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

#1
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

Well-Known Member
#2
meteomonk said:
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.
 
#4
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
 

jimbiggs

Well-Known Member
#6
callipygian said:
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

Well-Known Member
#7
callipygian said:
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

Well-Known Member
#8
Cherry7Up said:
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

Well-Known Member
#9
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

Well-Known Member
#10
callipygian said:
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).
 

SD Padres

Well-Known Member
#11
I think the whole thing is idiotic because they are trying to suggest you need to be a genius to count cards. I count cards and I'm far from a genius! :)
 

Sonny

Well-Known Member
#12
Let me try to explain this without using any math at all.

When you first pick door #1 (or any door for that matter) you are probably wrong. The car is more likely to be behind one of the other doors because there are more of them. After Monty opens a door you have more information. Why did he pick door #3 and not door #2? Either the car is behind door #1 and he could have chosen either door, which we have already said is unlikely, or he chose door #3 because the car is behind door #2. We now know that the car is most likely behind door #2. We picked door #1 which we know is probably wrong and Monty showed us that it is not behind door #3. Door #2 is the best choice.

-Sonny-
 

HarryKuntz

Well-Known Member
#13
Sonny said:
Let me try to explain this without using any math at all.

When you first pick door #1 (or any door for that matter) you are probably wrong. The car is more likely to be behind one of the other doors because there are more of them. After Monty opens a door you have more information. Why did he pick door #3 and not door #2? Either the car is behind door #1 and he could have chosen either door, which we have already said is unlikely, or he chose door #3 because the car is behind door #2. We now know that the car is most likely behind door #2. We picked door #1 which we know is probably wrong and Monty showed us that it is not behind door #3. Door #2 is the best choice.

-Sonny-
And if this still doesn't make sense,

imagine there are 100 doors, you pick a door and Monty eliminates all the others except one, where is the car more likely to be?
 
#14
Sonny said:
Let me try to explain this without using any math at all.

When you first pick door #1 (or any door for that matter) you are probably wrong. The car is more likely to be behind one of the other doors because there are more of them. After Monty opens a door you have more information. Why did he pick door #3 and not door #2? Either the car is behind door #1 and he could have chosen either door, which we have already said is unlikely, or he chose door #3 because the car is behind door #2. We now know that the car is most likely behind door #2. We picked door #1 which we know is probably wrong and Monty showed us that it is not behind door #3. Door #2 is the best choice.

-Sonny-
If that dosent explain it nothing will. Well said sony.
 

Lonesome Gambler

Well-Known Member
#15
HarryKuntz said:
And if this still doesn't make sense,

imagine there are 100 doors, you pick a door and Monty eliminates all the others except one, where is the car more likely to be?
The key is that you must be shown a goat after choosing your door (in this case, door 1). This presents three scenarios:

1. The car is in door 1 (your choice). A goat is shown in either door 2 or 3. Either way, if you switch, you get the goat.
2. A goat is in door 1. A goat is shown in door 2. Door 3 has the car, so if you switch, you get the car. Switching gets the car.
3. A goat is in door 1. A goat is shown in door 3. Door 2 has the car, so switching, again, gets you the car.

So you see, 2/3 (66.67%) odds favor switching.
 

gibsonlp33stl

Well-Known Member
#16
If you really don't trust it...make a picture of a car and 2 goats. Have a friend be monty...play the game 20 times. Never switch...see how often you get a car and how often you get a goat. Then play 20 times where you always switch...again see how often you get a car and how often you get a goat. If the great explanations don't work, then go to pure examples.

Simply put, is b/c you gain new information during the process...your odds increase drastically to 66% from the original 33%.
 

callipygian

Well-Known Member
#17
Cherry7Up said:
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).
They don't have equal probability because that's how conditional probability works. I don't know how else to explain it; the probability of X given Y is the probability of X given nothing divided by the probability of Y given nothing.

Let's try reversing the approach. Let's say, as you claim, that each of the scenarios has equal probability in the absence of any information. In that case, before Monty opens a door, there are 4 scenarios - (1a), (1b), (2), and (3). If each of them are equally likely, then the probability you guessed right is (2)+(3), or 1/2. Can you explain why (1b) would be half as likely as (2) before Monty opens the door, but equally likely after Monty opens a door?

gibsonlp33stl said:
play the game 20 times
You'll need far more than 40 trials to differentiate between a 50% and 67% of being correct. As it stands, you'd expect 10 wins if the probability is 50% and 13 wins if the probability is 67%. That's a razor-thin margin.
 

Cherry7Up

Well-Known Member
#18
callipygian said:
They don't have equal probability because that's how conditional probability works. I don't know how else to explain it; the probability of X given Y is the probability of X given nothing divided by the probability of Y given nothing.

...

Can you explain why (1b) would be half as likely as (2) before Monty opens the door, but equally likely after Monty opens a door?
The likelihood of (1b) compared to (2) "changes" not as a result of Monty opening the door, but as a result of the increased information we get when we know what door Monty has opened--you gave the explanation yourself, it is the nature of conditional probabilities (the probability of X and the probability of Y are different than the probabilities of X given Z and Y given Z).

We know initially that the car is equally likely to be behind any one of the three doors-- 1/3, 1/3, 1/3.

Monty opening door 2 tells us it isn't behind that door--but doesn't alter the underlying scenario, instead we just get 1/2, 0, 1/2. Switching gains us nothing as the two unopened doors remain equivalent--nothing has changed them, only the door that was opened has been "changed" by the added information.

I don't deny that the writers of the scene could have created an interesting puzzle that hinges on the "ins and outs" of conditional probability--they simply failed to do so.
 

sagefr0g

Well-Known Member
#19
goats, cars & probability

i think the whole Monty, car and goats thing is a pretty good lesson in just how wonderful probability is but how difficult and complex it is living and dealing in a world where it becomes necessary to rely on probability
 

callipygian

Well-Known Member
#20
Cherry7Up said:
the probability of X and the probability of Y are different than the probabilities of X given Z and Y given Z
Yes, but the ratio of X/Y doesn't change.

The probability of X given Z is P(X)/P(Z), the probability of Y given Z is P(Y)/P(Z). The ratio of P(X)/P(Y) doesn't change.

Cherry7Up said:
We know initially that the car is equally likely to be behind any one of the three doors-- 1/3, 1/3, 1/3.

Monty opening door 2 tells us it isn't behind that door--but doesn't alter the underlying scenario, instead we just get 1/2, 0, 1/2.
No, Monty opening door 3 tells us that either the car is behind door 2 (scenario (2)) or that the car is behind door 1 AND he chose door 3 to open when he could have chosen door 2 (scenario (1b)).

Initial: 1a: 1/6, 1b: 1/6, 2: 1/3, 3: 1/3
After door 3 is opened: 1a: 0, 1b: 1/6, 2: 1/3, 3:0

Cherry7Up said:
I don't deny that the writers of the scene could have created an interesting puzzle that hinges on the "ins and outs" of conditional probability--they simply failed to do so.
I argue they did, and that people are confused about the answer (whether you believe you're right or I'm right, it's very clear one of us is puzzled) is evidence enough that it's a good problem.
 
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