هل هذه "محاكاة" جيدة أو سيئة لقاعة مونتي؟ كيف ذلك؟
https://stackoverflow.com/questions/1247863
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12-09-2019 - |
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من خلال محاولة لشرح مشكلة مونتي هول لصديق أثناء الفصل أمس، انتهى بنا الأمر ترميزه في بيثون لإثبات أنه إذا كنت دائما مبادلة، فسوف تفوز 2/3 مرات. لقد توصلنا إلى هذا:
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"
اعتقد صديقي أن هذه طريقة جيدة للذهاب (وهي محاكاة جيدة لذلك)، لكن لدي شكوكي ومخاوفي. هل هو في الواقع عشوائي بما فيه الكفاية؟
المشكلة لدي معها هي أن جميع الخيارات هي نوع من الترميز بجد.
هل هذه "محاكاة" جيدة أو سيئة لمشكلة مونتي قاعة؟ كيف ذلك؟
يمكنك التوصل إلى نسخة أفضل؟
المحلول
الحل الخاص بك على ما يرام، ولكن إذا كنت تريد محاكاة أكثر صرامة للمشكلة كما هو موضح (وبيثون أعلى جودة إلى حد ما ؛-)، حاول:
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
الإخراج النموذجي هو:
Changing has 66721 wins and 33279 losses
IOW, by changing you win 66.7% of the time
نصائح أخرى
لقد ذكرت أن جميع الخيارات هي صغار. ولكن إذا كنت تبدو أقرب، فستلاحظ أن ما تعتقد أنه "الخيارات" ليست في الواقع خيارات على الإطلاق. قرار مونتي هو دون فقدان العمومية لأنه يختار دائما الباب مع الماعز وراء ذلك. يتم تحديد المبادلة الخاصة بك دائما عن طريق ما يختار مونتي، ومنذ أن "اختيار مونتي" لم يكن في الواقع لا يختار ولا لك. محاكاة الخاص بك يعطي النتائج الصحيحة ..
أنا أحب شيء مثل هذا.
#!/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
لم أسمع عن مشكلة مونتي قاعة قبل تعثرت عبر هذا السؤال. اعتقدت أنه كان مثيرا للاهتمام، لذلك قرأت عن ذلك وخلق محاكاة AC #. إنه نوع من الأبله لأنه يحاكي عرض اللعبة وليس فقط المشكلة.
لقد نشرت المصدر والافراج عن Codeplex:
إليك نسختي ...
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)) + '%'
فيما يلي نسخة تفاعلية:
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
حلاي مع الفهم القائمة لمحاكاة المشكلة
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))
طباعة محاكاة (ن)
ليس لي عينة
# -*- 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
لقد وجدت أن هذا هو الطريقة الأكثر بديهية لحل المشكلة.
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)
هذا يخرج شيئا مثل
If the participant knows_bayes she wins 66 % of the time.
If the participant does NOT knows_bayes she wins 33 % of the time.
قراءة فصل حول مشكلة مونتي الشهير مشكلة اليوم. هذا الحل بلدي.
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()
بعد "دليل آخر"، هذه المرة مع بيثون 3. لاحظ استخدام المولدات لتحديد 1) يفتح الباب مونتي الباب، و 2) أي باب يبحث اللاعب عليه.
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))
مونتي لا يفتح الباب مع السيارة - هذه هي الفترة القصيرة في المعرض (ليس صديقك لديه معرفة ما وراء كل باب)
تتوفر عينة رمز آخر في: http://standardwisdom.com/softwarejournal/code-samples/monty-hall-python/
يعد الرمز لفترة أطول قليلا وقد لا يستخدم بعض الميزات الرائعة في بيثون، لكنني آمل أن يكون مقروءا جيدا. استخدم بيثون على وجه التحديد لأنني لم يكن لدي أي تجربة فيها، لذلك يتم تقدير الملاحظات.
هنا هو مختلف مختلف أجدها أكثر بديهية. أتمنى أن يساعدك هذا!
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)
لا يزال التعلم هو نفسه، أن التحول يزيد من فرصك في الفوز، 0.665025416127 مقابل 0.33554730611 من المدى أعلاه.
هنا واحد صنعت في وقت سابق:
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)
كان لدي أيضا رمز لتشغيل اللعبة عدة مرات، وتخزين هذا وعينة الناتج في هذا الرميل.
هنا هو حلا لي لمشكلة مونتيكل المنفذة في بيثون.
هذا الحل يستفيد من numpy للسرعة، كما يسمح لك أيضا بتغيير عدد الأبواب.
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
لقد وجدت للتو أن النسبة العالمية للفوز هي 50٪ والنسبة من الخسارة هي 50٪ ... فمن الطريقة التي تناسب النسبة على الفوز أو الخسارة بناء على الخيار النهائي المحدد.
- WINS (البقاء): 16.692
- WINS (التبديل): 33.525
- خسائر (البقاء): 33.249
- خسائر (تبديل): 16.534
هنا هو رمزي، الذي يختلف عنك + مع تعليقات علق حتى تتمكن من تشغيله مع التكرارات الصغيرة:
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))
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