我需要编写一个基于旧电视节目的python程序，让我们做一笔交易。我得到程序打印出游戏的数量，如果用户应该切换或停留。现在我想弄清楚如何打印用户应该留下并完全切换的次数百分比。
以下是测试输入：
25
7
exit
这是程序应该输出的内容：
Game 1
Doors : [ ’G’, ’C’, ’G’ ]
Player Selects Door 1
Monty Selects Door 3
Player should switch to win.
Game 2
Doors : [ ’C’, ’G’, ’G’ ]
Player Selects Door 2
Monty Selects Door 3
Player should switch to win.
Game 3
Doors : [ ’G’, ’C’, ’G’ ]
Player Selects Door 1
Monty Selects Door 3
Player should switch to win.
Game 4
Doors : [ ’C’, ’G’, ’G’ ]
Player Selects Door 2
Monty Selects Door 3
Player should switch to win.
Game 5
Doors : [ ’G’, ’G’, ’C’ ]
Player Selects Door 3
Monty Selects Door 1
Player should stay to win.
Game 6
Doors : [ ’G’, ’C’, ’G’ ]
Player Selects Door 2
Monty Selects Door 1
Player should stay to win.
Game 7
Doors : [ ’G’, ’G’, ’C’ ]
Player Selects Door 2
Monty Selects Door 1
Player should switch to win.
Stay Won 28.6% of the time.
Switch Won 71.4% of the time.
How many tests should we run?
Thank you for using this program.
这是我的程序输出的内容：
Enter Random Seed:
25
Welcome to Monty Hall Analysis
Enter 'exit' to quit
How many tests should we run?
7
Game 1
Doors: ['G', 'C', 'G']
Player Selects Door 1
Monty Selects Door 3
Player should switch to win.
Game 2
Doors: ['G', 'C', 'G']
Player Selects Door 2
Monty Selects Door 1
Player should stay to win.
Game 3
Doors: ['C', 'G', 'G']
Player Selects Door 1
Monty Selects Door 3
Player should stay to win.
Game 4
Doors: ['G', 'G', 'C']
Player Selects Door 3
Monty Selects Door 2
Player should stay to win.
Game 5
Doors: ['G', 'G', 'C']
Player Selects Door 3
Monty Selects Door 2
Player should stay to win.
Game 6
Doors: ['G', 'C', 'G']
Player Selects Door 3
Monty Selects Door 1
Player should switch to win.
Game 7
Doors: ['C', 'G', 'G']
Player Selects Door 2
Monty Selects Door 3
Player should switch to win.
How many tests should we run?
这是我制作的代码：
import random
import sys
try:
randSeed = int(input('Enter Random Seed:\n'))
random.seed(randSeed)
except ValueError:
sys.exit("Seed is not a number!")
print('Welcome to Monty Hall Analysis')
print("Enter 'exit' to quit")
while True:
testNum = input('How many tests should we run?\n')
valid = False
while not valid:
try:
if testNum == "exit":
sys.exit("Thank you for using this program.")
else:
testNum = int(testNum)
valid = True
except ValueError:
testNum = input('Please enter a number:\n')
pStay = 0
pChange = 0
numGame = 0
for numGame in range(1, testNum + 1):
doorList = ['C', 'G', 'G']
random.shuffle(doorList)
print('Game', numGame)
print('Doors:', doorList)
playerDoor = random.randint(0,2)
montyDoor = random.randint(0,2)
print('Player Selects Door', playerDoor+1)
while montyDoor == playerDoor or doorList[montyDoor] == 'C':
montyDoor = random.randint(0,2)
print('Monty Selects Door', montyDoor+1)
if doorList[playerDoor] == 'C':
var = 0
else:
var = 1
if var == 0:
pStay += 1
print('Player should stay to win.')
pStay += 1
if var == 1:
print('Player should switch to win.')
很抱歉，如果我的代码看起来不正确或令人困惑。这是我第一次编程谢谢。
答案 0 :(得分：0)
好吧，你要记录玩家应该留下的次数。因此留下的百分比只是“（pStay / float（testNum））* 100）”然后简单地从100减去该数字以获得改变的百分比（因为它们必须加起来为100％）
我想我应该提供更多信息。公式是将游戏数量从游戏总数中扣除。您将乘以100以将十进制值转换为百分比。
所以，如果你应该留在1场比赛中，并且你打了10场比赛，它将是1/10，即.1，倍数100，是10％。
自1/10你应该留下来，这意味着你应该改变9/10。因此，您可以减去停留百分比以获得更改百分比，即100％ - 10％= 90％
我在代码中放置float（）转换的原因是因为在python2中，如果将整数除以整数，则不会计算小数部分。它只是向下舍入到整数值。所以1/10会给你0，而不是.1。在python3中它实际上产生了一个小数值，但由于我不知道你使用的是哪个版本，所以将它转换为浮点数以获得预期结果是安全的
答案 1 :(得分：0)
看到下面添加的评论，你很接近。但是，您错过了pSwitch的总计数变量。希望这可以帮助。
ATTR_KEEP_ROWS_TOGETHER答案 2 :(得分：-1)
这是使用集的通用版Make a Deal的Python 3.6模拟。在“达成交易”的通用版本中，可以打开的门和门的数量可以变化。请参阅以供参考：
https://math.stackexchange.com/questions/608957/monty-hall-problem-extended
如果在doors = 3且doors_to_open = 1的情况下运行，则如果未选择切换门的选项，则预期结果为33％，当选择门时为66％。
#!/usr/bin/env python
''' application of Make a deal statistics
application is using sets {}
for reference see:
https://math.stackexchange.com/questions/608957/monty-hall-problem-extended
'''
import random
def Make_a_Deal(doors, doors_to_open):
''' Generalised function of Make_a_Deal. Logic should be self explanatory
Returns win_1 for the option when no change is made in the choice of
door and win_2 when the option to change is taken.
'''
win_1, win_2 = False, False
doors = set(range(1, doors+1))
price = set(random.sample(doors, 1))
choice1 = set(random.sample(doors, 1))
open = set(random.sample(doors.difference(price).
difference(choice1), doors_to_open))
choice2 = set(random.sample(doors.difference(open).
difference(choice1), 1))
win_1 = choice1.issubset(price)
win_2 = choice2.issubset(price)
return win_1, win_2
def main():
''' input:
- throws: number of times to Make_a_Deal (must be > 0)
- doors: number of doors to choose from (must be > 2)
- doors_to_open: number of doors to be opened before giving the
option to change the initial choice (must be > 0 and <= doors-2)
'''
try:
throws = int(input('how many throws: '))
doors = int(input('how many doors: '))
doors_to_open = int(input('how many doors to open: '))
if (throws < 1) or (doors < 3) or \
(doors_to_open > doors-2) or (doors_to_open < 1):
print('invalid input')
return
except Exception as e:
print('invalid input: ', e)
return
number_of_wins_1, number_of_wins_2, counter = 0, 0, 0
while counter < throws:
win_1, win_2 = Make_a_Deal(doors, doors_to_open)
if win_1:
number_of_wins_1 += 1
if win_2:
number_of_wins_2 += 1
counter += 1
print('completion is {:.2f}%'.
format(100*counter/throws), end='\r')
print('number of wins option 1 is {:.2f}%: '.
format(100*number_of_wins_1/counter))
print('number of wins option 2 is {:.2f}%: '.
format(100*number_of_wins_2/counter))
if __name__ == '__main__':
main()