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July 27, 2008 at 9:04 PM #248229July 28, 2008 at 10:50 PM #248475
recordsclerk
ParticipantYou have three doors
One of the doors is a winner (car)
Each original door has a 33% or 1/3 chance of being the winning door.
So if you choose two doors then your chances of winning are 66% or 2/3.
If one of your doors are revealed and it’s not the winning door, in theory your odds are still the same. You still have a 66% chance of winning the car from the original circumstances. So in reverse if you choose one door you have a 33% or 1/3 chance of winning and the other two doors have a 66% or 2/3 chance of winning. By revealing one of the other doors, you still have a 66% chance of winning with the remaining door, so changing to that door gives you a probability of 66% or 2/3 chance of winning.
BTW if you have two doors, your chances are 50/50July 28, 2008 at 10:50 PM #248633recordsclerk
ParticipantYou have three doors
One of the doors is a winner (car)
Each original door has a 33% or 1/3 chance of being the winning door.
So if you choose two doors then your chances of winning are 66% or 2/3.
If one of your doors are revealed and it’s not the winning door, in theory your odds are still the same. You still have a 66% chance of winning the car from the original circumstances. So in reverse if you choose one door you have a 33% or 1/3 chance of winning and the other two doors have a 66% or 2/3 chance of winning. By revealing one of the other doors, you still have a 66% chance of winning with the remaining door, so changing to that door gives you a probability of 66% or 2/3 chance of winning.
BTW if you have two doors, your chances are 50/50July 28, 2008 at 10:50 PM #248638recordsclerk
ParticipantYou have three doors
One of the doors is a winner (car)
Each original door has a 33% or 1/3 chance of being the winning door.
So if you choose two doors then your chances of winning are 66% or 2/3.
If one of your doors are revealed and it’s not the winning door, in theory your odds are still the same. You still have a 66% chance of winning the car from the original circumstances. So in reverse if you choose one door you have a 33% or 1/3 chance of winning and the other two doors have a 66% or 2/3 chance of winning. By revealing one of the other doors, you still have a 66% chance of winning with the remaining door, so changing to that door gives you a probability of 66% or 2/3 chance of winning.
BTW if you have two doors, your chances are 50/50July 28, 2008 at 10:50 PM #248696recordsclerk
ParticipantYou have three doors
One of the doors is a winner (car)
Each original door has a 33% or 1/3 chance of being the winning door.
So if you choose two doors then your chances of winning are 66% or 2/3.
If one of your doors are revealed and it’s not the winning door, in theory your odds are still the same. You still have a 66% chance of winning the car from the original circumstances. So in reverse if you choose one door you have a 33% or 1/3 chance of winning and the other two doors have a 66% or 2/3 chance of winning. By revealing one of the other doors, you still have a 66% chance of winning with the remaining door, so changing to that door gives you a probability of 66% or 2/3 chance of winning.
BTW if you have two doors, your chances are 50/50July 28, 2008 at 10:50 PM #248705recordsclerk
ParticipantYou have three doors
One of the doors is a winner (car)
Each original door has a 33% or 1/3 chance of being the winning door.
So if you choose two doors then your chances of winning are 66% or 2/3.
If one of your doors are revealed and it’s not the winning door, in theory your odds are still the same. You still have a 66% chance of winning the car from the original circumstances. So in reverse if you choose one door you have a 33% or 1/3 chance of winning and the other two doors have a 66% or 2/3 chance of winning. By revealing one of the other doors, you still have a 66% chance of winning with the remaining door, so changing to that door gives you a probability of 66% or 2/3 chance of winning.
BTW if you have two doors, your chances are 50/50July 29, 2008 at 12:52 AM #248206pepsi
ParticipantI wrote my own program and, to my surprise, 66% is the right answer. ( I thought it was 50% chance)
Then I reviewed my own program and I finanlly understand why 66% is the right answer.
The reason is at the door the host opened.
In 2/3 of chance, the door he opened is not a random pick.
In your first guess, you have 2/3 chance to guess wrong, so the door number the host must opened is limited to a certain door. (He can not opened the other door).By doing the switch, you are saying: “the first door I chose is wrong”, which is 2/3 of chance.
And that is why you must switch !
July 29, 2008 at 12:52 AM #248362pepsi
ParticipantI wrote my own program and, to my surprise, 66% is the right answer. ( I thought it was 50% chance)
Then I reviewed my own program and I finanlly understand why 66% is the right answer.
The reason is at the door the host opened.
In 2/3 of chance, the door he opened is not a random pick.
In your first guess, you have 2/3 chance to guess wrong, so the door number the host must opened is limited to a certain door. (He can not opened the other door).By doing the switch, you are saying: “the first door I chose is wrong”, which is 2/3 of chance.
And that is why you must switch !
July 29, 2008 at 12:52 AM #248365pepsi
ParticipantI wrote my own program and, to my surprise, 66% is the right answer. ( I thought it was 50% chance)
Then I reviewed my own program and I finanlly understand why 66% is the right answer.
The reason is at the door the host opened.
In 2/3 of chance, the door he opened is not a random pick.
In your first guess, you have 2/3 chance to guess wrong, so the door number the host must opened is limited to a certain door. (He can not opened the other door).By doing the switch, you are saying: “the first door I chose is wrong”, which is 2/3 of chance.
And that is why you must switch !
July 29, 2008 at 12:52 AM #248427pepsi
ParticipantI wrote my own program and, to my surprise, 66% is the right answer. ( I thought it was 50% chance)
Then I reviewed my own program and I finanlly understand why 66% is the right answer.
The reason is at the door the host opened.
In 2/3 of chance, the door he opened is not a random pick.
In your first guess, you have 2/3 chance to guess wrong, so the door number the host must opened is limited to a certain door. (He can not opened the other door).By doing the switch, you are saying: “the first door I chose is wrong”, which is 2/3 of chance.
And that is why you must switch !
July 29, 2008 at 12:52 AM #248434pepsi
ParticipantI wrote my own program and, to my surprise, 66% is the right answer. ( I thought it was 50% chance)
Then I reviewed my own program and I finanlly understand why 66% is the right answer.
The reason is at the door the host opened.
In 2/3 of chance, the door he opened is not a random pick.
In your first guess, you have 2/3 chance to guess wrong, so the door number the host must opened is limited to a certain door. (He can not opened the other door).By doing the switch, you are saying: “the first door I chose is wrong”, which is 2/3 of chance.
And that is why you must switch !
July 29, 2008 at 4:52 PM #248789drunkle
Participantit’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.
July 29, 2008 at 4:52 PM #248947drunkle
Participantit’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.
July 29, 2008 at 4:52 PM #248955drunkle
Participantit’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.
July 29, 2008 at 4:52 PM #249016drunkle
Participantit’s much easier to think about with higher numbers. ie., 1000 doors; your odds of picking the winner is 1:1000. the odds of the car being behind any one of the remaining doors is 999:1000. eliminating all of the other incorrect doors means that the chance of the car being in the door you didn’t pick is 999:1000.
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