Originally Posted by Toolish
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I see and understand the logic you are trying to use, but I think it is wrong.
The fact one door was preselected has NO bearing on which door has the car behind it.
If you walk into a room with 3 doors, 1 is open showing a goat, and the other 2 are closed, you then have to pick a door knowing a car is behind 1, what is your chance of picking the car?
I say 50%...
This is exactly the same scenario, the fact that there was a previous action and door reveal does not change the probabilites related the this event...like I said look up Mutually exclusive...
It is like a coin toss, if you toss a coin and get a head 5 times in a row there is still a 50% chance the next will be a head, the previous events don't change the probabilities involved in this event.
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This called the Monty Hall Problem or Dillema. It is an example what is logical is not always instinctive.
Think of it like a process tree
Goat 1 - Goat 2 is shown - switching gets car.
Goat 2 - Goat 1 is shown - switching gets car.
Car - Either goat is shown - switching loses car.
Therefore you get into a situation where you have a choice between the two doors and two times out of three it is the other door than the one you originally picked. Therefore you double your chances by switching.
If you would like to bet $4.50 (you give me when I win) to my $5.50 (I give you when I lose) then I would be willing to accept (repeated 100 times to counteract any variance). If your right and its a cointoss - you will be in positive expectation. If I'm right I will be...lol