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inspired by this post [webmasterworld.com], here's a famous poser for you.

You're on "Let's Make a Deal".

Monty shows you three doors. He tells you that behind two of the doors you'll find a goat. Behind the other door, there's a car. Now you choose a door: 1, 2, or 3.

Monty then opens one of the doors you didn't pick, and shows you a goat. Now you know that the car is either behind your door, or the other door.

Monty gives you a choice: do you want to stay with the door you chose, or switch?

well?

Do you "stick" or "switch"?

There is a correct answer...

An effective demonstration of this problem [math.ucsd.edu]

To make things complicated:

The answer is that there is not enough information in the problem as stated to answer the question.

Here is why:

The answer is "switch" only if it is assumed that the host:

1. knows what is behind the other doors

2. always opens a door and shows the contestant a goat, regardless of whether the contestant has already picked the "correct" door or not.

If one of the conditions is not met the answer to the question changes.

Thats the same with those IQ tests.I always get mad when I see those tests because the questions always have some inherent assumptions, such as a dice for example has numbers ranging from 1-6. With most IQ test you really have to be very narrow minded to achieve a good result.

the questions always have some inherent assumptions, such as a dice for example has numbers ranging from 1-6

Since the definition of a die (dice is plural) is: "a small cube marked on each face with from one to six spots and used usually in pairs in various games and in gambling by being shaken and thrown to come to rest at random on a flat surface", that would be a great assumption.

There is not such a thing as a universal definition of a die (or any other thing).

For example in german style board games, strategy board games, role playing games the 1-6 dice is only one of dozens of possible variations of cubicle and non cubicle dice.

Like you can for example see here:

[trollandtoad.com...]

Even if you have 6 sided dice it is not necessary that the numbers from 1-6 are printed on it since there are d3 dice (1,2,3,1,2,3), doubling dice (2,4,8,16,32,64), average dice (2,3,3,4,4,5) and many other variants.

Thats one of the reasons why IQ test results (involving dice questions or not) are so dependend on individual and cultural background. (And the main reason why I never finished one until the end cause I always get upset by their narrow-mindedness)

[**edited by**: lawman at 12:05 pm (utc) on Jan. 28, 2006]

[edit reason] Delete Associate Id From Link [/edit]

Here is an example from the same page as above where you can see the difference between:

1. host knows

2. host doesn't know

[math.ucsd.edu...]

I think this is the reason why the debate sometimes can get a bit agitated when you do not choose your words precisly when posing the question.

@lawman

Thats one of the reasons I am self-employed since I left university. If I had to pass one of those intelligence or aptitude tests in a big company I would run out screaming. ;-)

The answer to the problem as stated above is SWITCH.

To jecasc.

First, the problem states "You're on "Let's Make a Deal".

This is a game, where the rules are well known. That's why variations of this problem have been called "The Monty Hall Problem".

If someone gives me a backgammon problem, I will make an assumption that they are following standard backgammon rules and that they haven't replaced one of the dice with a Dungeons and Dragons die.

Second, even if Monty Hall (born in 1921) is having "senior moments" and does not always recall the correct door, I would still switch.

First scenario: Host knows.

Monty's mind is clear and he follows the rules of the game.

Switching wins 66.7%

Not Switching wins 33.3%

Advantage Switching.

Second scenario: Host doesn't know.

Monty has no idea where the car is located or that his show is no longer on the air.

Switching wins 50%

Not switching wins 50%

No Advantage to either switching or not switching.

Third scenario: Host may or may not know.

Monty randomly displays signs of dementia.

Switching wins between 50% and 67%.

Not switching wins between 33.3% and 50%.

Advantage Switching.

The odds may change, but the correct answer SWITCH does not.

The host knows:

You will select the correct door 1/3 of the time and by switching you will lose.

The other 2/3 of the time you have originally selected the wrong door and the host by showing you the goat has told you where the car actually is located.

The host does not know:

You will select the correct door 1/3 of the time and by switching you will lose.

The other 2/3 of the time you have originally selected the wrong door and the host has a 50% chance of revealing the car, not meeting the criteria of the original problem which states that the host "shows you a goat". This changes the probability of success.

On a side note:

All of this is based on that you value the car more than the goat. ;)

Misguided Perfectionism at work:

Correction on my original post (unable to edit)

Third scenario:

Switching wins between 50% and 66.7%

I would switch and I don't care what the host knows.

My chance of picking a door with a goat were 2 in 3 when I started the game correct? The odds were against me. Now that there are only two doors to choose from my chances of winning a car go up if I switch doors.

The only information that is needed in order to make the decision to switch is that the odds were against you for your first choice. It was more likely that you would pick a goat.

A simple way to understand this problem is to imagine ten doors with goats behind nine of them.

You pick a door and then the host opens eight other doors and shows you eight goats. You are left with the door you originally picked and one other door.

Are you going to switch? Of course.

A simple way to understand this problem is to imagine ten doors with goats behind nine of them.

No, this doesn't change the final position one bit. It doesn't matter if you start with 3 doors, 10 doors, or 100 doors. At the last step (2 doors left unopened), you still have a 50% chance of being right.

At the last step (2 doors left unopened), you still have a 50% chance of being right.

At the very start you have a 1/3 chance of being correct.

You choose your door and there's a 1/3 chance of a car behind it. There's a 2/3 chance that the car is behind another door. The total amount of the other doors is 2. The host reduces the number of other doors to 1.

So there's a 1/3 chance of finding a car behind the door you've first chosen and 2/3 chance if you switch to the other door.

I think this is about as clearly as I can put it but if you're not satisfied try performing an experiment!

I would switch and I don't care what the host knows.

If the host doesn't know where the car is, there's a chance he might open the car.

If that means I win the car, I'm OK with that.

It doesn't matter if you start with 3 doors, 10 doors, or 100 doors

OK, what if Monty started with 10,000 doors, one car, and 9,999 goats?

you pick one, then he opens 9,998 goats. Now there's your door and one other door. The car is under one of them, of that you're sure.

That sounds like a 50% chance that you're right, but it's not.

The car does not have an equal probability of being under your door. It is more likely to be under the other door. I'll explain why.

Look at your choice: Your door was chosen randomly (barring any psychic gifts) from 1/10000. Thus, it is very unlikely that you picked the car. You probably have a goat. That makes it very likely that the host now has to reveal ALL the goats, and the one door left will have the car under it.

He has no choice. You chose a Goat, so his door must have a car behind it.

Monty has made it easy. The last two doors have a goat and a car. You are so likely to have the goat, you would be foolish not to switch to what is very probably the car.

Apply the same logic to 2/3 and it works the same.

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