The Monty Hall problem: You are playing a game show, in which there are prizes b

Question

The Monty Hall problem: You are playing a game show, in which there are prizes behind three doors. One prize is good (eg a car), and the others aren't (eg. a goat and a rabbit). Once you pick a door, the host (named Monty Hall) will open one of the other doors, showing you a bad prize. And he will ask you if you want to switch your choice. Should you?

The argument against is that since he always opens a door, he hasn't really given you any information that would favor one door over the other. So both remaining doors are equally likely to be correct (conditional prob 1/2 each), and there is no point in switching. The argument for switching is that you have no new information about your original choice. So the probability you picked correctly in the first place is unchanged at 1/3, and the (conditional) probability the other door is correct is now 2/3. Which argument is right

PLease say which argument is right.. and why?

Explanation / Answer

The argument for switching is that you have no new information about your original choice. So the probability you picked correctly in the first place is unchanged at 1/3, and the (conditional) probability the other door is correct is now 2/3. This argument is right. P(a)=1/3=P(b)=P(c) You chose one of a,b, or c. Say you picked a. P(a)=1/3. You are then told that it is not c. P(not a)=2/3. Not c. there not a is b. So P(not a)=p(b)=2/3

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