Wouldn't you have a 50% chance of winning before the game show host picked a door since he was going to always choose a door with a goat from the beginning? I don't think that was communicated in the movie but from my research of this question one door was always going to be elimenated so basically you only had 2 choices to pick the car.
There are three doors. You have equal opportunity to choose door #1, as you have to choose door #2, or as door #3. So that's three (not two) equal choices. Hence odds of 33:33:33, and not 50:50. That's from your (= contestant's) viewpoint.
From the car's viewpoint: The car could "choose" to hide behind your first chosen door (say door #1), or it could hide behind the other remaining, non-chosen door (say door #3). That seems like two potential events, but those two events are not equally likely and hence don't have the same probability weight.
Event 1: The car chose to hide behind the non-chosen door (door #3); in that case the host had no choice but to reveal the goat behind door #2.
Event 2: The car chose to hide behind your first pick (door #1); in that case the host could have chosen to reveal the goat behind door #2, thereby resulting in the above scenario where you're left with the dilemma to stay with door #1 or switch to door #3; or the host could equally likely have revealed the goat behind door #3, thereby resulting in a scenario where you'd be choosing between staying with #1 or switching to #2. But it was given that the remaining non-chosen door was door #3, and not #2. So only
half of event 2 applies.
In other words: event 2 has the same probability as event 1; but event 2
where also the non-chosen door is door #3 (and not door #2) has only *half* of the probability of event 1. (And in event 1, the non-chosen door is *always* door #3.)
Therefore, given that your first pick is door #1 and your remaining, non-chosen door is door #3, and comparing the probabilities of where the car would end up between those two doors, the probability of ending up behind door #3 (the non-chosen door, event 1) is twice the probablity of ending up behind door #1 (the first pick door, event 2). This translates to odds of 66:33 (= 2:1).
I kinda get the what the the movie was saying about it to going 66.7% if he switched after researching it. What I don't get is why couldn't door #1 get the extra 33.3% as opposed to the other door he could have switched to.
Draw an event tree. From all possible (weighted) situations in which door #1 was first chosen and then a goat door was opened by the host, only 1/3 of them has the car behind door #1. And that's because the chance that you would pick the door with the car as your first choice, was 1 out of 3.
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Joe Satriani - "Always With Me, Always With You"
http://youtu.be/VI57QHL6ge0
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