Monty Hall Problem

Hey guys,

Here's there link to what I was talking about.

http://mathforum.org/dr/math/faq/faq.monty.hall.html

Scroll down to this,

"If you're still not convinced that 2/3 is the correct probability, here are two more ways to think about the problem.

1. It seems to make sense that you have a 1/3 chance of picking the correct door. This means, however, that since the probabilities must add up to one - and the car has to be somewhere - you also have a 2/3 chance of not picking the correct door. In other words, you are more likely not to win the car than to win it. Imagine that Monty opens a door and shows that there's only a goat behind it. Consider that the car is more likely to be behind a door other than the one you choose. Monty has just shown that one of those two doors - which together have the greater probability of concealing the car - actually conceals a goat. This means that you should definitely switch doors, because the remaining door now has a 2/3 chance of concealing the car. Why? Well, your first choice still has a 1/3 probability of being the correct door, so the additional 2/3 probability must be somewhere else. Since you know that one of the two doors that previously shared the 2/3 probability does not hide the car, you should switch to the other door, which still has a 2/3 chance of concealing the car.
   
2. What if there were 1,000 doors? You would have a 1/1,000 chance of picking the correct door. If Monty opens 998 doors, all of them with goats behind them, the door that you chose first will still have a 1/1,000 chance of being the one that conceals the car, but the other remaining door will have a 999/1,000 probability of being the door that is concealing the car. Here switching sounds like a pretty good idea."

Also, check out the next post for visual aids.

Merianna