Does anyone know how to solve the Monty Hall Paradox?

Question

I would really appreciate an expanation. Thanks!

Answer 1

i don't know if there is a "solution" to the paradox. seems liek there is only and "understanding" of the paradox.

hope this helps in your understanding:

http://en.wikipedia.org/wiki/Monty_Hall_problem

Answer 2

When you play the game with three doors, it seems like a paradox. It seems like you should have a 50% shot at winning rather than a 67% chance.

But it is easier to understand if the number of doors is bigger. Suppose that there are 1000 doors. 999 have nothing, and one door has a million dollars. I know which door has the million. Now -- you choose a number between 1 and 1000. For the rest of this discussion, I'm going to assume that you chose something other than 673. Now, suppose that I open 998 different doors showing you that there is nothing behind any of those doors. The two doors left are the one you chose and door 673. I offer to let you switch. Do you really think that there is a 50% chance that you were right?

You had a 1/1000 chance of being right when you picked it. You already knew that at least 998 of the remaining doors were empty. I showed them to you -- but I didn't really give you any more information. So -- your chances never changed.

If you had randomly picked the doors to open, then it would increase your odds of being right because you are getting new information (like on DEAL OR NO DEAL). But when I am the one picking them -- it gives you no new information.

Answer 3

Here's a good way to look at it: You don't pick one door among three. You are picking among two sets of doors:

- Is the prize behind door #1?
- Or is the prize behind doors #2 and 3? Which one doesn't matter because you can get Monty to point you to the right one.

The second choice has a 2/3 probability, so that's what you should pick. How to pick it? Choose door #1 and get Monty to show you which of #2 or #3 you should pick. That's the legal way to pick both doors #2 and #3.

Answer 4

yes, the answer has to do with the fact that the host KNOWS what door the prize is behind.

so 3 doors and i pick one... my chance of the prize is 1/3.

the host opens one of the other two doors and he knows it doesn't have the prize. Now, out of the two doors that he had a choice there are two scenerios. One is that i had the prize and he could pick either door. The other being that i didn't have the prize and the host was forced to pick the door the one remaining door without the prize.

because of this fact when he ask me if i want to switch my probability of picking the door that he did not choose and the prize being behind it is 1/2, which is better odds than my first choice of 1/3.

Answer 5

If you have 3 doors, the probability for choosing the right door is 1/3, and of cause the probability for choosing one of the wrong doors is 2/3
After one of the doors is open, the probability for, that you have chosen the right door is still 1/3, and the probability for you have chosen the wrong door is still 2/3.
Because you only have to doors to choose between, you will interchange the probability if you change your choice.

Answer 6

Hi,

There is no solution, it's already solved. The easiest way I understood it was using Bayes theorem to calculate the probabilities that each outcome will occur.

To tell you the truth, this is one that I still don't believe even though the math tells me that it's true. I can't shake the feeling that I'm being tricked. That's probably what you're feeling too. The reference that cirric gave was a good one, read that page and you'll feel better.

Hope that helps,
Matt

Answer 7

Hi. I've read about it (changing your mind after a door fails to win results in better odds) but I believe it's just an imaginary effect. Read about it here : http://www.answers.com/main/ntquery?s=Monty%20Hall%20Paradox&gwp=16