# Monty Hall Problem

• Mar 4th 2010, 11:09 AM
helpwithassgn
Monty Hall Problem
The original problem is that there are three doors (2 with goats and 1 with a car behind it). A person chooses one of the three doors then the host picks another with a goat behind it. The person can then choose to pick the other unopened door or stick with his initial choice.

I am given an extension of this problem and I need help with it.
So the question is there are N doors, C cars, D doors that are opened, and G goats where G = N-C

Now I need to find

the probability that the person wins the car given he switches and
the probability that he wins the car given he does not switch

I am not quite sure how to do this. I think Bayes theorem is needed, but I don't know what to do.
• Mar 5th 2010, 03:50 AM
CaptainBlack
Quote:

Originally Posted by helpwithassgn
The original problem is that there are three doors (2 with goats and 1 with a car behind it). A person chooses one of the three doors then the host picks another with a goat behind it. The person can then choose to pick the other unopened door or stick with his initial choice.

I am given an extension of this problem and I need help with it.
So the question is there are N doors, C cars, D doors that are opened, and G goats where G = N-C

Now I need to find

the probability that the person wins the car given he switches and
the probability that he wins the car given he does not switch

I am not quite sure how to do this. I think Bayes theorem is needed, but I don't know what to do.

The second part is trivial, it is the same as the probability of winning if no doors were opened.

The first part I would draw a contigency tree for with two main branches, the player switches when there is a car behind their door, and the player switches when there is no car behind their door. We already know the probabilities for these two branches and the winning probability is now the chance that a door is picked with a car from the remaining doors and cars.

CB