Question

我正在尝试模拟Monty Hall问题，其中有人选择了一扇门，并且随机移除了一扇门 - 最后它必须是一辆有车和一辆没有，其中一人必须选择。虽然我现在不需要模拟/询问使用该程序的人他们想要哪个门，但我实际上在设置计算时遇到了麻烦。当我运行代码时，它输出0，其中应该是大约66％

import random

doors=[0,1,2]
wins=0

car=random.randint(0,2)
player=random.randint(0,2)

#This chooses the random door removed
if player==car:
    doors.remove.random.randint(0,2)
else:
    doors.remove(car)
    doors.remove(player)

for plays in range(100):
    if car == player:
        wins=wins+1

print(wins)

Answer 1

您需要将代码放在循环中以实际让它每次运行。您还需要确保您第二次只允许有效选择（他们不能选择拆除的门）并且您只是移除有效的门（您无法将车门或播放器移开 - 选择的门）。

import random

wins = 0

for plays in range(100):
    doors = [0,1,2]
    car = random.choice(doors)
    player = random.choice(doors)

    # This chooses the random door removed
    doors.remove(random.choice([d for d in doors if d != car and d != player]))

    # Player chooses again (stay or switch)
    player = random.choice(doors)
    if player == car:
        wins += 1

print(wins)

但是出于Monty Hall问题的目的，你甚至不需要追踪门。

win_if_stay = 0
win_if_switch = 0
for i in range(100):
    player = random.randint(0, 2)
    car = random.randint(0, 2)
    if player == car:
        win_if_stay += 1
    else:
        win_if_switch += 1

Monty Hall Python模拟计算

1 个答案: