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Monty Hall Problem and Intuitive Solutions
Jan 1, 2005

Foreword

Dear DES group: This problem bothered me all night, but I finally came to grips with it during my swim this morning.

Classical Problem

There are three doors. Two hide nothing. One hides a car. In phase 1, you pick a door. Monty (knowing where the car is) opens a door that does not hide a car and that you did not pick. In phase 2, you may stay with your chosen door (StayStrategy) or you may change your guess to the one remaining door (SwapStrategy). If you've picked the door that hides the car, you win the car.

The Truth

In the Classical Monty Hall problem, the StayStrategy succeeds with 1/3 probability, and the SwapStrategy succeeds with 2/3 probability.

Hand Wavy Justification

Imagine that there are 100 doors. You choose one. Knowing where the car is, Monty Hall opens 98 doors (other than your pick) that don't have the car. It is intuitively clear that your door has low probability of success, while the swap door has high probability of success.

Modified Problem

Imagine the same game, except Monty Hall doesn't know where the car is. After you pick a door in phase 1, he randomly reveals one of the remaining two doors. Now consider the situation where (by chance) he does not reveal the car. Should you stay or swap? Note: this problem only considers the situation when Monty Hall (by chance) has opened a door that does not hide the car.

The Truth

In the Modified Monty Hall problem, the StayStrategy succeeds with 1/2 probability, and the SwapStrategy succeeds with 1/2 probability.

Hand Wavy Anti-Justification

Imagine that there are 100 doors. You choose one. Not knowing where the car is, Monty Hall opens 98 doors (other than your pick) that don't have the car. This situation is highly unlikely, but possible. Originally, you knew that your door had a 1/100 probability of hiding the car, so it must reasonably be true that all the other doors combined had a 99/100 probability of hiding the car, and since they have been narrowed to one selection, the swap door must have a higher probability of success.

Proof For The Classical Problem

The Classical Monty Hall problem is exactly equivalent to the following problem. You pick a door with a 1/3 probability of success. There is a 2/3 probability that one of the other two doors hides the car. You are given the option of opening and winning the contents of just your selected door (StayStrategy) or both the other doors (SwapStrategy). Because Monty Hall never reveals the car, the (SwapStrategy) is exactly equivalent to having both the other doors, and therefore 2/3 probability of success. It is important to remember that there are only two possible outcomes to this game: the stay door hid the car or the swap door hid the car. The Monty door never hides the car. This proof is particularly evident when examining the source code for a simulation of the game. In the simulation it is not necessary to determine which door Monty reveals. All you need to do is randomly generate a car on [1..3] and randomly generate a guess on [1..3]. If the car equals the guess then the (StayStrategy) wins, otherwise the (SwapStrategy) wins.

Proof For The Modified Problem

The Modified Monty Hall problem is actually three problems considered in isolation. It is true that your original pick has 1/3 probability of success. It is true that Monty Hall is not allowed to reveal your door. This does not change the fact that the door Monty Hall picks has a 1/3 probability of hiding the car. Since neither your nor Monty have knowledge of which door hides the car, regardless of the picks, each door has a 1/3 probability of hiding the car. Hence in the millions of times that the game is played, 1/3 of the games end with Monty Hall accidentally revealing the car. In the remaining 2/3 of the games played there is a 50% probability that the (StayStrategy) wins, and a 50% probability that the (SwapStrategy) wins. That is to say that 1/3 of all games end with Monty Hall accidentally revealing the car, 1/3 of all games end with your original pick hiding the car, and 1/3 of all games end with the remaining door (picked by neither you nor Monty) hiding the car.

If You Don't Believe Me

Java Simulation Source Code
Java Simulation Executable Jar File
Java Simulation Results


Michael Chalkboards Notepads Blogs Public Monty Hall Revisited Monty Hall Problem... Puzzles Punchclocks Death Counters Photos Travel Services Member Login Visitor Login	Public Monty Hall Problem and Intuitive Solutions Jan 1, 2005 Foreword Dear DES group: This problem bothered me all night, but I finally came to grips with it during my swim this morning. Classical Problem There are three doors. Two hide nothing. One hides a car. In phase 1, you pick a door. Monty (knowing where the car is) opens a door that does not hide a car and that you did not pick. In phase 2, you may stay with your chosen door (StayStrategy) or you may change your guess to the one remaining door (SwapStrategy). If you've picked the door that hides the car, you win the car. The Truth In the Classical Monty Hall problem, the StayStrategy succeeds with 1/3 probability, and the SwapStrategy succeeds with 2/3 probability. Hand Wavy Justification Imagine that there are 100 doors. You choose one. Knowing where the car is, Monty Hall opens 98 doors (other than your pick) that don't have the car. It is intuitively clear that your door has low probability of success, while the swap door has high probability of success. Modified Problem Imagine the same game, except Monty Hall doesn't know where the car is. After you pick a door in phase 1, he randomly reveals one of the remaining two doors. Now consider the situation where (by chance) he does not reveal the car. Should you stay or swap? Note: this problem only considers the situation when Monty Hall (by chance) has opened a door that does not hide the car. The Truth In the Modified Monty Hall problem, the StayStrategy succeeds with 1/2 probability, and the SwapStrategy succeeds with 1/2 probability. Hand Wavy Anti-Justification Imagine that there are 100 doors. You choose one. Not knowing where the car is, Monty Hall opens 98 doors (other than your pick) that don't have the car. This situation is highly unlikely, but possible. Originally, you knew that your door had a 1/100 probability of hiding the car, so it must reasonably be true that all the other doors combined had a 99/100 probability of hiding the car, and since they have been narrowed to one selection, the swap door must have a higher probability of success. Proof For The Classical Problem The Classical Monty Hall problem is exactly equivalent to the following problem. You pick a door with a 1/3 probability of success. There is a 2/3 probability that one of the other two doors hides the car. You are given the option of opening and winning the contents of just your selected door (StayStrategy) or both the other doors (SwapStrategy). Because Monty Hall never reveals the car, the (SwapStrategy) is exactly equivalent to having both the other doors, and therefore 2/3 probability of success. It is important to remember that there are only two possible outcomes to this game: the stay door hid the car or the swap door hid the car. The Monty door never hides the car. This proof is particularly evident when examining the source code for a simulation of the game. In the simulation it is not necessary to determine which door Monty reveals. All you need to do is randomly generate a car on [1..3] and randomly generate a guess on [1..3]. If the car equals the guess then the (StayStrategy) wins, otherwise the (SwapStrategy) wins. Proof For The Modified Problem The Modified Monty Hall problem is actually three problems considered in isolation. It is true that your original pick has 1/3 probability of success. It is true that Monty Hall is not allowed to reveal your door. This does not change the fact that the door Monty Hall picks has a 1/3 probability of hiding the car. Since neither your nor Monty have knowledge of which door hides the car, regardless of the picks, each door has a 1/3 probability of hiding the car. Hence in the millions of times that the game is played, 1/3 of the games end with Monty Hall accidentally revealing the car. In the remaining 2/3 of the games played there is a 50% probability that the (StayStrategy) wins, and a 50% probability that the (SwapStrategy) wins. That is to say that 1/3 of all games end with Monty Hall accidentally revealing the car, 1/3 of all games end with your original pick hiding the car, and 1/3 of all games end with the remaining door (picked by neither you nor Monty) hiding the car. If You Don't Believe Me Java Simulation Source Code Java Simulation Executable Jar File Java Simulation Results
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