蒙蒂四门 python

Question

我正在编写一个程序来计算 Monty Hall 问题中有 4 扇门的概率，但未打印更改选择的概率。

import random

door = ['a', 'b', 'c', 'd']
nc = 0 # When not changed
c = 0  # When changed

tr = 100 #Total number of iterations
for a in range(tr):
    car = random.randint(0, 3) #Inquiry number with car behind

    pc = random.randint(0, 3) #Inquiry number chosen by the participant

    ed = [] #Empty door

for i in range(4):
    if i != pc and i != car:
        ed.append(door[i])


com = random.sample(ed, 2)

if pc == car:
    nc += 1


def list_remover(the_list, val):
    while val in the_list:
        the_list.remove(val)

list_remover(ed, com)


if not ed:
    c += 1

print((nc / tr) * 100, "%")
print((c / tr) * 100, "%")

Answer 1

最初发布的代码不清楚（可能是缩进问题，我没有遵循代码逻辑）。

但是，对于任意数量的门的 Monty Hall 模拟如下。

代码

from random import choice

# Number of doors
N_doors = int(input("How many doors (>= 3? "))

wins_change = 0                    # number of wins when we change doors
wins_no_change = 0                 # number of wins without changing doors
doors = list(range(1, N_doors+1))  # doors as list [1, 2, ... N_doors]
N_Trials = 10000                   # number of times to run simulation
for K in range(1, N_Trials): 
    # Host randomly chooses a door for answer
    host_pick = choice(doors) 
    
    # Player randomly chooses a door for their answer
    player_pick = choice(doors)
    
    # Host picks a door to show
    # that's not host_pick or player_pick
    # i.e. door won't be host_pick or player_pick
    show = choice([i for i in doors if i != host_pick and i != player_pick])
    
    # Update win count if player doesn't change selection now that a door is shown
    if host_pick == player_pick:
      wins_no_change += 1

    # Player changes selection 
    # i.e. picks door that's not shown and wasn't their original pick
    player_pick = choice([i for i in doors if i != show and i != player_pick])

    # Player wins if player_pick equals host_pick
    if player_pick == host_pick:
      wins_change += 1
    
# show results
print(f'Probablity winning not changing selection {wins_no_change/N_Trials:.2%}')
print(f'Probablity winning after changing selection {wins_change/N_Trials:.2%}')

测试

测试 3 扇门

How many doors (assume >= 3)? 3
Probablity winning not changing selection 33.29%
Probablity winning after changing selection 66.70%

测试 4 扇门

How many doors? 4
Probablity winning not changing selection 25.73%
Probablity winning after changing selection 36.90%

注：

3门答案同意Monty Hall problem
4门答案符合数学推导monty hall question with 4 doors

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