3 Door Question

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.

______
Joe Satriani - "Always With Me, Always With You"
http://youtu.be/VI57QHL6ge0

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"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.

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."

The brain-busting part of this problem is precisely that notion: that you have a two choices: switch or don't switch.

You actually have SIX choices: pick one of three doors, THEN choose to switch from that first choice.

If you don't switch, you have a 1/3 chance of winning. That means if you DO switch, you MUST have a 2/3 chance of winning, because 1/3 + 2/3 = 1

There have been many many excellent explanations on this board as well as the rest of the Internet, but to me, it's easiest once you realize how many possible outcomes there actually are. Not two, but six.

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But always remember: the movie writers themselves didn't understand the problem, and screwed it up by having Kevin Spacey throw out the notion that the host might be "messing with you."

Sturgess says it doesn't matter. But it absolutely matters.

The Monty Hall Dilemma only works if the host ALWAYS opens a goat door after your initial choice. It's an unspoken assumption of the dilemma, like the assumption that the host knows where the car and goats are.

If there's ANY possibility that he WON'T open a goat door (because, just for example, he's "messing with you"), the analysis fails.

Once the host's free will enters the equation, switching no longer automatically results in 2/3 chance of winning the car.

Suppose, for example, that the host REALLY doesn't want you to win the car. Such a host might ONLY open the goat door & offer a switch when you HAVE chosen the car.

If that's the case, you should NEVER switch. By offering you the choice, the host is effectively telling you that you DID choose the car.

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Start with 100 doors. 99 have a goat. 1 has a car.

Player A gets to pick 1 door.

Player B gets the other 99 doors.

Q.1 Who has the better odds of winning the car at this point? (B 99%; A 1%)

Now, open 98 of B's 99 doors showing the goats.

Q.2 Who still has the better odds of winning the car? (B 99%; A 1%)

Q.3 Finally, if A is given the choice to switch with B, should A do it?
(Yes, based upon the above statistics.)

It doesn't matter if the game starts with 3 doors or 3 million doors. The answer to Q.3 should be yes if A wants to change the odds to his favour.

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