#### satch

##### Footballguy

Monty Hall Problem

Here's a quick summary:

There are 3 doors. Behind one of the doors is a car, behind the other 2 doors are goats. The contestant must try to pick the door that has the car behind it. Let's say the contestant chooses door #1. Before the host reveals what's behind door #1, he opens door #3 revealing a goat. This leaves doors #1 and #2 still unopened, one of them must have a car behind it. The host then gives the contestant the opportunity to change his pick from door #1 to door #2. What should the contestant do? Stay with his original choice of door #1, or switch to door #2? Does it matter?

Supposedly, the correct answer is: The contestant should choose door #2, because door #2 has a higher probability of having the car behind it than door #1. This has been "proven" by computer simulations.

There are many web pages, articles written about this. I've read a few of them and I just can't understand how this could be true. The contestant has no idea which door the car is behind. When given 3 choices, each door has a 33% chance of being right. When one of the three options is eliminated, each the 2 remaining doors now have a 50% chance of being right. It should make absolutely no difference which door the contestant picked originally. They're just doors. His picking one doesn't have any effect on what's actually behind it, and should have no effect on probability.

Can anyone explain this in a way that makes sense? Specifically, explain how after door #3 has been eliminated, the 2 remaining doors

Here's a quick summary:

There are 3 doors. Behind one of the doors is a car, behind the other 2 doors are goats. The contestant must try to pick the door that has the car behind it. Let's say the contestant chooses door #1. Before the host reveals what's behind door #1, he opens door #3 revealing a goat. This leaves doors #1 and #2 still unopened, one of them must have a car behind it. The host then gives the contestant the opportunity to change his pick from door #1 to door #2. What should the contestant do? Stay with his original choice of door #1, or switch to door #2? Does it matter?

Supposedly, the correct answer is: The contestant should choose door #2, because door #2 has a higher probability of having the car behind it than door #1. This has been "proven" by computer simulations.

There are many web pages, articles written about this. I've read a few of them and I just can't understand how this could be true. The contestant has no idea which door the car is behind. When given 3 choices, each door has a 33% chance of being right. When one of the three options is eliminated, each the 2 remaining doors now have a 50% chance of being right. It should make absolutely no difference which door the contestant picked originally. They're just doors. His picking one doesn't have any effect on what's actually behind it, and should have no effect on probability.

Can anyone explain this in a way that makes sense? Specifically, explain how after door #3 has been eliminated, the 2 remaining doors

*do not*have a 50/50 chance of being right. 2 doors, one car, one choice.
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