 3) or {exit(1)} // Choose a door
for doors[shown] == 1  shown == guess {
shown = rand.int_in_range(0, 3) or {exit(1)} // Shown door
}
stay_wins += doors[guess]
switch_wins += doors[3  guess  shown]
for clear in 0..3 {if doors[clear] != 0 {doors[clear] = 0}}
}
println("Simulating ${games} games:")
println("Staying wins ${stay_wins} times at ${(f32(stay_wins) / f32(games) * 100):.2}% of games")
println("Switching wins ${switch_wins} times at ${(f32(switch_wins) / f32(games) * 100):.2}% of games")
}
 Output:
Simulating 1000000 games: Staying wins 332518 times at 33.25% of games Switching wins 667482 times at 66.75% of games
Wren
import "random" for Random
var montyHall = Fn.new { games
var rand = Random.new()
var switchWins = 0
var stayWins = 0
for (i in 1..games) {
var doors = [0] * 3 // all zero (goats) by default
doors[rand.int(3)] = 1 // put car in a random door
var choice = rand.int(3) // choose a door at random
var shown = 0
while (true) {
shown = rand.int(3) // the shown door
if (doors[shown] != 1 && shown != choice) break
}
stayWins = stayWins + doors[choice]
switchWins = switchWins + doors[3choiceshown]
}
System.print("Simulating %(games) games:")
System.print("Staying wins %(stayWins) times")
System.print("Switching wins %(switchWins) times")
}
montyHall.call(1e6)
 Output:
Sample output:
Simulating 1000000 games: Staying wins 333970 times Switching wins 666030 times
X++
//Evidence of the Monty Hall solution in Dynamics AX (by Wessel du Plooy  HiGH Software).
int changeWins = 0;
int noChangeWins = 0;
int attempts;
int picked;
int reveal;
int switchdoor;
int doors[];
for (attempts = 0; attempts < 32768; attempts++)
{
doors[1] = 0; //0 is a goat, 1 is a car
doors[2] = 0;
doors[3] = 0;
doors[(xGlobal::randomPositiveInt32() mod 3) + 1] = 1; //put a winner in a random door
picked = (xGlobal::randomPositiveInt32() mod 3) + 1; //pick a door, any door
do
{
reveal = (xGlobal::randomPositiveInt32() mod 3) + 1;
}
while (doors[reveal] == 1  reveal == picked); //don't show the winner or the choice
if (doors[picked] == 1)
noChangeWins++;
else
changeWins++;
}
print strFmt("Switching wins %1 times.", changeWins);
print strFmt("Staying wins %1 times.", noChangeWins);
pause;
Output:
Switching wins 21699 times. Staying wins 11069 times.
XPL0
def NGames = 10000; \number of games simulated
int Game, NWins;
include c:\cxpl\codes;
func int IsGameWon(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
return Player = Car;
];
[Format(2,1);
NWins:= 0;
for Game:= 0 to NGames1 do
if IsGameWon(false) then NWins:= NWins+1;
Text(0, "NOT switching doors wins car in ");
RlOut(0, float(NWins)/float(NGames)*100.0);
Text(0, "% of games.^M^J");
NWins:= 0;
for Game:= 0 to NGames1 do
if IsGameWon(true) then NWins:= NWins+1;
Text(0, "But switching doors wins car in ");
RlOut(0, float(NWins)/float(NGames)*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.
Zig
const std = @import("std");
const number_of_simulations: u32 = 10_000_000;
pub fn main() !void {
var stick_wins: u32 = 0;
var switch_wins: u32 = 0;
var doors = [3]bool{ true, false, false };
var t = std.rand.DefaultPrng.init(42);
const r = t.random();
var guess: u8 = undefined;
var door_shown: u8 = undefined;
for (0..number_of_simulations) _ {
std.rand.Random.shuffle(r, bool, &doors);
guess = r.intRangeAtMost(u8, 0, 2);
door_shown = r.intRangeAtMost(u8, 0, 2);
while (!doors[door_shown] and door_shown != guess) door_shown = r.intRangeAtMost(u8, 0, 2);
if (doors[guess]) {
stick_wins += 1;
} else {
switch_wins += 1;
}
}
std.debug.print("After {} simulations:\nStick wins: {}\nSwitch wins: {}\n", .{ number_of_simulations, stick_wins, switch_wins });
}
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%)
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