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Monty Hall problem
From Rosetta Code
Monty Hall problem is a programming task. Visitors like you are encouraged to solve it according to the task description, using any language they may happen to know.
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
- Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice? (Krauss and Wang 2003:10)
Note that the player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
Contents |
[edit] ActionScript
package {
import flash.display.Sprite;
public class MontyHall extends Sprite
{
public function MontyHall()
{
var iterations:int = 30000;
var switchWins:int = 0;
var stayWins:int = 0;
for (var i:int = 0; i < iterations; i++)
{
var doors:Array = [0, 0, 0];
doors[Math.floor(Math.random() * 3)] = 1;
var choice:int = Math.floor(Math.random() * 3);
var shown:int;
do
{
shown = Math.floor(Math.random() * 3);
} while (doors[shown] == 1 || shown == choice);
stayWins += doors[choice];
switchWins += doors[3 - choice - shown];
}
trace("Switching wins " + switchWins + " times. (" + (switchWins / iterations) * 100 + "%)");
trace("Staying wins " + stayWins + " times. (" + (stayWins / iterations) * 100 + "%)");
}
}
}
Output:
Switching wins 18788 times. (62.626666666666665%) Staying wins 11212 times. (37.37333333333333%)
[edit] Ada
-- Monty Hall Game
with Ada.Text_Io; use Ada.Text_Io;
with Ada.Float_Text_Io; use Ada.Float_Text_Io;
with ada.Numerics.Discrete_Random;
procedure Monty_Stats is
Num_Iterations : Positive := 100000;
type Action_Type is (Stay, Switch);
type Prize_Type is (Goat, Pig, Car);
type Door_Index is range 1..3;
package Random_Prize is new Ada.Numerics.Discrete_Random(Door_Index);
use Random_Prize;
Seed : Generator;
Doors : array(Door_Index) of Prize_Type;
procedure Set_Prizes is
Prize_Index : Door_Index;
Booby_Prize : Prize_Type := Goat;
begin
Reset(Seed);
Prize_Index := Random(Seed);
Doors(Prize_Index) := Car;
for I in Doors'range loop
if I /= Prize_Index then
Doors(I) := Booby_Prize;
Booby_Prize := Prize_Type'Succ(Booby_Prize);
end if;
end loop;
end Set_Prizes;
function Play(Action : Action_Type) return Prize_Type is
Chosen : Door_Index := Random(Seed);
Monty : Door_Index;
begin
Set_Prizes;
for I in Doors'range loop
if I /= Chosen and Doors(I) /= Car then
Monty := I;
end if;
end loop;
if Action = Switch then
for I in Doors'range loop
if I /= Monty and I /= Chosen then
Chosen := I;
exit;
end if;
end loop;
end if;
return Doors(Chosen);
end Play;
Winners : Natural;
Pct : Float;
begin
Winners := 0;
for I in 1..Num_Iterations loop
if Play(Stay) = Car then
Winners := Winners + 1;
end if;
end loop;
Put("Stay : count" & Natural'Image(Winners) & " = ");
Pct := Float(Winners * 100) / Float(Num_Iterations);
Put(Item => Pct, Aft => 2, Exp => 0);
Put_Line("%");
Winners := 0;
for I in 1..Num_Iterations loop
if Play(Switch) = Car then
Winners := Winners + 1;
end if;
end loop;
Put("Switch : count" & Natural'Image(Winners) & " = ");
Pct := Float(Winners * 100) / Float(Num_Iterations);
Put(Item => Pct, Aft => 2, Exp => 0);
Put_Line("%");
end Monty_Stats;
Results
Stay : count 34308 = 34.31% Switch : count 65695 = 65.69%
[edit] ALGOL 68
Translation of: C
Works with: ALGOL 68 version Standard - no extensions to language used Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386 Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
INT trials=100 000;
PROC brand = (INT n)INT: 1 + ENTIER (n * random);
PROC percent = (REAL x)STRING: fixed(100.0*x/trials,0,2)+"%";
main:
(
INT prize, choice, show, not shown, new choice;
INT stay winning:=0, change winning:=0, random winning:=0;
INT doors = 3;
[doors-1]INT other door;
TO trials DO
# put the prize somewhere #
prize := brand(doors);
# let the user choose a door #
choice := brand(doors);
# let us take a list of unchoosen doors #
INT k := LWB other door;
FOR j TO doors DO
IF j/=choice THEN other door[k] := j; k+:=1 FI
OD;
# Monty opens one... #
IF choice = prize THEN
# staying the user will win... Monty opens a random port#
show := other door[ brand(doors - 1) ];
not shown := other door[ (show+1) MOD (doors - 1 ) + 1]
ELSE # no random, Monty can open just one door... #
IF other door[1] = prize THEN
show := other door[2];
not shown := other door[1]
ELSE
show := other door[1];
not shown := other door[2]
FI
FI;
# the user randomly choose one of the two closed doors
(one is his/her previous choice, the second is the
one not shown ) #
other door[1] := choice;
other door[2] := not shown;
new choice := other door[ brand(doors - 1) ];
# now let us count if it takes it or not #
IF choice = prize THEN stay winning+:=1 FI;
IF not shown = prize THEN change winning+:=1 FI;
IF new choice = prize THEN random winning+:=1 FI
OD;
print(("Staying: ", percent(stay winning), new line ));
print(("Changing: ", percent(change winning), new line ));
print(("New random choice: ", percent(random winning), new line ))
)
Sample output:
Staying: 33.62% Changing: 66.38% New random choice: 50.17%
[edit] AWK
#!/bin/gawk -f
# Monty Hall problem
BEGIN {
srand()
doors = 3
iterations = 10000
# Behind a door:
EMPTY = "empty"; PRIZE = "prize"
# Algorithm used
KEEP = "keep"; SWITCH="switch"; RAND="random";
#
}
function monty_hall( choice, algorithm ) {
# Set up doors
for ( i=0; i<doors; i++ ) {
door[i] = EMPTY
}
# One door with prize
door[int(rand()*doors)] = PRIZE
chosen = door[choice]
del door[choice]
#if you didn't choose the prize first time around then
# that will be the alternative
alternative = (chosen == PRIZE) ? EMPTY : PRIZE
if( algorithm == KEEP) {
return chosen
}
if( algorithm == SWITCH) {
return alternative
}
return rand() <0.5 ? chosen : alternative
}
function simulate(algo){
prizecount = 0
for(j=0; j< iterations; j++){
if( monty_hall( int(rand()*doors), algo) == PRIZE) {
prizecount ++
}
}
printf " Algorithm %7s: prize count = %i, = %6.2f%%\n", \
algo, prizecount,prizecount*100/iterations
}
BEGIN {
print "\nMonty Hall problem simulation:"
print doors, "doors,", iterations, "iterations.\n"
simulate(KEEP)
simulate(SWITCH)
simulate(RAND)
}
Sample output:
bash$ ./monty_hall.awk
Monty Hall problem simulation:
3 doors, 10000 iterations.
Algorithm keep: prize count = 3411, = 34.11%
Algorithm switch: prize count = 6655, = 66.55%
Algorithm random: prize count = 4991, = 49.91%
bash$
[edit] BASIC
Works with: QuickBasic version 4.5 Translation of: Java
RANDOMIZE TIMER
DIM doors(3) '0 is a goat, 1 is a car
CLS
switchWins = 0
stayWins = 0
FOR plays = 0 TO 32767
winner = INT(RND * 3) + 1
doors(winner) = 1'put a winner in a random door
choice = INT(RND * 3) + 1'pick a door, any door
DO
shown = INT(RND * 3) + 1
'don't show the winner or the choice
LOOP WHILE doors(shown) = 1 OR shown = choice
stayWins = stayWins + doors(choice) 'if you won by staying, count it
switchWins = switchWins + doors(3 - choice - shown) 'could have switched to win
doors(winner) = 0 'clear the doors for the next test
NEXT plays
PRINT "Switching wins"; switchWins; "times."
PRINT "Staying wins"; stayWins; "times."
Output:
Switching wins 21805 times. Staying wins 10963 times.
[edit] C
#include <stdio.h>
#include <stdlib.h>
#define TRIALS 10000
int brand(int n)
{
return ( n * ((double)rand()/((double)RAND_MAX + (double)1.0 ) ) );
}
#define PERCENT(X) ( 100.0*((float)(X))/((float)TRIALS) )
int main()
{
int prize, choice, show, notshown, newchoice;
int staywinning=0, changewinning=0, randomwinning=0;
int odoor[2];
int i,j,k;
for(i=0; i < TRIALS; i++)
{
/* put the prize somewhere */
prize = brand(3);
/* let the user choose a door */
choice = brand(3);
/* let us take a list of unchoosen doors */
for (j=0, k=0; j<3; j++)
{
if (j!=choice) odoor[k++] = j;
}
/* Monty opens one... */
if ( choice == prize )
{ /* staying the user will win... Monty opens a random
port*/
show = odoor[ brand(2) ];
notshown = odoor[ (show+1)%2 ];
} else { /* no random, Monty can open just one door... */
if ( odoor[0] == prize )
{
show = odoor[1];
notshown = odoor[0];
} else {
show = odoor[0];
notshown = odoor[1];
}
}
/* the user randomly choose one of the two closed doors
(one is his/her previous choice, the second is the
one notshown */
odoor[0] = choice;
odoor[1] = notshown;
newchoice = odoor[ brand(2) ];
/* now let us count if it takes it or not */
if ( choice == prize ) staywinning++;
if ( notshown == prize ) changewinning++;
if ( newchoice == prize ) randomwinning++;
}
printf("Staying: %.2f\n", PERCENT(staywinning) );
printf("Changing: %.2f\n", PERCENT(changewinning) );
printf("New random choice: %.2f\n", PERCENT(randomwinning) );
}
Output of one run:
Staying: 33.41 Changing: 66.59 New random choice: 49.67
[edit] Common Lisp
(defun make-round ()
(let ((array (make-array 3
:element-type 'bit
:initial-element 0)))
(setf (bit array (random 3)) 1)
array))
(defun show-goat (initial-choice array)
(loop for i = (random 3)
when (and (/= initial-choice i)
(zerop (bit array i)))
return i))
(defun won? (array i)
(= 1 (bit array i)))
CL-USER> (progn (loop repeat #1=(expt 10 6)
for round = (make-round)
for initial = (random 3)
for goat = (show-goat initial round)
for choice = (loop for i = (random 3)
when (and (/= i initial)
(/= i goat))
return i)
when (won? round (random 3))
sum 1 into result-stay
when (won? round choice)
sum 1 into result-switch
finally (progn (format t "Stay: ~S%~%" (float (/ result-stay
#1# 1/100)))
(format t "Switch: ~S%~%" (float (/ result-switch
#1# 1/100))))))
Stay: 33.2716%
Switch: 66.6593%
[edit] C++
#include <iostream>
#include <cstdlib>
#include <ctime>
int randint(int n)
{
return (1.0*n*std::rand())/(1.0+RAND_MAX);
}
int other(int doorA, int doorB)
{
int doorC;
if (doorA == doorB)
{
doorC = randint(2);
if (doorC >= doorA)
++doorC;
}
else
{
for (doorC = 0; doorC == doorA || doorC == doorB; ++doorC)
{
// empty
}
}
return doorC;
}
int check(int games, bool change)
{
int win_count = 0;
for (int game = 0; game < games; ++game)
{
int const winning_door = randint(3);
int const original_choice = randint(3);
int open_door = other(original_choice, winning_door);
int const selected_door = change?
other(open_door, original_choice)
: original_choice;
if (selected_door == winning_door)
++win_count;
}
return win_count;
}
int main()
{
std::srand(std::time(0));
int games = 10000;
int wins_stay = check(games, false);
int wins_change = check(games, true);
std::cout << "staying: " << 100.0*wins_stay/games << "%, changing: " << 100.0*wins_change/games << "%\n";
}
Sample output:
staying: 33.73%, changing: 66.9%
[edit] D
import std.stdio, std.random;
void main() {
Random gen = Random(unpredictableSeed);
uint switchWins = 0, stayWins = 0;
while(switchWins + stayWins < 100_000) {
uint carPos = dice(gen, 1, 1, 1); // Which door is car behind?
uint pickPos = dice(gen, 1, 1, 1); // Contestant's initial pick.
uint 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 = dice(gen, 1, 1, 1);
} while(openPos == pickPos || openPos == carPos);
uint switchPos = 0;
// 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 wins: %d Staying wins: %d", switchWins, stayWins);
}
Output:
Switching wins: 66673 Staying wins: 33327
[edit] Forth
include random.fs
variable stay-wins
variable switch-wins
: trial ( -- )
3 random 3 random ( prize choice )
= if 1 stay-wins +!
else 1 switch-wins +!
then ;
: trials ( n -- )
0 stay-wins ! 0 switch-wins !
dup 0 do trial loop
cr stay-wins @ . [char] / emit dup . ." staying wins"
cr switch-wins @ . [char] / emit . ." switching wins" ;
1000 trials
or in iForth:
0 value stay-wins
0 value switch-wins
: trial ( -- )
3 choose 3 choose ( -- prize choice )
= IF 1 +TO stay-wins exit ENDIF
1 +TO switch-wins ;
: trials ( n -- )
CLEAR stay-wins
CLEAR switch-wins
dup 0 ?DO trial LOOP
CR stay-wins DEC. ." / " dup DEC. ." staying wins,"
CR switch-wins DEC. ." / " DEC. ." switching wins." ;
With output:
FORTH> 100000000 trials 33336877 / 100000000 staying wins, 66663123 / 100000000 switching wins. ok
[edit] Fortran
Works with: Fortran version 90 and later
PROGRAM MONTYHALL
IMPLICIT NONE
INTEGER, PARAMETER :: trials = 10000
INTEGER :: i, choice, prize, remaining, show, staycount = 0, switchcount = 0
LOGICAL :: door(3)
REAL :: rnum
CALL RANDOM_SEED
DO i = 1, trials
door = .FALSE.
CALL RANDOM_NUMBER(rnum)
prize = INT(3*rnum) + 1
door(prize) = .TRUE. ! place car behind random door
CALL RANDOM_NUMBER(rnum)
choice = INT(3*rnum) + 1 ! choose a door
DO
CALL RANDOM_NUMBER(rnum)
show = INT(3*rnum) + 1
IF (show /= choice .AND. show /= prize) EXIT ! Reveal a goat
END DO
SELECT CASE(choice+show) ! Calculate remaining door index
CASE(3)
remaining = 3
CASE(4)
remaining = 2
CASE(5)
remaining = 1
END SELECT
IF (door(choice)) THEN ! You win by staying with your original choice
staycount = staycount + 1
ELSE IF (door(remaining)) THEN ! You win by switching to other door
switchcount = switchcount + 1
END IF
END DO
WRITE(*, "(A,F6.2,A)") "Chance of winning by not switching is", real(staycount)/trials*100, "%"
WRITE(*, "(A,F6.2,A)") "Chance of winning by switching is", real(switchcount)/trials*100, "%"
END PROGRAM MONTYHALL
Sample Output
Chance of winning by not switching is 32.82% Chance of winning by switching is 67.18%
[edit] 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
I had a very polite suggestion that I simulate Monty's "pick" so I'm putting in a version that does that. I compare the outcome with my original outcome and, unsurprisingly, show that this is essentially a noop that has no bearing on the output, but I (kind of) get where the request is coming from so here's that version...
let montySlower nSims =
let rnd = new Random()
let MontyPick winner pick =
if pick = winner then
[0..2] |> Seq.filter (fun i -> i <> pick) |> Seq.nth (rnd.Next(0,2))
else
3 - pick - winner
let SwitchGame() =
let winner, pick = rnd.Next(0,3), rnd.Next(0,3)
let monty = MontyPick winner pick
let pickFinal = 3 - monty - pick
// Show that Monty's pick has no effect...
if (winner <> pick) <> (pickFinal = winner) then
printfn "Monty's selection actually had an effect!"
if pickFinal = winner then 1 else 0
let StayGame() =
let winner, pick = rnd.Next(0,3), rnd.Next(0,3)
let monty = MontyPick winner pick
// This one's even more obvious than the above since pickFinal
// is precisely the same as pick
let pickFinal = pick
if (winner = pick) <> (winner = pickFinal) then
printfn "Monty's selection actually had an effect!"
if winner = pickFinal then 1 else 0
let Wins (f:unit -> int) = seq {for i in [1..nSims] -> f()} |> Seq.sum
printfn "Stay: %d wins out of %d - Switch: %d wins out of %d" (Wins StayGame) nSims (Wins SwitchGame) nSims
[edit] 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)
Library: mtl With a State monad, we can avoid having to explicitly pass around the StdGen so often. play and cars can be rewritten as follows:
import Control.Monad.State
play :: Bool -> State StdGen Door
play switch = do
i <- rand
let d1 = [Car, Goat, Goat] !! i
return $ case switch of
False -> d1
True -> case d1 of
Car -> Goat
Goat -> Car
where rand = do
g <- get
let (v, new_g) = randomR (0, 2) g
put new_g
return v
cars :: Int -> Bool -> StdGen -> (Int, StdGen)
cars n switch g = (numcars, new_g)
where numcars = length $ filter (== Car) prize_list
(prize_list, new_g) = runState (replicateM n (play switch)) g
Sample output (for either implementation):
The switch strategy succeeds 67% of the time.
The stay strategy succeeds 34% of the time.
[edit] J
NB. Monty Hall Simulation (a tacit version)
'SIZE CAR STAY MONTY SWITCH DOORS ALL'=. i.7 NB. Setting mnemonics for boxes
f=. &({::) NB. Fetching the contents of a box
o=. @: NB. Composing verbs (functions)
pick=. (? @ #) { ] NB. Picking randomly an element from a vector
freq=. +/ % #
PickDoors=. (SIZE f ?@$ DOORS f) NB. Picking doors randomly
boxes=. < , a: $~ 6: NB. Appending 6 (empty) boxes to the input
doors=. < o 3: DOORS} ] NB. 3 doors
car=. < o PickDoors CAR} ] NB. Randomizing car door positions
stay=. < o PickDoors STAY} ] NB. Randomizing stay door selections
all=. < o (i. o (DOORS f)) ALL} ] NB. All doors set
Monty=. < o ((pick&>) o (ALL f (< @ -.)"1 (CAR f ,"0 STAY f))) MONTY } ]
NB. Calculating Monty's selections
switch=. < o (, o ((ALL f) -."1 (MONTY f) ,"0 (STAY f))) SWITCH } ]
NB. Calculating switching selections
StayDisplay=. 'Stay: ' , ": o (CAR f freq o = STAY f)
SwitchDisplay=. 'Switch: ' , ": o (CAR f freq o = SWITCH f)
sim=. (StayDisplay ; SwitchDisplay) o switch o Monty o all o stay o car o doors o boxes f.
Example:
sim 1000000
+--------------+----------------+
|Stay: 0.333143|Switch: 0.666857|
+--------------+----------------+
[edit] Java
import java.util.Random;
public class Monty{
public static void main(String[] args){
int switchWins = 0;
int stayWins = 0;
Random gen = new Random();
for(int plays = 0;plays < 32768;plays++ ){
int[] doors = {0,0,0};//0 is a goat, 1 is a car
doors[gen.nextInt(3)] = 1;//put a winner in a random door
int choice = gen.nextInt(3); //pick a door, any door
int shown; //the shown door
do{
shown = gen.nextInt(3);
//don't show the winner or the choice
}while(doors[shown] == 1 || shown == choice);
stayWins += doors[choice];//if you won by staying, count it
//the switched (last remaining) door is (3 - choice - shown), because 0+1+2=3
switchWins += doors[3 - choice - shown];
}
System.out.println("Switching wins " + switchWins + " times.");
System.out.println("Staying wins " + stayWins + " times.");
}
}
Output:
Switching wins 21924 times. Staying wins 10844 times.
[edit] Lua
function playgame(player)
local car = math.random(3)
local pchoice = player.choice()
local function neither(a, b) --slow, but it works
local el = math.random(3)
return (el ~= a and el ~= b) and el or neither(a, b)
end
local el = neither(car, pchoice)
if(player.switch) then pchoice = neither(pchoice, el) end
player.wins = player.wins + (pchoice == car and 1 or 0)
end
for _, v in ipairs{true, false} do
player = {choice = function() return math.random(3) end,
wins = 0, switch = v}
for i = 1, 20000 do playgame(player) end
print(player.wins)
end
[edit] MAXScript
fn montyHall choice switch =
(
doors = #(false, false, false)
doors[random 1 3] = true
chosen = doors[choice]
if switch then chosen = not chosen
chosen
)
fn iterate iterations switched =
(
wins = 0
for i in 1 to iterations do
(
if (montyHall (random 1 3) switched) then
(
wins += 1
)
)
wins * 100 / iterations as float
)
iterations = 10000
format ("Stay strategy:%\%\n") (iterate iterations false)
format ("Switch strategy:%\%\n") (iterate iterations true)
Output:
Stay strategy:33.77%
Switch strategy:66.84%
[edit] OCaml
let trials = 10000
type door = Car | Goat
let play switch =
let n = Random.int 3 in
let d1 = [|Car; Goat; Goat|].(n) in
if not switch then d1
else match d1 with
Car -> Goat
| Goat -> Car
let cars n switch =
let total = ref 0 in
for i = 1 to n do
let prize = play switch in
if prize = Car then
incr total
done;
!total
let () =
let switch = cars trials true
and stay = cars trials false in
let msg strat n =
Printf.printf "The %s strategy succeeds %f%% of the time.\n"
strat (100. *. (float n /. float trials)) in
msg "switch" switch;
msg "stay" stay
[edit] Perl
#! /usr/bin/perl
use strict;
my $trials = 10000;
my $stay = 0;
my $switch = 0;
foreach (1 .. $trials)
{
my $prize = int(rand 3);
# let monty randomly choose a door where he puts the prize
my $chosen = int(rand 3);
# let us randomly choose a door...
my $show;
do { $show = int(rand 3) } while $show == $chosen || $show == $prize;
# ^ monty opens a door which is not the one with the
# prize, that he knows it is the one the player chosen
$stay++ if $prize == $chosen;
# ^ if player chose the correct door, player wins only if he stays
$switch++ if $prize == 3 - $chosen - $show;
# ^ if player switches, the door he picks is (3 - $chosen - $show),
# because 0+1+2=3, and he picks the only remaining door that is
# neither $chosen nor $show
}
print "Stay win ratio " . (100.0 * $stay/$trials) . "\n";
print "Switch win ratio " . (100.0 * $switch/$trials) . "\n";
[edit] Perl 6
Works with: Rakudo version #22 "Thousand Oaks"
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.
sub remove (@a is rw, Int $i) {
my $temp = @a[$i];
@a = @a[0 ..^ $i], @a[$i ^..^ @a];
return $temp;
}
enum Prize <Car Goat>;
enum Strategy <Stay Switch>;
sub play (Strategy $strategy, Int :$doors = 3) returns Prize {
# Call the door with a car behind it door 0. Number the
# remaining doors starting from 1.
my Prize @doors = Car, Goat xx $doors - 1;
# The player chooses a door.
my Prize $initial_pick = remove @doors, @doors.keys().pick;
# Of the n doors remaining, the host chooses n - 1 that have
# goats behind them and opens them, removing them from play.
remove @doors, $_
for pick @doors.elems - 1, grep { @doors[$_] == Goat }, keys @doors;
# If the player stays, they get their initial pick. Otherwise,
# they get whatever's behind the remaining door.
return $strategy == Stay ?? $initial_pick !! @doors[0];
}
constant TRIALS = 100;
for 3, 10 -> $doors {
my %wins;
say "With $doors doors: ";
for Stay, 'Staying', Switch, 'Switching' -> $s, $name {
for ^TRIALS {
++%wins{$s} if play($s, doors => $doors) == Car;
}
say " $name wins ",
round(100*%wins{$s} / TRIALS),
'% of the time.'
}
}
Sample output:
With 3 doors: Staying wins 34% of the time. Switching wins 64% of the time. With 10 doors: Staying wins 14% of the time. Switching wins 94% of the time.
A hundred trials generally isn't enough to get good numbers, but as of release #22, Rakudo is quite slow and may segfault while executing a long-running program.
[edit] PHP
<?php
function montyhall($iterations){
$switch_win = 0;
$stay_win = 0;
foreach (range(1, $iterations) as $i){
$doors = array(0, 0, 0);
$doors[array_rand($doors)] = 1;
$choice = array_rand($doors);
do {
$shown = array_rand($doors);
} while($shown == $choice || $doors[$shown] == 1);
$stay_win += $doors[$choice];
$switch_win += $doors[3 - $choice - $shown];
}
$stay_percentages = ($stay_win/$iterations)*100;
$switch_percentages = ($switch_win/$iterations)*100;
echo "Iterations: {$iterations} - ";
echo "Stayed wins: {$stay_win} ({$stay_percentages}%) - ";
echo "Switched wins: {$switch_win} ({$switch_percentages}%)";
}
montyhall(10000);
?>
Output:
Iterations: 10000 - Stayed wins: 3331 (33.31%) - Switched wins: 6669 (66.69%)
[edit] PureBasic
Procedure MontyHall(Redecide)Output:
Dim Doors(2)
Doors(Random(2)) = 1
Open = Doors(0)
If Redecide
ProcedureReturn Doors(1-Open)
EndIf
ProcedureReturn Doors(2)
EndProcedure
OpenConsole()
#Tries = 1000000
For I = 0 To #Tries
Win1 + MontyHall(1)
Win2 + MontyHall(0)
Next
PrintN("Trial runs for each option: " + Str(#Tries))
PrintN("Wins when redeciding: " + Str(Win1) + " (" + StrD(Win1/#Tries*100, 2) + "% chance)")
PrintN("Wins when sticking: " + Str(Win2) + " (" + StrD(Win2/#Tries*100, 2) + "% chance)")
Input()
Trial runs for each option: 1000000 Wins when redeciding: 666425 (66.64% chance) Wins when sticking: 331930 (33.19% chance)
[edit] Python
'''
I could understand the explanation of the Monty Hall problem
but needed some more evidence
References:
http://www.bbc.co.uk/dna/h2g2/A1054306
http://en.wikipedia.org/wiki/Monty_Hall_problem especially:
http://en.wikipedia.org/wiki/Monty_Hall_problem#Increasing_the_number_of_doors
'''
from random import randrange
doors, iterations = 3,100000 # could try 100,1000
def monty_hall(choice, switch=False, doorCount=doors):
# Set up doors
door = [False]*doorCount
# One door with prize
door[randrange(doorCount)] = True
chosen = door[choice]
unpicked = door
del unpicked[choice]
# Out of those unpicked, the alternative is either:
# the prize door, or
# an empty door if the initial choice is actually the prize.
alternative = True in unpicked
if switch:
return alternative
else:
return chosen
print "\nMonty Hall problem simulation:"
print doors, "doors,", iterations, "iterations.\n"
print "Not switching allows you to win",
print sum(monty_hall(randrange(3), switch=False)
for x in range(iterations)),
print "out of", iterations, "times."
print "Switching allows you to win",
print sum(monty_hall(randrange(3), switch=True)
for x in range(iterations)),
print "out of", iterations, "times.\n"
Sample output:
Monty Hall problem simulation: 3 doors, 100000 iterations. Not switching allows you to win 33337 out of 100000 times. Switching allows you to win 66529 out of 100000 times.
[edit] R
# Since R is a vector based language that penalizes for loops, we will avoid
# for-loops, instead using "apply" statement variants (like "map" in other
# functional languages).
set.seed(19771025) # set the seed to set the same results as this code
N <- 10000 # trials
true_answers <- sample(1:3, N, replace=TRUE)
# We can assme that the contestant always choose door 1 without any loss of
# generality, by equivalence. That is, we can always relabel the doors
# to make the user-chosen door into door 1.
# Thus, the host opens door '2' unless door 2 has the prize, in which case
# the host opens door 3.
host_opens <- 2 + (true_answers == 2)
other_door <- 2 + (true_answers != 2)
## if always switch
summary( other_door == true_answers )
## if we never switch
summary( true_answers == 1)
## if we randomly switch
random_switch <- other_door
random_switch[runif(N) >= .5] <- 1
summary(random_switch == true_answers)
## To go with the exact parameters of the Rosetta challenge, complicating matters....
## Note that the player may initially choose any of the three doors (not just Door 1),
## that the host opens a different door revealing a goat (not necessarily Door 3), and
## that he gives the player a second choice between the two remaining unopened doors.
N <- 10000 #trials
true_answers <- sample(1:3, N, replace=TRUE)
user_choice <- sample(1:3, N, replace=TRUE)
## the host_choice is more complicated
host_chooser <- function(user_prize) {
# this could be cleaner
bad_choices <- unique(user_prize)
# in R, the x[-vector] form implies, choose the indices in x not in vector
choices <- c(1:3)[-bad_choices]
# if the first arg to sample is an int, it treats it as the number of choices
if (length(choices) == 1) { return(choices)}
else { return(sample(choices,1))}
}
host_choice <- apply( X=cbind(true_answers,user_choice), FUN=host_chooser,MARGIN=1)
not_door <- function(x){ return( (1:3)[-x]) } # we could also define this
# directly at the FUN argument following
other_door <- apply( X = cbind(user_choice,host_choice), FUN=not_door, MARGIN=1)
## if always switch
summary( other_door == true_answers )
## if we never switch
summary( true_answers == user_choice)
## if we randomly switch
random_switch <- user_choice
change <- runif(N) >= .5
random_switch[change] <- other_door[change]
summary(random_switch == true_answers)
Results: > ## if always switch > summary( other_door == true_answers ) Mode FALSE TRUE logical 3298 6702 > ## if we never switch > summary( true_answers == 1) Mode FALSE TRUE logical 6702 3298 > ## if we randomly switch > summary(random_switch == true_answers) Mode FALSE TRUE logical 5028 4972 > ## if always switch > summary( other_door == true_answers ) Mode FALSE TRUE logical 3295 6705 > ## if we never switch > summary( true_answers == user_choice) Mode FALSE TRUE logical 6705 3295 > ## if we randomly switch > summary(random_switch == true_answers) Mode FALSE TRUE logical 4986 5014
[edit] Ruby
n = 10_000 #number of times to play
stay = switch = 0 #sum of each strategy's wins
n.times do #play the game n times
#the doors reveal 2 goats and a car
doors = [ :goat , :goat , :goat ]
doors[rand(3)] = :car
#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
#staying with the initial guess wins if the initial guess is the car
stay += 1 if doors[guess] == :car
#switching guesses wins if the unshown door is the car
switch += 1 if doors[3-guess-shown] == :car
end
puts "Staying wins %.2f%% of the time." % (100.0 * stay / n)
puts "Switching wins %.2f%% of the time." % (100.0 * switch / n)
Sample Output:
Staying wins 33.84% of the time. Switching wins 66.16% of the time.
[edit] Scala
import scala.util.Random
object MontyHallSimulation {
def main(args: Array[String]) {
val samples = if (args.size == 1 && (args(0) matches "\\d+")) args(0).toInt else 1000
val doors = Set(0, 1, 2)
var stayStrategyWins = 0
var switchStrategyWins = 0
1 to samples foreach { _ =>
val prizeDoor = Random shuffle doors head;
val choosenDoor = Random shuffle doors head;
val hostDoor = Random shuffle (doors - choosenDoor - prizeDoor) head;
val switchDoor = doors - choosenDoor - hostDoor head;
(choosenDoor, switchDoor) match {
case (`prizeDoor`, _) => stayStrategyWins += 1
case (_, `prizeDoor`) => switchStrategyWins += 1
}
}
def percent(n: Int) = n * 100 / samples
val report = """|%d simulations were ran.
|Staying won %d times (%d %%)
|Switching won %d times (%d %%)""".stripMargin
println(report
format (samples,
stayStrategyWins, percent(stayStrategyWins),
switchStrategyWins, percent(switchStrategyWins)))
}
}
Sample:
1000 simulations were ran. Staying won 333 times (33 %) Switching won 667 times (66 %)
[edit] Scheme
(define (random-from-list list) (list-ref list (random (length list))))
(define (random-permutation list)
(if (null? list)
'()
(let* ((car (random-from-list list))
(cdr (random-permutation (remove car list))))
(cons car cdr))))
(define (random-configuration) (random-permutation '(goat goat car)))
(define (random-door) (random-from-list '(0 1 2)))
(define (trial strategy)
(define (door-with-goat-other-than door strategy)
(cond ((and (not (= 0 door)) (equal? (list-ref strategy 0) 'goat)) 0)
((and (not (= 1 door)) (equal? (list-ref strategy 1) 'goat)) 1)
((and (not (= 2 door)) (equal? (list-ref strategy 2) 'goat)) 2)))
(let* ((configuration (random-configuration))
(players-first-guess (strategy `(would-you-please-pick-a-door?)))
(door-to-show-player (door-with-goat-other-than players-first-guess
configuration))
(players-final-guess (strategy `(there-is-a-goat-at/would-you-like-to-move?
,players-first-guess
,door-to-show-player))))
(if (equal? (list-ref configuration players-final-guess) 'car)
'you-win!
'you-lost)))
(define (stay-strategy message)
(case (car message)
((would-you-please-pick-a-door?) (random-door))
((there-is-a-goat-at/would-you-like-to-move?)
(let ((first-choice (cadr message)))
first-choice))))
(define (switch-strategy message)
(case (car message)
((would-you-please-pick-a-door?) (random-door))
((there-is-a-goat-at/would-you-like-to-move?)
(let ((first-choice (cadr message))
(shown-goat (caddr message)))
(car (remove first-choice (remove shown-goat '(0 1 2))))))))
(define-syntax repeat
(syntax-rules ()
((repeat <n> <body> ...)
(let loop ((i <n>))
(if (zero? i)
'()
(cons ((lambda () <body> ...))
(loop (- i 1))))))))
(define (count element list)
(if (null? list)
0
(if (equal? element (car list))
(+ 1 (count element (cdr list)))
(count element (cdr list)))))
(define (prepare-result strategy results)
`(,strategy won with probability
,(exact->inexact (* 100 (/ (count 'you-win! results) (length results)))) %))
(define (compare-strategies times)
(append
(prepare-result 'stay-strategy (repeat times (trial stay-strategy)))
'(and)
(prepare-result 'switch-strategy (repeat times (trial switch-strategy)))))
;; > (compare-strategies 1000000)
;; (stay-strategy won with probability 33.3638 %
;; and switch-strategy won with probability 66.716 %)
[edit] Tcl
A simple way of dealing with this one, based on knowledge of the underlying probabilistic system, is to use code like this:
set stay 0; set change 0; set total 10000
for {set i 0} {$i<$total} {incr i} {
if {int(rand()*3) == int(rand()*3)} {
incr stay
} else {
incr change
}
}
puts "Estimate: $stay/$total wins for staying strategy"
puts "Estimate: $change/$total wins for changing strategy"
But that's not really the point of this challenge; it should add the concealing factors too so that we're simulating not just the solution to the game, but also the game itself. (Note that we are using Tcl's lists here to simulate sets.)
We include a third strategy that is proposed by some people (who haven't thought much about it) for this game: just picking at random between all the doors offered by Monty the second time round.
package require Tcl 8.5
# Utility: pick a random item from a list
proc pick list {
lindex $list [expr {int(rand()*[llength $list])}]
}
# Utility: remove an item from a list if it is there
proc remove {list item} {
set idx [lsearch -exact $list $item]
return [lreplace $list $idx $idx]
}
# Codify how Monty will present the new set of doors to choose between
proc MontyHallAction {doors car picked} {
set unpicked [remove $doors $picked]
if {$car in $unpicked} {
# Remove a random unpicked door without the car behind it
set carless [remove $unpicked $car]
return [list {*}[remove $carless [pick $carless]] $car]
# Expressed this way so Monty Hall isn't theoretically
# restricted to using 3 doors, though that could be written
# as just: return [list $car]
} else {
# Monty has a real choice now...
return [remove $unpicked [pick $unpicked]]
}
}
# The different strategies you might choose
proc Strategy:Stay {originalPick otherChoices} {
return $originalPick
}
proc Strategy:Change {originalPick otherChoices} {
return [pick $otherChoices]
}
proc Strategy:PickAnew {originalPick otherChoices} {
return [pick [list $originalPick {*}$otherChoices]]
}
# Codify one round of the game
proc MontyHallGameRound {doors strategy winCounter} {
upvar 1 $winCounter wins
set car [pick $doors]
set picked [pick $doors]
set newDoors [MontyHallAction $doors $car $picked]
set picked [$strategy $picked $newDoors]
# Check for win...
if {$car eq $picked} {
incr wins
}
}
# We're always using three doors
set threeDoors {a b c}
set stay 0; set change 0; set anew 0
set total 10000
# Simulate each of the different strategies
for {set i 0} {$i<$total} {incr i} {
MontyHallGameRound $threeDoors Strategy:Stay stay
MontyHallGameRound $threeDoors Strategy:Change change
MontyHallGameRound $threeDoors Strategy:PickAnew anew
}
# Print the results
puts "Estimate: $stay/$total wins for 'staying' strategy"
puts "Estimate: $change/$total wins for 'changing' strategy"
puts "Estimate: $anew/$total wins for 'picking anew' strategy"
This might then produce output like
Estimate: 3340/10000 wins for 'staying' strategy Estimate: 6733/10000 wins for 'changing' strategy Estimate: 4960/10000 wins for 'picking anew' strategy
Of course, this challenge could also be tackled by putting up a GUI and letting the user be the source of the randomness. But that's moving away from the letter of the challenge and takes a lot of effort anyway...
[edit] Ursala
This is the same algorithm as the Perl solution. Generate two lists of 10000 uniformly distributed samples from {1,2,3}, count each match as a win for the staying strategy, and count each non-match as a win for the switching strategy.
#import std
#import nat
#import flo
rounds = 10000
car_locations = arc{1,2,3}* iota rounds
initial_choices = arc{1,2,3}* iota rounds
staying_wins = length (filter ==) zip(car_locations,initial_choices)
switching_wins = length (filter ~=) zip(car_locations,initial_choices)
format = printf/'%0.2f'+ (times\100.+ div+ float~~)\rounds
#show+
main = ~&plrTS/<'stay: ','switch: '> format* <staying_wins,switching_wins>
Output will vary slightly for each run due to randomness.
stay: 33.95 switch: 66.05
[edit] Vedit macro language
Translation of: BASIC
Vedit macro language does not have random number generator, so one is implemented in subroutine RANDOM (the algorithm was taken from ANSI C library).
#90 = Time_Tick // seed for random number generator
#91 = 3 // random numbers in range 0 to 2
#1 = 0 // wins for "always stay" strategy
#2 = 0 // wins for "always switch" strategy
for (#10 = 0; #10 < 10000; #10++) { // 10,000 iterations
Call("RANDOM")
#3 = Return_Value // #3 = winning door
Call("RANDOM")
#4 = Return_Value // #4 = players choice
do {
Call("RANDOM")
#5 = Return_Value // #5 = door to open
} while (#5 == #3 || #5 == #4)
if (#3 == #4) { // original choice was correct
#1++
}
if (#3 == 3 - #4 - #5) { // switched choice was correct
#2++
}
}
Ins_Text("Staying wins: ") Num_Ins(#1)
Ins_Text("Switching wins: ") Num_Ins(#2)
return
//--------------------------------------------------------------
// Generate random numbers in range 0 <= Return_Value < #91
// #90 = Seed (0 to 0x7fffffff)
// #91 = Scaling (0 to 0xffff)
:RANDOM:
#92 = 0x7fffffff / 48271
#93 = 0x7fffffff % 48271
#90 = (48271 * (#90 % #92) - #93 * (#90 / #92)) & 0x7fffffff
return ((#90 & 0xffff) * #91 / 0x10000)
Sample output:
Staying winns: 3354 Switching winns: 6646
[edit] C#
Translation of: Java
using System;
namespace MontyHallProblem
{
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];
}
System.Console.Out.Write("Switching wins " + switchWins + " times.");
System.Console.Out.Write("Staying wins " + stayWins + " times.");
}
}
}
Sample output:
Staying winns: 333830 Switching winns: 666170
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