Quote:
Originally Posted by CyberSEO
Two doors can not have 66% chance because... because... 66 + 66 > 100. Go to the school already.
The things are easy:
3 closed doors: 33.3(3)% chance for each
2 closed doors: 50% chance for each
1 closed door: 100% chance
You don't have to be Einstein to understand the basics of probability theory. In Soviet Union we learned it in the school at 13-14 year old (7th grade).
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um, but did you go past 7th grade? lol this is a college level class.
https://en.wikipedia.org/wiki/Monty_Hall_problem
Quote:
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Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence does not hold. Before the host opens a door there is a 1/3 probability the car is behind each door. If the car is behind door 1 the host can open either door 2 or door 3, so the probability the car is behind door 1 AND the host opens door 3 is 1/3 * 1/2 = 1/6. If the car is behind door 2 (and the player has picked door 1) the host must open door 3, so the probability the car is behind door 2 AND the host opens door 3 is 1/3 * 1 = 1/3. These are the only cases where the host opens door 3, so if the player has picked door 1 and the host opens door 3 the car is twice as likely to be behind door 2. The key is that if the car is behind door 2 the host must open door 3, but if the car is behind door 1 the host can open either door.
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