Probablilty - Monty Hall problem

Question

A candidate chooses a door (which remains closed at first), so that he can win a car behind. Moderator opens n-2 other doors with goats. 2 doors remain closed.

We consider the goat problem for n=5 and assign the following numbers to doors: The candidate chooses a door, we call it door 1, the moderator opens 3 doors with goats behind it, which we call door 3,4,5:

G=Goat, C= Candidate, M = Moderator

     ? ? G G G
door 1 2 3 4 5
     C   M M M

A1 = "Car behind door 1" etc, M = "Moderator opens doors 3,4 and 5"

My questions are as follows.

1. Calculate P(A1|M)

   P(A1|M) = 1/5. How about if Moderator only opens doors 4 and 5, is the answer of P(A1|M) still 1/5?

2. Calculate P(A2|M)

   P(A2|M) = 4/5, may i know why?

   i thought that there are only 2 doors left after M and the answer was 1/2

3. Should the candidate stick with his original choice or change his mind for the other door?

   What should I calculate actually?

Thanks for the help!

Answer 1

Monty is the moderator.

1. Since you named the candidate's door 'Door 1', there's a flat $20 \%$ chance of the car being behind it regardless of what doors Monty opens.
2. There is a flat $20 \%$ chance of the car being behind Door $1$. In that case, the car is definitely not behind door $2$. If the car is not behind door $1$, as is $80 \%$ of times the case, it must be behind doors $2,3,4,5$ with equal probability. However, you renamed the doors Monty opens (which will be every door except the car door and Door 1) to $3,4,5$, so the car will always be behind door $2$. So in $20 \%$ of the cases, the chance is $0$, while in $80 \%$ of the cases, the chance is $1$. This yields a total probability of $\frac{4}{5}$.
3. Look at your calculated chances in $1$ and $2$. Does Door $1$ or Door $2$ offer the best chances of a car?

Answer 2

The difficulty (or rather what makes it so interesting) with the presentation is that while revealing the goats appears to provide new information, it actually doesn't.

Here's an alternative, logical but not so obviously mathematical, view of the problem.

Suppose, if instead of opening all but one of the other doors, Monty simply moved everything but your original choice to behind a single large door. You would not feel that he had provided any new information, nor would you feel that the situation had changed.

You now can choose your original small door (1 item), or you can choose to take anything you want from behind the large door (n-1 items).

Given that the car can be in any one of those n locations, what are the chances that it is behind the small door? 1/n.

And the chances that it is behind the large door? (n-1)/n.

Would the situation be any different if, after moving everything behind the large door, Monty had removed all the goats?

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A good demonstration of the same principle is to:

• Tell someone you want them to psychically find the Ace of Spades in a deck of cards.
• Tell them to choose a single card, the one they think is the Ace, from a deck and, without looking at it, place it face down on the table.
• Look through the remaining deck, move one card to the top, and place the deck face down beside the single card.
• State that the Ace of Spades is the top card of one of those two piles.
• Offer $1 for choosing the Ace.

How likely are people to stick with their original choice?

Would things be any different if you removed all but the top card from the large stack?

This is exactly the same situation as Monty's car and goats.