Probability Problem

Introduction

In the seventies, Monty Hall presented a television show Let's Make a Deal.
He offered, for instance, $100 for every paperclip you had with you or $50 for every hard-boiled egg you could hand him over.
One of the favourite games of the show was the three doors problem.

The participant is placed in front of three closed doors. Behind one of the doors is a significant prize (a car in the show), the two other doors offer nothing (a goat in the Monty Hall show). When the contestant points one of the doors, the game master does not open it yet. Instead, he opens another door behind which there is no prize*("let's first see what you did not win")*.
In the Let's Make a Deal show, participants were not allowed to change their choice.
It was only a means of building up excitement.
But what if they could have picked another choice?
What would be the best strategy: make a new choice or stick with the initial one?
Will the chances to win the prize become better, worse or remain the same if you changed your choice?

One of the earliest known appearances of this problem was in Joseph Bertrand's Calcul des probabilités (1889) where it was known as Bertrand's Box Paradox. It later reappeared in Martin Gardner's 1961 book, More Mathematical Puzzles and Diversions, as The Three Prisoner Problem.

In September 1990, Marilyn vos Savant puzzle columnist for the U.S. magazine Parade and listed in the Guinness Book of World Records as having the highest recorded score (228) in an IQ test, was sent this problem by a reader. The publication of her solution in her Ask Marilyn column has produced a huge debate amongst mathematicians and laymen alike ever since. She received an estimated 10,000 letters, many from learned academics, the vast majority of which told her she was wrong in no uncertain terms. The very serious New York Times opened its front-page for an article about the issue. And Byte's famous Chaos Manor columnist, Jerry Pournell also discussed the problem at length as an opponent to Marilyn's solution, to agree finally with her.

But, what do you think? Should you change your initial choice or not to get a better chance to pick the prize or wouldn't it change anything? After you have made up your mind, the simulation on the next page will show you whether you are right or wrong. And an explanation of the solution will also be given.

The participant is placed in front of three closed doors. Behind one of the doors is a significant prize (a car in the show), the two other doors offer nothing (a goat in the Monty Hall show). When the contestant points one of the doors, the game master does not open it yet. Instead, he opens another door behind which there is no prize

One of the earliest known appearances of this problem was in Joseph Bertrand's Calcul des probabilités (1889) where it was known as Bertrand's Box Paradox. It later reappeared in Martin Gardner's 1961 book, More Mathematical Puzzles and Diversions, as The Three Prisoner Problem.

In September 1990, Marilyn vos Savant puzzle columnist for the U.S. magazine Parade and listed in the Guinness Book of World Records as having the highest recorded score (228) in an IQ test, was sent this problem by a reader. The publication of her solution in her Ask Marilyn column has produced a huge debate amongst mathematicians and laymen alike ever since. She received an estimated 10,000 letters, many from learned academics, the vast majority of which told her she was wrong in no uncertain terms. The very serious New York Times opened its front-page for an article about the issue. And Byte's famous Chaos Manor columnist, Jerry Pournell also discussed the problem at length as an opponent to Marilyn's solution, to agree finally with her.

But, what do you think? Should you change your initial choice or not to get a better chance to pick the prize or wouldn't it change anything? After you have made up your mind, the simulation on the next page will show you whether you are right or wrong. And an explanation of the solution will also be given.

A Simulation

In this simulation, you don't have to make up your mind whether or not to change your choice.
You will be able to do both at the same time: stick with your initial choice and change it.
The simulation panel shows two rows of three doors:

At the top, nine lights (three for each column) indicate what happened:

For each column of doors, counters indicate the proportion of wins, allowing you to verify that each column has equal chances. The sum of these proportions being equal to 1, each door should approximate 0.333 wins.

Underneath each door, one light indicates what happened in the previous run:

You better start with some single steps in order to understand the mechanism before going for a run (preferably at least a hundred tries) to see which strategy works the best.

- for the top row of doors, your initial choice will always be maintained,
- on the bottom row, you will systematically change your initial choice.

At the top, nine lights (three for each column) indicate what happened:

- becomes when that door was the first choice
- becomes when that door was opened by the game master
- becomes when the prize was behind that door

For each column of doors, counters indicate the proportion of wins, allowing you to verify that each column has equal chances. The sum of these proportions being equal to 1, each door should approximate 0.333 wins.

Underneath each door, one light indicates what happened in the previous run:

- becomes when that door was opened by the player
- the combination indicates that the player got a win

You better start with some single steps in order to understand the mechanism before going for a run (preferably at least a hundred tries) to see which strategy works the best.

Door A

Door B

Door C

Tries

0

Maintain first choice

proportion of wins

proportion of wins

0.000

Switch first choice

proportion of wins

proportion of wins

0.000

0.000

0.000

0.000

Zero

Run

Stop

Step

An Explanation

The 3 doors problem is a very swell example of the non-intuitive nature of probabilities.
Most people believe there is no advantage in changing their initial choice.
Their reasoning is that after the opening of a door with no prize, there is a 50 % chance that the prize will be behind one of the two other doors and that switching their choice will change nothing.
In fact, this reasoning is false and the fault lies in the fact that the game host knows more than you do.

When you make your initial choice, you have 1 chance out of 3 to pick the prize. When the game master opens a door, you are down to one win and one loose. Not changing at this point does not change anything; you still have 1 chance out of 3 to pick the prize. But when you switch at this point, your initial odds reverse completely. If your initial choice was a loose, changing it will now get you a win. And if your initial choice was a win, changing will get you a loose. You should look at it as a second chance of 1/3 to pick the prize. Since your initial chance to get a win was only 1 out of 3, changing will give you then 2 chances out of 3 to get a win.

Or you could look at it in another way:

(*) For the second step of the first row of the scheme, the game master has two choices: he can open door B or door C without revealing the prize. But this doesn't change anything: for the third step, the player still has only one possibility to switch to. So that won't alter any chances.

If you want to, you can make the scheme for doors B and C being the winning doors and you will see that the results are the same:

When you make your initial choice, you have 1 chance out of 3 to pick the prize. When the game master opens a door, you are down to one win and one loose. Not changing at this point does not change anything; you still have 1 chance out of 3 to pick the prize. But when you switch at this point, your initial odds reverse completely. If your initial choice was a loose, changing it will now get you a win. And if your initial choice was a win, changing will get you a loose. You should look at it as a second chance of 1/3 to pick the prize. Since your initial chance to get a win was only 1 out of 3, changing will give you then 2 chances out of 3 to get a win.

Or you could look at it in another way:

- the probability that the player initially chose the correct door is 1 out of 3, since there are three doors each having equal chances to conceal the prize.
- the probability that the door the game master chooses conceals the prize is 0, since he never chooses the door that contains the prize.
- since the sum of the three probabilities is 1, the probability that the prize is behind the other door is 1 - (1/3 + 0), which equals 2/3.

First step |
||||||||||

You choose door A | You choose door B | You choose door C | ||||||||

Second step |
||||||||||

The game master opens door C (*) |
The game master opens door C |
The game master opens door B |
||||||||

Third step : no change |
||||||||||

You stay with door A | You stay with door B | You stay with door C | ||||||||

and you win |
and you loose |
and you loose |
||||||||

Third step : switch doors |
||||||||||

As A was your first choice and C was opened, you switch to B |
As B was your first choice and C was opened, you switch to A |
As C was your first choice and B was opened, you switch to A |
||||||||

and you loose |
and you win |
and you win |

(*) For the second step of the first row of the scheme, the game master has two choices: he can open door B or door C without revealing the prize. But this doesn't change anything: for the third step, the player still has only one possibility to switch to. So that won't alter any chances.

If you want to, you can make the scheme for doors B and C being the winning doors and you will see that the results are the same:

- without changing, you only get 1 win out of 3,
- while with the change strategy, you get 2 wins out of 3.