Probability and Intuition
The Monty-Hall-Problem in Figures

The Monty Hall Problem, originally posed in a letter by Steve Selvin to the American Statistician in 1975, is loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem is about strategies and chances of a player, who may win a big price (a new car) by the following procedure:

• There are three closed doors (Door #1, Door #2, Door #3).
• Behind one of the doors is the big price, while there are just goats behind the other two.
• The player picks one of the doors.
• The game host then opens one of the two other doors to reveal a goat.
• The player is given a chance to alter the initial choice by switching to the third, remaining door.

Obviously, there are three basic strategies:

• Sticking to the first choice.
• Switching to the remaining door.
• Chosing one of the above by random.

Is there a difference?

The problem became famous in the form of a reader's letter quoted in Marilyn vos Savant's “Ask Marilyn” column in Parade magazine in 1990:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

Compare: https://en.wikipedia.org/wiki/Monty_Hall_problem.

Putting it to a Test

Here are computed results of a simulation of 1000 random games:

This is because the winning condition for strategy #2 is the first choice being a goat (p = 2:3).
Since Monty Hall already disclosed a goat, the player will win the price, if her/his first choice was a door with a goat, leaving only the door with the price to choose. Notably, there are two goats and just a single price. While sticking with the first choice, the player has only a chance of 1:3 of picking the price.

Switching the choice effectively inverts the original winning condition (p₂ = ¬p₁):

p₂ = ¬p₁   ⇒   p₂ = 1 − p₁   ⇒   p₂ = 1 − 1/3   ⇒   p₂ = 2/3

So, in fact there is no second choice made with strategy #2.
We're just betting now on the first choice being a goat, which is twice the chance of it being the price.
If a real second choice is made — as with random strategy #3 —, p equals 1/2 as might be intuitively expected (p = (p₁ + p₂) / 2 = 1/2).

Thus, the probablity of winning using strategy #2 (switching, 2:3) is twice the one of sticking to the first choice (1:3) as the player wins every time using strategy #2, when there'd be a loss with strategy #1, and still 150% as compared to a pure random strategy (1:2).

Should this is still be contrary to your intuition, a closer look at simulated results may help:

Protocol of 1000 Simulated Games Rerun

Results per Strategy:

#1 – Stick with first choice:
#2 – Switch choice strictly:
#3 – Random strategy:

Statistics:

L ..................	lose
W ..................	win
G ..................	door with goat
P ..................	door with price
green ..................	player's first choice
red ..................	door with goat disclosed by Monty Hall
blue ..................	altered, second choice (if applicable)
grey ..................	ignored, still closed (if applicable)

Each round we use the same random allocation of goats and price and determine an initial choice per random, which is used for each of the three strategies. Then we either stick with the choice or switch it according to the respective strategy, resulting in a final pick (underlined) and evaluate the results.