Let’s Make A Deal!
July 24, 2009
Since the host always reveals a goat, the only thing that matters is whether the contestant’s initial pick was an auto or a goat; if the initial pick was an auto, staying wins, and if the initial pick was a goat, switching wins. The monty function (named after the original host, Monty Hall), plays n games and reports the number of wins when switching and the number of wins when staying:
(define (monty n)
(let monty ((n n) (switch 0) (stay 0))
(let ((auto (randint 3)) (pick (randint 3)))
(cond ((zero? n) (values switch stay))
((= auto pick) (monty (- n 1) switch (+ stay 1)))
(else (monty (- n 1) (+ switch 1) stay))))))
The variables switch and stay count the number of wins for each strategy. (randint n) returns a non-negative integer less than n; variable auto records the location of the automobile, and variable pick records the location of the contestant’s initial pick.
Running monty for 100,000 contests shows a clear winner:
> (monty 100000)
66824
33176
You can run the program at http://programmingpraxis.codepad.org/54PDMDSv, including rand and randint, which are taken from the Standard Prelude.
When the contestant first makes a choice, he wins the automobile one-third of the time. The other two doors hide the automobile the other two-thirds of the time. When the host reveals a goat, the probabilities don’t change: the door first chosen by the contestant wins the automobile one-third of the time, and wins a goat two-thirds of the time. But since the other two doors hid the automobile two-thirds of the time, and we now know that one of the doors certainly does not hide the automobile, the other door must hide the automobile two-thirds of the time. It is better to switch than stay.
When this puzzle appeared in Marilyn vos Savant’s column in Parade Magazine in 1990, it generated more mail than any previous column; over a thousand people with Ph.D.s thought she was wrong.
#lang scheme (define DOOR-COUNT 3) (define (car? n lst) (eq? (list-ref lst n) 'car)) (define (get-door-layout n) (let loop ([n n] [doors (list)]) (cond [(= 0 n) doors] [(or (member 'car doors) (not (= 0 (random n)))) (loop (sub1 n) (cons 'goat doors))] [else (loop (sub1 n) (cons 'car doors))]))) (define (get-random-door door-count criteria-function) (let ([door (random door-count)]) (if (criteria-function door) door (get-random-door door-count criteria-function)))) (define (get-revealed-door doors contestant-selection) (get-random-door (length doors) (lambda (n) (and (not (= n contestant-selection)) (eq? (list-ref doors n) 'goat))))) (define (get-switch-door door-count unavailable-doors) (get-random-door door-count (lambda (n) (not (member n unavailable-doors))))) (define (make-a-deal n (show-debug-info #f)) (let ([success-count-stick 0] [success-count-switch 0]) (for ((trial (in-range n))) (let ([doors (get-door-layout DOOR-COUNT)] [selected-door (random DOOR-COUNT)]) (let ([revealed-door (get-revealed-door doors selected-door)]) (let ([selected-door-switch (get-switch-door DOOR-COUNT (list selected-door revealed-door))]) (when show-debug-info (printf "DOORS: ~A~%" doors) (printf "SELECTED DOOR: ~A~%" selected-door) (printf "REVEALED DOOR: ~A~%" revealed-door) (printf "SWITCH DOOR: ~A~%" selected-door-switch)) (cond [(car? selected-door doors) (set! success-count-stick (add1 success-count-stick))] [(car? selected-door-switch doors) (set! success-count-switch (add1 success-count-switch))]))))) (printf "Success Rate of Sticking: ~A%~%" (exact->inexact (* 100 (/ success-count-stick n)))) (printf "Success Rate of Switching: ~A%~%" (exact->inexact (* 100 (/ success-count-switch n)))) (if (> success-count-switch success-count-stick) (printf "Better off Switching~%") (printf "Better off Sticking~%")))) (make-a-deal 100000)I already knew that the correct answer is the probability of winning by switching is 2/3, but:
#!r6rs (import (rnrs) (srfi :27)) (define (create-game) ; false = goat, true = car (let ((v (make-vector 3 #f))) (vector-set! v (random-integer 3) #t) v)) (define (player-choice) (random-integer 3)) (define (host-choice g p) ; host always chooses a goat (case p ((0) (if (vector-ref g 1) 2 1)) ((1) (if (vector-ref g 0) 2 0)) ((2) (if (vector-ref g 0) 1 0)))) (define (switch p h) (case (+ p h) ((1) 2) ((2) 1) ((3) 0))) (define (run-test n) (let loop ((wins 0) (total 0)) (if (= total n) (/ wins total) (let* ((g (create-game)) (p (player-choice)) (h (host-choice g p)) (s (switch p h))) (if (vector-ref g s) (loop (+ wins 1) (+ total 1)) (loop wins (+ total 1))))))) (display (inexact (run-test 1000000)))=> 0.66756
Ugly, but it works.
#!/usr/bin/env python # code for the monty hall problem # written by Mark VandeWettering as a challenge on # the programming praxis website. import random NTRIALS=10000 def trial(switch=False): # pick a random door... door = random.randint(1, 3) guess = random.randint(1, 3) if not switch: return door == guess else: # slightly clever, but the only way that you # lose is if you were lucky enough (or unlucky # enough) to guess the right door in the first # place. This gives away the entire problem. return door != guess cnt_noswitch = 0 cnt_switch = 0 for x in range(NTRIALS): if trial(): cnt_noswitch = cnt_noswitch + 1 if trial(switch=True): cnt_switch = cnt_switch + 1 print "By not switching, you won %d/%d rounds." % (cnt_noswitch, NTRIALS) print "By switching, you won %d/%d rounds." % (cnt_switch, NTRIALS) if cnt_switch > cnt_noswitch: print "It appears you should switch." else: print "It appears you should not switch."My Haskell version: http://hpaste.org/fastcgi/hpaste.fcgi/view?id=7478#a7478
Lots of fun – thanks!
-Andy
I suggest also reading a nice treatment of this exact problem in “The man who loved only numbers” by Paul Hoffman.
I’m just now learning Python and figured I’d play around with its list capabilities for this one…
Originally I was using list comprehension to generate a list of random numbers (to indicate which door the car is behind) and keeping the contestant’s choice constant (always picking door number 1), but I realized it might be slightly faster to randomize the contestant’s choice and assume the car is always located behind door number 1 – simply using the list as a looping mechanism.
[…] Let’s Make A Deal! « Programming Praxis. […]
I made an attempt in PL/SQL (Oracle). I made the number of doors configurable (assuming that the host opens all doors except yours or the prize, or a random single door if you have chosen the prize door initially).
declare Runs integer := 10000; Doors integer := 3; function RunMonteHaul ( Runs in integer, MaxDoor in integer, Swap in boolean ) return integer is Prize integer; OtherDoor integer; MyDoor integer; Wins integer := 0; i integer; j integer; begin for i in 1 .. Runs loop Prize := dbms_random.value(1,MaxDoor); MyDoor := dbms_random.value(1,MaxDoor); if Prize = MyDoor then if Prize = MaxDoor then OtherDoor := 1; else OtherDoor := MaxDoor; end if; else OtherDoor := Prize; end if; if Swap then MyDoor := OtherDoor; end if; if MyDoor = Prize then Wins := Wins + 1; end if; end loop; return Wins; end RunMonteHaul; begin dbms_output.put_line('Runs ' || Runs); dbms_output.put_line('Doors ' || Doors); dbms_output.put_line('Swap Wins ' || RunMonteHaul(Runs,Doors,true)); dbms_output.put_line('No Swap Wins ' || RunMonteHaul(Runs,Doors,false)); end;This produces the output
# Monty Hall Paradox simulation / Let's make a deal # Simulates a number of games and output the total number of wins and looses, # and the percentage of each, to determine if switching doors after the host has opened one # is a better option. from random import random,randint def scrambleDoors(doors): """ Simulates the scramble of doors by choosing one to contain a Car instead of a Goat """ doors[randint(1,3)] = "Car" return doors def pick(): """ Choose a door between 1 and 3 """ return int(random() * 100) % 3 + 1 def openDoor(door_list,door): """ The host opens a door, wich must not reveal the price, and also is not the one the player has choosen """ return [i for i in door_list.keys() if (door_list[i] != "Car") & (i != door)][0] def switchDoor(door_list,door): return [i for i in door_list.keys() if i != door] def play(): # 1 - Scramble doors doors = scrambleDoors({1:"Goat",2:"Goat",3:"Goat"}) # 2 - Pick a door door = pick() # 3 - Open a door, and remove it from door list od = openDoor(doors,door) del doors[od] # 4 - Switch door newDoor = switchDoor(doors,door) # 5 - Won? return doors[newDoor[0]] == "Car" def game(n = 100): """ n: Number of games to simulate """ win = 0 loose = 0 for i in range(0,n): if (play()): win +=1 else: loose += 1 porcWin = (float(win) / n) * 100 porcLoose = (float(loose) / n) * 100 print """ Runs : % s Win : %s Percentage win : %s Loose : %s Percentage Loose: %s """ % (n,win,porcWin, loose,porcLoose)Which produces the following output
Yes, but when the game was played, did the host always open a door? Or did the host, for example, have the discretion to let the original choice go through. Unless we know this, it is meaningless to claim that we know the best strategy for the player.