Python-Monty-Hall模拟给出了交换和n的相等概率

Question

我在Python中创建了一个（相当复杂的）MontyHall模拟，但是，运行时，切换和不切换的几率相等，而不是，当我知道，实际上，情况并非如此。怎么了？

import math
import random

Right = 0
def TestWithSwitch():
    global Right
    wdoor = math.floor(random.random() * 3)
    doors = [0,0,0]
    doors[wdoor] = 1
    Ldoors = [0,0]
    i=0
    ##Declare winning door to be the winning door in the door Array

    for x in range(0, 3):
        if(x!=3):
            if(doors[x] != 1):
                Ldoors[i] = x
                i+=1
    ##Chose the losing doors to be the doors that aren't the winning door
    choice = math.floor(random.random() * 3)
    DoorOut = 0
    ##Pick a Choice
    LChose = False
    for y in range(0, 2):
        if(y!= 2):
            if(Ldoors[y] == choice):
                DoorOut = Ldoors[(y+1)%2]
                LChose = True
    if(LChose == False):
        DoorOut = Ldoors[math.floor(random.random() * 2)]
    Reserved = [DoorOut, choice]

##DoorOut is chosen from any of the losing doors we didn't pick as our choice, and is the door the computer is told doesn't have the prize
    for z in range(0, 3):
        if(z!= 3):
            if(z in Reserved == False):
                choice = z
                ##Make our new choice the other choice that we didn't previously choose
    if(choice == wdoor):
        Right+=1

def TestNoSwitch():
    global Right
    wdoor = math.floor(random.random() * 3)
    doors = [0,0,0]
    doors[wdoor] = 1
    Ldoors = [0,0]
    i=0


    for x in range(0, 3):
        if(x!=3):
            if(doors[x] != 1):
                Ldoors[i] = x
                i+=1
    choice = math.floor(random.random() * 3)
    if(choice == wdoor):
        Right+=1



for j in range(1, 10000):
    ## TestWithSwitch() and TestNoSwitch() both result in about 1/3. You can test by putting either function in.
    if(j == 9999):
        print(Right/10000)

我知道，转换应该返回66%的几率，而不应该返回33%的几率。我所得到的几率甚至不到100%，而是大约三分之二，这是不可能的概率。

Answer 1

问题在于：

if(z in Reserved == False):

这将被计算为chained comparison，结果总是错误的。在

^{pr2}$

使用括号创建右运算符顺序：

^{3}$

或者最好使用明确的“不在”运算符：

if(z not in Reserved):

Answer 2

首先

for x in range(0, 3):
    if(x!=3):          # redundant: x will never be 3

另一方面，通过全局变量收集函数结果有点邪恶。在

我可以这样做：

^{pr2}$

产生的输出

^{3}$

注意：您可以通过任意地将guess = pick()替换为guess = "A"来加快它的速度（不损失通用性）。在