三門問題模擬：換門跟不換門的中獎機率

threedoors.py

import random

ROUNDS = 10000  # The times to re-play
COUNT_CHANGE = 0  # If we change the chosen door, the times receiving the car
COUNT_NOT_CHANGE = 0  # If we don't change the chosen door, the times receiving the car

# Play the game, and choose to change the door after first picked
for _ in range(ROUNDS):
    # Create 3 doors
    doors = {0: 0, 1: 0, 2: 0}
    # Mark one door as the grand prize
    doors[random.randint(0, 2)] = 1
    # print("The answer is door #{}".format(str(doors)))

    # Pick a door
    pick = random.randint(0, 2)
    # print("I pick door #{}".format(str(pick)))
    doors.pop(pick)

    # Open a door has a goat and it's not my pick
    goat = list(filter(lambda k: doors[k]==0, doors))[0]
    # print("Another door #{} has a goat!".format(str(goat)))
    doors.pop(goat)

    last_door = list(doors.keys())[0]
    # print("I change my mind and pick another door #{}".format(str(last_door)))

    if doors[last_door] == 0:
        pass
        # print("The last door #{} has a goat. Booooooooo".format(str(last_door)))
    else:
        COUNT_CHANGE += 1
        # print("The last door #{} has a Car. YAAAAAAAA!".format(str(last_door)))

# Play the game, and not to change the picked door
for _ in range(ROUNDS):
    # Create 3 doors
    doors = {0: 0, 1: 0, 2: 0}
    # Mark one door as the grand prize
    doors[random.randint(0, 2)] = 1
    # print("The answer is door #{}".format(str(doors)))

    # Pick a door
    pick = random.randint(0, 2)
    
    # I'm not gonna change my mind, so just verify the answer
    if doors[pick] == 0:
        pass
        # print("The door #{} has a goat. Booooooooo".format(str(pick)))
    else:
        COUNT_NOT_CHANGE += 1
        # print("The door #{} has a Car. YAAAAAAAA!".format(str(last_door)))

# Print the times we get the car
print(COUNT_CHANGE, COUNT_NOT_CHANGE)  # Change my mind / Not change my mind
print(COUNT_CHANGE / COUNT_NOT_CHANGE)  # Print ratio

threedoors_v2.py

import random

ROUNDS = 10000  # The times to re-play
COUNT_CHANGE = 0  # If we change the chosen door, the times receiving the car
COUNT_NOT_CHANGE = 0  # If we don't change the chosen door, the times receiving the car

# Play the game, and choose to change the door after first picked
for _ in range(ROUNDS):
    # Create 3 doors
    doors = {0: 0, 1: 0, 2: 0}
    # Mark one door as the grand prize
    doors[random.randint(0, 2)] = 1

    # Pick a door
    pick = doors.pop(0)
    if pick == 1:  # If the first pick is a car, change a door must get a goat, just do nothing.
        pass
    else:
        goat = list(filter(lambda k: doors[k]==0, doors))[0]  # Filter out a remaining door that has a goat
        doors.pop(goat)  # Remove that door

        last_door = list(doors.keys())[0]  # The last door
        if doors[last_door] == 0:  # If the last door has a goat
            pass
        else:  # If the last door has the car
            COUNT_CHANGE += 1

# Play the game, and not to change the picked door
for _ in range(ROUNDS):
    # Create 3 doors
    doors = {0: 0, 1: 0, 2: 0}
    # Mark one door as the grand prize
    doors[random.randint(0, 2)] = 1

    # Pick a door, and I won't change my mind, so just verify the answer
    if doors[pick] == 0:
        pass
    else:
        COUNT_NOT_CHANGE += 1

# Print the times we get the car
print(COUNT_CHANGE, COUNT_NOT_CHANGE)  # Change my mind / Not change my mind
print(COUNT_CHANGE / COUNT_NOT_CHANGE)  # Print ratio

v2 跟原版的差別：選擇的門 (pick) 不用做隨機。因為已經隨機指定一個門裡有車子了，接下來可以直接指定一個門就好，再隨機也沒有多大的意義。

也就是：

1. 隨機指定一個門裡有車
2. 隨機選一個門

這兩件事情只做一件就好 😄

原版的做法比較直覺，就是照著綜藝節目的流程運作，而 v2 的方法就是單純求出結果。

最終：有換門而得到汽車的機率，將是不換門的兩倍，可以得到汽車的機率分別是 2/3 跟 1/3。

Oh my hands slipped and accidentally made this public 💫 😵 💫