蒙蒂霍尔问题说明了一个令人惊讶的概率事实。这个问题大致基于老游戏节目《让我们做个交易》和它的主持人蒙蒂·霍尔。在蒙蒂大厅问题中，你可以选择三扇门中的一扇门。一扇门后有一个奖品：一辆新车。另外两扇门都通向一只没用的山羊。假设你选了 1 号门。在你选择的门打开之前，主人打开了另一扇门（2 号或 3 号），这导致了一只山羊。您可以选择打开您最初选择的门，或者切换到另一扇未打开的门。
看起来你是否换门并不重要，但是如果你换了门，你的机会就会增加！这个程序通过让你重复做实验来演示蒙蒂霍尔问题。
为了理解为什么你的机会增加了，考虑一个有一千个门而不是三个门的蒙蒂大厅问题的版本。你挑一扇门，然后主持人打开 998 扇门，里面都是山羊。仅有的两个未打开的门是您选择的门和另一个门。如果你一开始就选对了车门（1/1000 的概率），那么主人会让一个随机的山羊车门关着。如果你选了一个山羊车门（1000 分之 999 的概率），主人会特别选择车门保持关闭。选择打开哪扇门不是随机的；主人知道让车门关着。几乎可以肯定的是，你一开始就没有选车门，所以你应该换另一个车门。
另一种想法是，你有 1000 个盒子，一个盒子里装着奖品。你猜猜奖品在哪个盒子里，主持人把它放到你手里。你认为奖品在你的盒子里还是在其他 999 个盒子里？你不需要主持人打开 999 个不含奖品的盒子中的 998 个；选择的数量与 1000 扇门相同。你一开始猜对的几率是 1/1000，而猜错的几率（奖品在其他盒子里）几乎是 999/1000。
更多关于蒙蒂霍尔问题的信息可以在en.wikipedia.org/wiki/Monty_Hall_problem找到。
当您运行montyhall.py时，输出将如下所示：
The Monty Hall Problem, by Al Sweigart email@protected
`--snip--`
+------+ +------+ +------+
| | | | | |
| 1 | | 2 | | 3 |
| | | | | |
| | | | | |
| | | | | |
+------+ +------+ +------+
Pick a door 1, 2, or 3 (or "quit" to stop):
> 1
+------+ +------+ +------+
| | | | | (( |
| 1 | | 2 | | oo |
| | | | | /_/|_|
| | | | | | |
| | | | |GOAT|||
+------+ +------+ +------+
Door 3 contains a goat!
Do you want to swap doors? Y/N
> y
+------+ +------+ +------+
| (( | | CAR! | | (( |
| oo | | __| | oo |
| /_/|_| | _/ | | /_/|_|
| | | | /_ __| | | |
|GOAT||| | O | |GOAT|||
+------+ +------+ +------+
Door 2 has the car!
You won!
Swapping: 1 wins, 0 losses, success rate 100.0%
Not swapping: 0 wins, 0 losses, success rate 0.0%
Press Enter to repeat the experiment...
`--snip--`ASCII 艺术画门的多行字符串存储在几个常量变量中，比如ALL_CLOSED、FIRST_GOAT和FIRST_CAR_OTHERS_GOAT。使用这些常量的代码，比如第 125 行的print(FIRST_GOAT)，即使我们更新了图形也保持不变。通过将多行字符串一起放在源代码文件的顶部，我们将更容易比较它们，以确保图形是一致的。
"""The Monty Hall Problem, by Al Sweigart email@protected
A simulation of the Monty Hall game show problem.
More info at https://en.wikipedia.org/wiki/Monty_Hall_problem
This code is available at https://nostarch.com/big-book-small-python-programming
Tags: large, game, math, simulation"""
import random, sys
ALL_CLOSED = """
+------+ +------+ +------+
| | | | | |
| 1 | | 2 | | 3 |
| | | | | |
| | | | | |
| | | | | |
+------+ +------+ +------+"""
FIRST_GOAT = """
+------+ +------+ +------+
| (( | | | | |
| oo | | 2 | | 3 |
| /_/|_| | | | |
| | | | | | |
|GOAT||| | | | |
+------+ +------+ +------+"""
SECOND_GOAT = """
+------+ +------+ +------+
| | | (( | | |
| 1 | | oo | | 3 |
| | | /_/|_| | |
| | | | | | |
| | |GOAT||| | |
+------+ +------+ +------+"""
THIRD_GOAT = """
+------+ +------+ +------+
| | | | | (( |
| 1 | | 2 | | oo |
| | | | | /_/|_|
| | | | | | |
| | | | |GOAT|||
+------+ +------+ +------+"""
FIRST_CAR_OTHERS_GOAT = """
+------+ +------+ +------+
| CAR! | | (( | | (( |
| __| | oo | | oo |
| _/ | | /_/|_| | /_/|_|
| /_ __| | | | | | |
| O | |GOAT||| |GOAT|||
+------+ +------+ +------+"""
SECOND_CAR_OTHERS_GOAT = """
+------+ +------+ +------+
| (( | | CAR! | | (( |
| oo | | __| | oo |
| /_/|_| | _/ | | /_/|_|
| | | | /_ __| | | |
|GOAT||| | O | |GOAT|||
+------+ +------+ +------+"""
THIRD_CAR_OTHERS_GOAT = """
+------+ +------+ +------+
| (( | | (( | | CAR! |
| oo | | oo | | __|
| /_/|_| | /_/|_| | _/ |
| | | | | | | /_ __|
|GOAT||| |GOAT||| | O |
+------+ +------+ +------+"""
print('''The Monty Hall Problem, by Al Sweigart email@protected
In the Monty Hall game show, you can pick one of three doors. One door
has a new car for a prize. The other two doors have worthless goats:
{} 77\. Say you pick Door #1.
Before the door you choose is opened, another door with a goat is opened:
{} 80\. You can choose to either open the door you originally picked or swap
to the other unopened door.
It may seem like it doesn't matter if you swap or not, but your odds
do improve if you swap doors! This program demonstrates the Monty Hall
problem by letting you do repeated experiments.
You can read an explanation of why swapping is better at
https://en.wikipedia.org/wiki/Monty_Hall_problem
'''.format(ALL_CLOSED, THIRD_GOAT))
input('Press Enter to start...')
swapWins = 0
swapLosses = 0
stayWins = 0
stayLosses = 0
while True: # Main program loop.
# The computer picks which door has the car:
doorThatHasCar = random.randint(1, 3)
# Ask the player to pick a door:
print(ALL_CLOSED)
while True: # Keep asking the player until they enter a valid door.
print('Pick a door 1, 2, or 3 (or "quit" to stop):')
response = input('> ').upper()
if response == 'QUIT':
# End the game.
print('Thanks for playing!')
sys.exit()
if response == '1' or response == '2' or response == '3':
break
doorPick = int(response)
# Figure out which goat door to show the player:
while True:
# Select a door that is a goat and not picked by the player:
showGoatDoor = random.randint(1, 3)
if showGoatDoor != doorPick and showGoatDoor != doorThatHasCar:
break
# Show this goat door to the player:
if showGoatDoor == 1:
print(FIRST_GOAT)
elif showGoatDoor == 2:
print(SECOND_GOAT)
elif showGoatDoor == 3:
print(THIRD_GOAT)
print('Door {} contains a goat!'.format(showGoatDoor))
# Ask the player if they want to swap:
while True: # Keep asking until the player enters Y or N.
print('Do you want to swap doors? Y/N')
swap = input('> ').upper()
if swap == 'Y' or swap == 'N':
break
# Swap the player's door if they wanted to swap:
if swap == 'Y':
if doorPick == 1 and showGoatDoor == 2:
doorPick = 3
elif doorPick == 1 and showGoatDoor == 3:
doorPick = 2
elif doorPick == 2 and showGoatDoor == 1:
doorPick = 3
elif doorPick == 2 and showGoatDoor == 3:
doorPick = 1
elif doorPick == 3 and showGoatDoor == 1:
doorPick = 2
elif doorPick == 3 and showGoatDoor == 2:
doorPick = 1
# Open all the doors:
if doorThatHasCar == 1:
print(FIRST_CAR_OTHERS_GOAT)
elif doorThatHasCar == 2:
print(SECOND_CAR_OTHERS_GOAT)
elif doorThatHasCar == 3:
print(THIRD_CAR_OTHERS_GOAT)
print('Door {} has the car!'.format(doorThatHasCar))
# Record wins and losses for swapping and not swapping:
if doorPick == doorThatHasCar:
print('You won!')
if swap == 'Y':
swapWins += 1
elif swap == 'N':
stayWins += 1
else:
print('Sorry, you lost.')
if swap == 'Y':
swapLosses += 1
elif swap == 'N':
stayLosses += 1
# Calculate success rate of swapping and not swapping:
totalSwaps = swapWins + swapLosses
if totalSwaps != 0: # Prevent zero-divide error.
swapSuccess = round(swapWins / totalSwaps * 100, 1)
else:
swapSuccess = 0.0
totalStays = stayWins + stayLosses
if (stayWins + stayLosses) != 0: # Prevent zero-divide.
staySuccess = round(stayWins / totalStays * 100, 1)
else:
staySuccess = 0.0
print()
print('Swapping: ', end='')
print('{} wins, {} losses, '.format(swapWins, swapLosses), end='')
print('success rate {}%'.format(swapSuccess))
print('Not swapping: ', end='')
print('{} wins, {} losses, '.format(stayWins, stayLosses), end='')
print('success rate {}%'.format(staySuccess))
print()
input('Press Enter repeat the experiment...') 试着找出下列问题的答案。尝试对代码进行一些修改，然后重新运行程序，看看这些修改有什么影响。
- 如果把第 100 行的
doorThatHasCar = random.randint(1, 3)改成doorThatHasCar = 1会怎么样？ - 如果把第 124 到 129 行换成
print([FIRST_GOAT, SECOND_GOAT, THIRD_GOAT][showGoatDoor - 1])会怎么样？