The Monty Hall paradox is named after the host of the TV game show “Let’s Make a Deal”. In the game, players are presented with three doors. One door has a new car behind it while the other two each have a goat behind them. The player is instructed to select a door. After the player makes their selection, Monty opens one of the two remaining doors to reveals a goat. The player is then given the choice to change their selection to the remaining unopened door or stay with their previous selection.

The paradox was introduced in an 1990 Parade Magazine article written by Marilyn vos Savant, Guinness Book of Records holder of highest recorded IQ. To almost everyone who first saw the article, including some very smart PhDs who wrote to Parade after the article was published, the remaining two closed doors after Monty’s reveal appear to each have an equal probability of containing a car.

Instead, after Monty opens one of the unselected doors, revealing a goat, the remaining unselected closed door has a ⅔ probability of containing the car while the door initially selected by the contestant has a ⅓ probability of containing the car.

The basic idea of the paradox is that the uniform probability of a car being behind each door is altered by the decision Monty makes when he picks one of the unselected doors to open, the criteria for which door to open being that it doesn’t have a car behind it. The remaining closed door, given that it has stood up to the test of “between these two doors, open a door that doesn’t contain a car” gives it a higher probability of containing a car.

But wait! What if you initially select the door with the car and the remaining two doors are goats?! Monty’s decision doesn’t show anything about the remaining unselected closed door because Monty could have selected either of the unselected doors to open.

This is exactly why the remaining unselected closed door has a ⅔ probability of containing a car. The case mentioned above will occur ⅓ of the time, therefore the unselected unopened door will contain a goat ⅓ of the time, leaving ⅔ when it contains a car.

If it weren’t for the case where you initially select the door hiding the car, then the remaining unselected closed door after Monty’s reveal would be a car every time, which would make the game very boring.

As you can see, by making sure you take the remaining unselected closed door each time Monty reveals a goat, you can double your chances of getting the car (from a 33.33% chance to a 66.66% chance).

Still not convinced? Experiment for yourself. The python script bellow will play 1 million games using the strategy of changing the selection to the remaining closed unselected door each time Monty reveals a goat. The win rate will be close to 66.66%.

import random

# number of games to play
num_games = 1000000

# keeps track of wins that occur when using
# the selection-change-after-reveal strategy
reselect_win_tally = 0

for i in range(num_games):

# there are three doors, and a random
# one has the car behind it
doors = [False, False, False]
cars_door = random.randrange(len(doors))
doors[cars_door] = True

# player selects a random door
selected_door = random.randrange(len(doors))

# if the selected door is not a car, then the
# strategy of changing the selection after the
# reveal will result in a win
if not doors[selected_door]:
reselect_win_tally+=1

print(str(reselect_win_tally) + " wins, " +
str(num_games - reselect_win_tally) + " losses")