Monty Hall question

[daigo]

There are three doors and one of the doors with a goat are removed. Which of the two remaining doors should you choose to have a higher chance of getting the car?

There are ten doors and eight of the doors with goats are removed. Which of the two remaining doors should you choose to have a higher chance of getting the car?

There are two doors. Which of the two doors should you choose to have a higher chance of getting the car?

Are all of these the exact same version of the Monty Hall problem?
If so, I don't understand why it's not 50/50 in all cases.
If not, I don't understand how these are any different from the Monty Hall problem.

 

[CaptainBlack]

Are all of these the exact same version of the Monty Hall problem?
If so, I don't understand why it's not 50/50 in all cases.
If not, I don't understand how these are any different from the Monty Hall problem.

Please post the entire question

CB

 

[daigo]

Which entire question? I am just wanting to compare the aforementioned 3 scenarios to the Monty Hall/3 Prisoners/Bertrand's Box problem

 

[mfb]

If these quotes contain the whole problem, they are different from the Monty Hall problem because:

1. you do not begin by choosing a door (which cannot be reveailed in the next step)
2. the algorithm of the door-opening is not given (could it have relealed a car?) <- not relevant here due to 1, but important in Monty Hall.

 

[daigo]

Why does it make a difference if you begin by choosing a door? In the end, don't you only have two choices either way in both the quoted problems and the Monty Hall? (Door #1 or Door #2)

Also I'm not sure what you mean by your second point.

 

[Jameson]

It makes a huge difference if you choose before and are asked to switch, which is the set up of the Monty Hall problem. The best way to understand the Monty Hall problem is there are two ways to switch:

1) Correct door to incorrect door
2) Incorrect door to correct door

You cannot go incorrect to incorrect or correct to correct. So the question is how often do you choose a correct door? Well that is just 1/3 of the time. How often do you choose an incorrect door? 2/3 of the time. So if you always switch, 2/3 of the time you will switch to the correct door.

If you are just given two doors and the prize is behind one of them, then your guess is 50/50. The Monty Hall situation is different than that.

 

[mfb]

Why does it make a difference if you begin by choosing a door?

It does influence Mr. Monty Hall, who has to open an empty door, but may not use your door.

 

[zzephod]

XKCD on the Monty Hall Problem

​

 

[Jameson]

I love this problem for yet another reason now.

Earlier this year in one of my classes on probability, this problem was introduced and it was fun as always to watch people get stuck and even disagree with the solution very stubbornly. So I suppose this thread is a good place to write down some more comments as it pertains to the OP still.

If I were teaching this concept, I would do two things to help convince the students:

1) Run a test program with a large number of iterations to show that switching converges to a 66.7% success rate.

2) Reword the problem such that instead of going from 3 doors to 2 doors, go from 1,000,000 doors to 2 doors. Now it should become very obvious to most everyone that the door they original chose is almost never the correct door.

 

[MarkFL]

...
2) Reword the problem such that instead of going from 3 doors to 2 doors, go from 1,000,000 doors to 2 doors. Now it should become very obvious to most everyone that the door they original chose is almost never the correct door.

I have used this tactic with success in converting "nonbelievers."

 

[Jameson]

It's a very nice modification to the problem to account for our very poor statistical intuition. Even so, I think the answer can be so puzzling and counter-intuitive to some that this example still won't convince them. That's why I suggested the computer simulation, which confirms that the theory at work isn't some hypothetical mind trick.