Question

from random import randint

numberOfDoors = 3

success = 0
attempts = 0

while True:
    try:
        doors = [0] * numberOfDoors
        doors[randint(0, numberOfDoors - 1)] = 1

        chosen = randint(0, numberOfDoors - 1)

        while numberOfDoors > 2:
            notIn = -1
            while notIn == -1:
                index = randint(0, numberOfDoors - 1)
                if doors[index] == 0 and index != chosen:
                    notIn = index

            if notIn < chosen:
                chosen -= 1
            del doors[notIn]
            numberOfDoors -= 1

        # doors is 2, so not chosen (0 or 1) will return the opposite (1 or 0)
        success += doors[not chosen]
        attempts += 1
        if attempts % 1000000 == 0:
            print float(success) / float(attempts)
    except KeyboardInterrupt:
        print float(success) / float(attempts)
        break

经过几个小时的模拟后，我的结果几乎完全是50％ - 我做了一些特别错误的事情吗？

理论上你选择的门在1/3赔率和2/3赔率之间，所以至少你应该得到高于50的门。

This回答似乎与我做同样的事情（忽略了他对monty的选择没有做任何事情 - 我想说明这个概念）。

Answer 1

您忘记重置numberOfDoors（仍然关闭的门的数量，对吗？）回到3.由于第一个while True:的每次迭代代表一个新的游戏节目运行，该节目从最初关闭的所有三扇门开始。

...
while True:
    numberOfDoors = 3
    try:
        doors = [0] * numberOfDoors
        doors[randint(0, numberOfDoors - 1)] = 1
...

下次尝试添加print语句以帮助您进行调试。在这种情况下，在分配汽车后立即添加print doors表示doors在第一次迭代后只有两个元素。

Answer 2

我不久前写了一篇Monty Hall模拟问题。也许它会帮助您使用代码 - 特别是注释可能很有用：

from __future__ import division
import random

results = [] # a list containing the results of the simulations, either 'w' or 'l', win or lose

count = 0

while count <200: #number of simulations to perform
    l = [] 

    prize = random.randint(1, 3) #choose the prize door

    initialchoice = random.randint(1, 3) #make an initial choice (the door the contestant chooses)

    exposed = random.randint(1, 3) #choose the exposed door (the door the host chooses)

    while exposed == initialchoice or exposed == prize: #make sure exposed is not same as prize or the initial choice
        exposed = random.randint(1, 3)

    if initialchoice != prize:
        results.append('l') #if the initial choice was incorrect, append 'l'
    else:
        results.append('w') #if the initial choice was correct, append 'w'

    count += 1
    print 'prize door:', prize, 'initial choice:',initialchoice, 'exposed door:',exposed #print the results of the simulation
    print

w = 0
for i in results:
    if i == 'w':
        w += 1
print w/len(results) #fraction of times sticking with the original door was successful

蒙蒂大厅模拟返回50％赔率？

2 个答案: