Quote:
Originally Posted by woj
1. you pick a door, 1/3rd of the time it's a winner... 2/3rds it's not...
2a. IF the chosen door is a winner (1/3rd chance) and you switch = you lose
2b. IF the chosen door is not a winner (2/3rd chance) and you switch = you win
so by using the "switching" strategy, you win 2/3rds of the time...
|
1. What is the probability of winning the car by always switching?
2. What is the probability of winning the car given the player has picked door 1 and the host has opened door 3?
The answer to the first question is 2/3, as is correctly shown by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1/2. This is because Monty's preference for rightmost doors means he opens door 3 if the car is behind door 1 (which it is originally with probability 1/3) or if the car is behind door 2 (also originally with probability 1/3). For this variation, the two questions yield different answers. However as long as the initial probability the car is behind each door is 1/3, it is never to the contestant's disadvantage to switch, as the conditional probability of winning by switching is always at least 1/2. (Morgan et al. 1991)
There is disagreement regarding whether vos Savant's formulation of the problem, as presented is asking the first or second question.
source: wikipedia
you are answering question number one so yes probability is 2/3 , but the way Frank worded it with door 1 already picked and door 3 opened it's question number two. 1/2