Est-ce une bonne ou une mauvaise « simulation » pour Monty Hall? Comment venir?
https://stackoverflow.com/questions/1247863
-
12-09-2019 - |
italiano
english
français
española
中国
日本の
العربية
Deutsch
한국어
Português
RussianQuestion
En essayant d'expliquer problème Monty Hall à un ami au cours de classe hier, nous avons fini le codage en Python pour prouver que si vous échangez toujours, vous gagnerez 2/3 fois. Nous sommes venus à ceci:
import random as r
#iterations = int(raw_input("How many iterations? >> "))
iterations = 100000
doors = ["goat", "goat", "car"]
wins = 0.0
losses = 0.0
for i in range(iterations):
n = r.randrange(0,3)
choice = doors[n]
if n == 0:
#print "You chose door 1."
#print "Monty opens door 2. There is a goat behind this door."
#print "You swapped to door 3."
wins += 1
#print "You won a " + doors[2] + "\n"
elif n == 1:
#print "You chose door 2."
#print "Monty opens door 1. There is a goat behind this door."
#print "You swapped to door 3."
wins += 1
#print "You won a " + doors[2] + "\n"
elif n == 2:
#print "You chose door 3."
#print "Monty opens door 2. There is a goat behind this door."
#print "You swapped to door 1."
losses += 1
#print "You won a " + doors[0] + "\n"
else:
print "You screwed up"
percentage = (wins/iterations) * 100
print "Wins: " + str(wins)
print "Losses: " + str(losses)
print "You won " + str(percentage) + "% of the time"
Mon ami a pensé que c'était une bonne façon de faire (et est une bonne simulation pour cela), mais je dois mes doutes et préoccupations. Est-il vraiment assez aléatoire?
Le problème que j'ai avec ce que tous les choix sont les types de codées en dur dans.
Est-ce une bonne ou une mauvaise 'simulation' pour le problème de Monty Hall? Comment ça se fait?
Pouvez-vous trouver une meilleure version?
La solution
Votre solution est très bien, mais si vous voulez une simulation plus stricte du problème posé (et un peu Python meilleure qualité ;-), essayez:
import random
iterations = 100000
doors = ["goat"] * 2 + ["car"]
change_wins = 0
change_loses = 0
for i in xrange(iterations):
random.shuffle(doors)
# you pick door n:
n = random.randrange(3)
# monty picks door k, k!=n and doors[k]!="car"
sequence = range(3)
random.shuffle(sequence)
for k in sequence:
if k == n or doors[k] == "car":
continue
# now if you change, you lose iff doors[n]=="car"
if doors[n] == "car":
change_loses += 1
else:
change_wins += 1
print "Changing has %s wins and %s losses" % (change_wins, change_loses)
perc = (100.0 * change_wins) / (change_wins + change_loses)
print "IOW, by changing you win %.1f%% of the time" % perc
une sortie typique est:
Changing has 66721 wins and 33279 losses
IOW, by changing you win 66.7% of the time
Autres conseils
Vous avez mentionné que tous les choix sont en dur dans. Mais si vous regardez de plus près, vous remarquerez que ce que vous pensez sont « choix » sont en fait pas des choix du tout. La décision de Monty est sans perte de généralité car il choisit toujours la porte avec la chèvre derrière elle. Votre swapping est toujours déterminée par ce que Monty choisit, et depuis le « choix » de Monty était en fait pas un choix, ni est le vôtre. Votre simulation donne des résultats corrects ..
J'aime quelque chose comme ça.
#!/usr/bin/python
import random
CAR = 1
GOAT = 0
def one_trial( doors, switch=False ):
"""One trial of the Monty Hall contest."""
random.shuffle( doors )
first_choice = doors.pop( )
if switch==False:
return first_choice
elif doors.__contains__(CAR):
return CAR
else:
return GOAT
def n_trials( switch=False, n=10 ):
"""Play the game N times and return some stats."""
wins = 0
for n in xrange(n):
doors = [CAR, GOAT, GOAT]
wins += one_trial( doors, switch=switch )
print "won:", wins, "lost:", (n-wins), "avg:", (float(wins)/float(n))
if __name__=="__main__":
import sys
n_trials( switch=eval(sys.argv[1]), n=int(sys.argv[2]) )
$ ./montyhall.py True 10000
won: 6744 lost: 3255 avg: 0.674467446745
Je ne l'avais pas entendu parler de la problème de Monty Hall avant trébuché à travers cette question. Je pensais qu'il était intéressant, donc je l'ai lu et créé une simulation c #. Il est un peu loufoque car il simule le jeu-show et pas seulement le problème.
Je publiais la source et la libération sur CodePlex:
Voici ma version ...
import random
wins = 0
for n in range(1000):
doors = [1, 2, 3]
carDoor = random.choice(doors)
playerDoor = random.choice(doors)
hostDoor = random.choice(list(set(doors) - set([carDoor, playerDoor])))
# To stick, just comment out the next line.
(playerDoor, ) = set(doors) - set([playerDoor, hostDoor]) # Player swaps doors.
if playerDoor == carDoor:
wins += 1
print str(round(wins / float(n) * 100, 2)) + '%'
Voici une version interactive:
from random import shuffle, choice
cars,goats,iters= 0, 0, 100
for i in range(iters):
doors = ['goat A', 'goat B', 'car']
shuffle(doors)
moderator_door = 'car'
#Turn 1:
selected_door = choice(doors)
print selected_door
doors.remove(selected_door)
print 'You have selected a door with an unknown object'
#Turn 2:
while moderator_door == 'car':
moderator_door = choice(doors)
doors.remove(moderator_door)
print 'Moderator has opened a door with ', moderator_door
#Turn 3:
decision=raw_input('Wanna change your door? [yn]')
if decision=='y':
prise = doors[0]
print 'You have a door with ', prise
elif decision=='n':
prise = selected_door
print 'You have a door with ', prise
else:
prise = 'ERROR'
iters += 1
print 'ERROR:unknown command'
if prise == 'car':
cars += 1
elif prise != 'ERROR':
goats += 1
print '==============================='
print ' RESULTS '
print '==============================='
print 'Goats:', goats
print 'Cars :', cars
Ma solution avec la compréhension de la liste pour simuler le problème
from random import randint
N = 1000
def simulate(N):
car_gate=[randint(1,3) for x in range(N)]
gate_sel=[randint(1,3) for x in range(N)]
score = sum([True if car_gate[i] == gate_sel[i] or ([posible_gate for posible_gate in [1,2,3] if posible_gate != gate_sel[i]][randint(0,1)] == car_gate[i]) else False for i in range(N)])
return 'you win %s of the time when you change your selection.' % (float(score) / float(N))
Simuler d'impression (N)
Non échantillon de la mine
# -*- coding: utf-8 -*-
#!/usr/bin/python -Ou
# Written by kocmuk.ru, 2008
import random
num = 10000 # number of games to play
win = 0 # init win count if donot change our first choice
for i in range(1, num): # play "num" games
if random.randint(1,3) == random.randint(1,3): # if win at first choice
win +=1 # increasing win count
print "I donot change first choice and win:", win, " games"
print "I change initial choice and win:", num-win, " games" # looses of "not_change_first_choice are wins if changing
Je trouve que ce soit la façon la plus intuitive de résoudre le problème.
import random
# game_show will return True/False if the participant wins/loses the car:
def game_show(knows_bayes):
doors = [i for i in range(3)]
# Let the car be behind this door
car = random.choice(doors)
# The participant chooses this door..
choice = random.choice(doors)
# ..so the host opens another (random) door with no car behind it
open_door = random.choice([i for i in doors if i not in [car, choice]])
# If the participant knows_bayes she will switch doors now
if knows_bayes:
choice = [i for i in doors if i not in [choice, open_door]][0]
# Did the participant win a car?
if choice == car:
return True
else:
return False
# Let us run the game_show() for two participants. One knows_bayes and the other does not.
wins = [0, 0]
runs = 100000
for x in range(0, runs):
if game_show(True):
wins[0] += 1
if game_show(False):
wins[1] += 1
print "If the participant knows_bayes she wins %d %% of the time." % (float(wins[0])/runs*100)
print "If the participant does NOT knows_bayes she wins %d %% of the time." % (float(wins[1])/runs*100)
répondre quelque chose comme
If the participant knows_bayes she wins 66 % of the time.
If the participant does NOT knows_bayes she wins 33 % of the time.
Lire un chapitre sur le fameux problème de Monty Hall aujourd'hui. Ceci est ma solution.
import random
def one_round():
doors = [1,1,0] # 1==goat, 0=car
random.shuffle(doors) # shuffle doors
choice = random.randint(0,2)
return doors[choice]
#If a goat is chosen, it means the player loses if he/she does not change.
#This method returns if the player wins or loses if he/she changes. win = 1, lose = 0
def hall():
change_wins = 0
N = 10000
for index in range(0,N):
change_wins += one_round()
print change_wins
hall()
Encore une autre « preuve », cette fois avec Python 3. Notez l'utilisation de générateurs pour sélectionner 1) qui porte Monty ouvre, et 2) qui porte les commutateurs du lecteur à.
import random
items = ['goat', 'goat', 'car']
num_trials = 100000
num_wins = 0
for trial in range(num_trials):
random.shuffle(items)
player = random.randrange(3)
monty = next(i for i, v in enumerate(items) if i != player and v != 'car')
player = next(x for x in range(3) if x not in (player, monty))
if items[player] == 'car':
num_wins += 1
print('{}/{} = {}'.format(num_wins, num_trials, num_wins / num_trials))
Monty ouvre jamais la porte avec la voiture - c'est le point de l'ensemble du spectacle (il n'est pas votre un ami a une connaissance de ce qui est derrière chaque porte)
Un autre exemple de code est disponible à: http://standardwisdom.com/ softwarejournal / code-échantillons / monty-python-salle /
Le code est un peu plus long et ne peut pas utiliser certaines des fonctionnalités intéressantes de Python, mais je l'espère, il est bien lisible. précisément Python utilisé parce que je n'ai aucune expérience, donc apprécié des commentaires.
Voici différentes variantes que je trouve la plus intuitive. Espérons que cela aide!
import random
class MontyHall():
"""A Monty Hall game simulator."""
def __init__(self):
self.doors = ['Door #1', 'Door #2', 'Door #3']
self.prize_door = random.choice(self.doors)
self.contestant_choice = ""
self.monty_show = ""
self.contestant_switch = ""
self.contestant_final_choice = ""
self.outcome = ""
def Contestant_Chooses(self):
self.contestant_choice = random.choice(self.doors)
def Monty_Shows(self):
monty_choices = [door for door in self.doors if door not in [self.contestant_choice, self.prize_door]]
self.monty_show = random.choice(monty_choices)
def Contestant_Revises(self):
self.contestant_switch = random.choice([True, False])
if self.contestant_switch == True:
self.contestant_final_choice = [door for door in self.doors if door not in [self.contestant_choice, self.monty_show]][0]
else:
self.contestant_final_choice = self.contestant_choice
def Score(self):
if self.contestant_final_choice == self.prize_door:
self.outcome = "Win"
else:
self.outcome = "Lose"
def _ShowState(self):
print "-" * 50
print "Doors %s" % self.doors
print "Prize Door %s" % self.prize_door
print "Contestant Choice %s" % self.contestant_choice
print "Monty Show %s" % self.monty_show
print "Contestant Switch %s" % self.contestant_switch
print "Contestant Final Choice %s" % self.contestant_final_choice
print "Outcome %s" % self.outcome
print "-" * 50
Switch_Wins = 0
NoSwitch_Wins = 0
Switch_Lose = 0
NoSwitch_Lose = 0
for x in range(100000):
game = MontyHall()
game.Contestant_Chooses()
game.Monty_Shows()
game.Contestant_Revises()
game.Score()
# Tally Up the Scores
if game.contestant_switch and game.outcome == "Win": Switch_Wins = Switch_Wins + 1
if not(game.contestant_switch) and game.outcome == "Win": NoSwitch_Wins = NoSwitch_Wins + 1
if game.contestant_switch and game.outcome == "Lose": Switch_Lose = Switch_Lose + 1
if not(game.contestant_switch) and game.outcome == "Lose": NoSwitch_Lose = NoSwitch_Lose + 1
print Switch_Wins * 1.0 / (Switch_Wins + Switch_Lose)
print NoSwitch_Wins * 1.0 / (NoSwitch_Wins + NoSwitch_Lose)
L'apprentissage est toujours le même, que la commutation augmente vos chances de gagner, 0,665025416127 vs ,33554730611 de la course ci-dessus.
Voici que je disais plus tôt:
import random
def game():
"""
Set up three doors, one randomly with a car behind and two with
goats behind. Choose a door randomly, then the presenter takes away
one of the goats. Return the outcome based on whether you stuck with
your original choice or switched to the other remaining closed door.
"""
# Neither stick or switch has won yet, so set them both to False
stick = switch = False
# Set all of the doors to goats (zeroes)
doors = [ 0, 0, 0 ]
# Randomly change one of the goats for a car (one)
doors[random.randint(0, 2)] = 1
# Randomly choose one of the doors out of the three
choice = doors[random.randint(0, 2)]
# If our choice was a car (a one)
if choice == 1:
# Then stick wins
stick = True
else:
# Otherwise, because the presenter would take away the other
# goat, switching would always win.
switch = True
return (stick, switch)
Je devais également le code pour lancer le jeu plusieurs fois, et emmagasinés cela et la sortie de l'échantillon dans ce repostory.
Voici ma solution au problème MontyHall mis en œuvre en python.
Cette solution utilise numpy pour la vitesse, il vous permet également de modifier le nombre de portes.
def montyhall(Trials:"Number of trials",Doors:"Amount of doors",P:"Output debug"):
N = Trials # the amount of trial
DoorSize = Doors+1
Answer = (nprand.randint(1,DoorSize,N))
OtherDoor = (nprand.randint(1,DoorSize,N))
UserDoorChoice = (nprand.randint(1,DoorSize,N))
# this will generate a second door that is not the user's selected door
C = np.where( (UserDoorChoice==OtherDoor)>0 )[0]
while (len(C)>0):
OtherDoor[C] = nprand.randint(1,DoorSize,len(C))
C = np.where( (UserDoorChoice==OtherDoor)>0 )[0]
# place the car as the other choice for when the user got it wrong
D = np.where( (UserDoorChoice!=Answer)>0 )[0]
OtherDoor[D] = Answer[D]
'''
IfUserStays = 0
IfUserChanges = 0
for n in range(0,N):
IfUserStays += 1 if Answer[n]==UserDoorChoice[n] else 0
IfUserChanges += 1 if Answer[n]==OtherDoor[n] else 0
'''
IfUserStays = float(len( np.where((Answer==UserDoorChoice)>0)[0] ))
IfUserChanges = float(len( np.where((Answer==OtherDoor)>0)[0] ))
if P:
print("Answer ="+str(Answer))
print("Other ="+str(OtherDoor))
print("UserDoorChoice="+str(UserDoorChoice))
print("OtherDoor ="+str(OtherDoor))
print("results")
print("UserDoorChoice="+str(UserDoorChoice==Answer)+" n="+str(IfUserStays)+" r="+str(IfUserStays/N))
print("OtherDoor ="+str(OtherDoor==Answer)+" n="+str(IfUserChanges)+" r="+str(IfUserChanges/N))
return IfUserStays/N, IfUserChanges/N
Je viens de découvrir que le ratio global de gain est de 50% et le ratio de perdre 50% ... Il est ainsi que la proportion de gagner ou de perdre en fonction de l'option finale sélectionnée.
- % Wins (restant): 16,692
- % Victoires (commutation): 33,525
- % Pertes (restant): 33,249
- % Pertes (commutation): 16,534
Voici mon code, qui diffère de la vôtre + avec les commentaires de commentaires afin que vous puissiez exécuter avec de petites itérations:
import random as r
#iterations = int(raw_input("How many iterations? >> "))
iterations = 100000
doors = ["goat", "goat", "car"]
wins_staying = 0
wins_switching = 0
losses_staying = 0
losses_switching = 0
for i in range(iterations):
# Shuffle the options
r.shuffle(doors)
# print("Doors configuration: ", doors)
# Host will always know where the car is
car_option = doors.index("car")
# print("car is in Option: ", car_option)
# We set the options for the user
available_options = [0, 1 , 2]
# The user selects an option
user_option = r.choice(available_options)
# print("User option is: ", user_option)
# We remove an option
if(user_option != car_option ) :
# In the case the door is a goat door on the user
# we just leave the car door and the user door
available_options = [user_option, car_option]
else:
# In the case the door is the car door
# we try to get one random door to keep
available_options.remove(available_options[car_option])
goat_option = r.choice(available_options)
available_options = [goat_option, car_option]
new_user_option = r.choice(available_options)
# print("User final decision is: ", new_user_option)
if new_user_option == car_option :
if(new_user_option == user_option) :
wins_staying += 1
else :
wins_switching += 1
else :
if(new_user_option == user_option) :
losses_staying += 1
else :
losses_switching += 1
print("%Wins (staying): " + str(wins_staying / iterations * 100))
print("%Wins (switching): " + str(wins_switching / iterations * 100))
print("%Losses (staying) : " + str(losses_staying / iterations * 100))
print("%Losses (switching) : " + str(losses_switching / iterations * 100))
