
Title:
3801 Door_Chance

Description:

Let's have some fun.

This gentleman you probably don't know. His name is Monty Hall.

Starting in 1963, for many years Monty Hall ran a game show called Let's Make a Deal.

In the center of the game show was a puzzle

that puzzles statisticians to the present day.

In this unit, which is optional in the statistics class,

I'll let you solve the Monty Hall problem, and you'll get a chance to program it

to verify that the assertion, which is not obvious, is actually correct.

Here we go.

The essence of the game are three doors.

Behind one of those is a beautiful car, but Monty won't tell you where the car is.

Instead, all the doors are covered by curtains.

The subject of the game is as follows.

You get to place a bet. Say you pick door 2.

If behind the bet you place you'll win the car. Otherwise, you'll win nothing.

So far, so good.

Here's the caveat.

Obviously, between the two remaining doors there'd be one that doesn't contain the car.

Monty is going to be mean. He's going to show you.

He's going to lift the curtain from one of the other doors, and there is clearly no car.

After now closing the curtain again, he'll ask you do you want to switch.

That is, do you want to change your vote to the remaining curtain

in the hopes of having an increased chance of winning the car.

The reason why this problem is interesting is because when Monty Hall lifted the curtain

he really didn't tell you anything.

You knew in advance that one of the two curtains wouldn't contain the car.

The mere fact that he lifted the curtain should give you zero information

relative to where between the remaining doors the car is.

All you know now is that door 3 doesn't contain the car,

but why should door 1 now be more likely than door 2.

Here's the question.

What are the probabilities of the car being behind door 1, door 2, and door 3,

provided that your initial pick was door 2 and Monty then lifted door 3 to show you there's no car?