Goat or Car, Rediculous or don't I get it?

[meteomonk]

This simply does NOT make sense to me!

Would the odds not go from 33.3% to 50% after being revealed what was behind door 3? why on earth would changing his answer give him 66% odds?
there is no reason that after being shown there was a goat behind the door he DIDNT choose that his first answer would be right only 33% of the time!!!!

Is this total sillyness or am I just not an MIT genious?

[callipygian]

Originally Posted by meteomonk

Would the odds not go from 33.3% to 50% after being revealed what was behind door 3? why on earth would changing his answer give him 66% odds?

The key that you're missing here is that this problem is not a strict probability problem. It's a conditional probability problem, because the contestant has more information available to him than someone who walked in off the street.

As a matter of fact, this is a good place to make this point: if a stranger walked in off the street after door #3 had been opened (and had no idea which door had been originally chosen), that stranger would have a 50% chance of guessing correctly.

The contestant can do better than 50% because he has information that the stranger didn't: the contestant selected door #1, and whether or not the car was behind door #1, door #1 could not have been opened.

The question now is not "you have two random doors, which one has the car" - it is "given that you chose door #1 and Monty opened door #3, which one has the car".

Play out each scenario.

(1) Car is behind door #1.
=> Contestant chooses door #1, Monty opens either door #2 or door #3.
=> Contestant wins by staying, loses by switching.

(2) Car is behind door #2.
=> Contestant chooses door #1, Monty is forced to open door #3.
=> Contestant loses by staying, wins by switching.

(3) Car is behind door #3.
=> Contestant chooses door #1, Monty is forced to open door #2.
=> Contestant loses by staying, wins by switching.

The contestant has a better chance than blindly guessing because he forced Monty's actions in two of the three scenarios.

[KenSmith]

There's a really long thread about this just below:
http://www.blackjackinfo.com/bb/showthread.php?t=9691

[standard toaster]

who cares you could still win a goat! Long term it might be worth more! Lets take that goat and buy another. We can start a farm. Then we can eat some of them, milk some of them and sell some of them. Lets see a car do that

[nottooshabby]

I'm willing to bet U.S. goat dealers have been making more than U.S. auto makers lately . . .

[jimbiggs]

Originally Posted by callipygian

The key that you're missing here is that this problem is not a strict probability problem. It's a conditional probability problem, because the contestant has more information available to him than someone who walked in off the street.

As a matter of fact, this is a good place to make this point: if a stranger walked in off the street after door #3 had been opened (and had no idea which door had been originally chosen), that stranger would have a 50% chance of guessing correctly.

The contestant can do better than 50% because he has information that the stranger didn't: the contestant selected door #1, and whether or not the car was behind door #1, door #1 could not have been opened.

The question now is not "you have two random doors, which one has the car" - it is "given that you chose door #1 and Monty opened door #3, which one has the car".

Play out each scenario.

(1) Car is behind door #1.
=> Contestant chooses door #1, Monty opens either door #2 or door #3.
=> Contestant wins by staying, loses by switching.

(2) Car is behind door #2.
=> Contestant chooses door #1, Monty is forced to open door #3.
=> Contestant loses by staying, wins by switching.

(3) Car is behind door #3.
=> Contestant chooses door #1, Monty is forced to open door #2.
=> Contestant loses by staying, wins by switching.

The contestant has a better chance than blindly guessing because he forced Monty's actions in two of the three scenarios.

This is the first explanation that I've read that has made any sense to me. Thank you for breaking it down for us dummies.

[Cherry7Up]

Originally Posted by callipygian

The question now is not "you have two random doors, which one has the car" - it is "given that you chose door #1 and Monty opened door #3, which one has the car".

Play out each scenario.

...

(3) Car is behind door #3.
=> Contestant chooses door #1, Monty is forced to open door #2.
=> Contestant loses by staying, wins by switching.

Isn't this third scenario impossible given the conditional question you posed--Monty opened door #3, so no scenario in which Monty is forced to open door #2 can occur. If scenario 3 will not occur, that just leaves an equal probability of scenarios 1 and 2 meaning that there is no reason to switch doors (but also no harm in switching--the odds of winning remain the same).

Maybe I am still missing something, but I really do think this questioning in the movie is flawed.

[callipygian]

Originally Posted by Cherry7Up

Isn't this third scenario impossible given the conditional question you posed--Monty opened door #3, so no scenario in which Monty is forced to open door #2 can occur. If scenario 3 will not occur, that just leaves an equal probability of scenarios 1 and 2 meaning that there is no reason to switch doors (but also no harm in switching--the odds of winning remain the same).

No, because Scenario #1 contains a probability that Monty will open door #2 as well.

(1a) Car is behind door #1.
=> Contestant chooses door #1, Monty opens door #2.
=> Contestant wins by staying, loses by switching.

(1b) Car is behind door #1.
=> Contestant chooses door #1, Monty opens door #3.
=> Contestant wins by staying, loses by switching.

(2) Car is behind door #2.
=> Contestant chooses door #1, Monty is forced to open door #3.
=> Contestant loses by staying, wins by switching.

(3) Car is behind door #3.
=> Contestant chooses door #1, Monty is forced to open door #2.
=> Contestant loses by staying, wins by switching.

Given that the contestant chose door #1 and Monty opened door #3, the choice is between (1b) and (2), and (1b) is half as likely as (2).

[hawkeye]

It's somewhat strange that people still ask about this. I realize it's a somewhat confusing problem, but a pencil and paper will clear it up for anyone in about 3 minutes.

[Cherry7Up]

Originally Posted by callipygian

Given that the contestant chose door #1 and Monty opened door #3, the choice is between (1b) and (2), and (1b) is half as likely as (2).

You are right that (1b) and (2) are the only possible outcomes once Monty opens door #3, but I think you are wrong that (1b) is half as likely as (2) given that door #3 was opened.

I agree that (1b) is half as likely as (2) given no additional information (i.e. before Monty opens the door), but once we know door #3 is open, then (1b) and (2) are the only possible outcomes and I see no reason why they should not have equal probability at that point (and thus there is no incentive to switch doors).