You are on a game show. The host offers you a choice between three doors. Behind one is a new car. Behind the others there is nothing. You choose door one giving you a 33.3% of winning. The host opens door three which is empty. He always opens a empty door first as he always knows which door has the car. He always then gives the contestant a chance to change his/her mind and choose a different door. Does it matter which door you choose now and why?
Great Puzzle
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This is the famous 'Monty Hall' puzzle made famous because it is counter intuitive. Always choose the other door(ie change your choice).You double your chance of winning.
I don't get it. If one door is already opened, then the odds would be .5 (from the two remaining doors) regardless of whether you changed your answer or not.Joy said:
If you choose door two you have added door three's odds (as door three has been eliminated) to your original odds giving you a 66% chance. If you keep door one your odds are not 50% but remain at 33%.
This is the answer. The bigger question is why.
This is called the variable change effect. As new info is added so do your odds.
I will try, but struggle, to explain.
The key is that Monty knows which door has the car.
If the car is behind door one then it does not matter which door he opens first as they are both empty and no new information has been added.
If door one is empty then Monty has to open the remaining door that is also empty thus revealing with certainty which door has the car ( door two.) This is the new information that may be added as it is not a certainty this information will be added. It has a 50% chance of being added.
Can some wamker stop what he/she is doing and complete this answer as it resides as an abstaction in my brain that I cannot grasp well enough to articulate.
This is the answer. The bigger question is why.
This is called the variable change effect. As new info is added so do your odds.
I will try, but struggle, to explain.
The key is that Monty knows which door has the car.
If the car is behind door one then it does not matter which door he opens first as they are both empty and no new information has been added.
If door one is empty then Monty has to open the remaining door that is also empty thus revealing with certainty which door has the car ( door two.) This is the new information that may be added as it is not a certainty this information will be added. It has a 50% chance of being added.
Can some wamker stop what he/she is doing and complete this answer as it resides as an abstaction in my brain that I cannot grasp well enough to articulate.
Thankyou "a 1 player".
http://en.wikipedia.org/wiki/Monty_Hall_problem
offers the solution and a complete answer.
I did not know the answer actually. I was on the right track but just could not finish with a complete answer.
The problem is posed in "21" which is a movie about the M.I.T. students that perfected card counting in blackjack and beat Vegas. The movie is wrong in a lot of details. They got caught because a Casino hired a dectective to figure out why they were losing. The morons from M.I.T. all gave their real home addresses to the hotels. It did not take the detective long to figure it out when he realized all the big winners came from M.I.T.
The problem is posed in the movie but a complete answer not given. When I woke up this morning I could not stop thinking about it.
The explanation is:
Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.
Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3.
With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty.
The total probability of winning when switching is thus 2/3.
http://en.wikipedia.org/wiki/Monty_Hall_problem
offers the solution and a complete answer.
I did not know the answer actually. I was on the right track but just could not finish with a complete answer.
The problem is posed in "21" which is a movie about the M.I.T. students that perfected card counting in blackjack and beat Vegas. The movie is wrong in a lot of details. They got caught because a Casino hired a dectective to figure out why they were losing. The morons from M.I.T. all gave their real home addresses to the hotels. It did not take the detective long to figure it out when he realized all the big winners came from M.I.T.
The problem is posed in the movie but a complete answer not given. When I woke up this morning I could not stop thinking about it.
The explanation is:
Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.
Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3.
With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty.
The total probability of winning when switching is thus 2/3.
And if Monty`s backstage guys switch the car after your first pick, the whole thing changes.
I hafta find more to do, and you can read more about this puzzle—which perplexed the learned—by checking the TERB 2005 thread on the topic
I hafta find more to do, and you can read more about this puzzle—which perplexed the learned—by checking the TERB 2005 thread on the topic
It may depend on more than just the trickster dehind the doors oldjones.
Question
In the scenario we just played out, Monty always knew what door the car was behind. What if he does not know and chooses randomly, does this change anything?
Question
In the scenario we just played out, Monty always knew what door the car was behind. What if he does not know and chooses randomly, does this change anything?
Joy if I come to the Loco will you sit on my lap and talk variables and probabilities with me. Smart ladies are so sexy!!!!!!!!!!!!!!!!!!!!!Joy said:
red : this is getting more interesting
red
you are correct
"If the host does not know what lies behind the doors, and opens one at random without revealing the car (Granberg and Brown, 1995:712). Switching wins the car half of the time. "
However. I am struggling to understand why. Monty took a random chance here, he could have revealed the car but did not. If he had, clearly your odds have dropped to zero no matter what door you now choose. That he did not reveal the car, even though there was a 33% chance he could have, should not change our original plan of switching doors. Even though it has occurred by chance, and not by deliberate manipulation on Monty's part, our original scenario remains the same.
Apparently, I am wrong on this.
You seem to comprehend so please be so kind to say why this makes sense to you.
red
you are correct
"If the host does not know what lies behind the doors, and opens one at random without revealing the car (Granberg and Brown, 1995:712). Switching wins the car half of the time. "
However. I am struggling to understand why. Monty took a random chance here, he could have revealed the car but did not. If he had, clearly your odds have dropped to zero no matter what door you now choose. That he did not reveal the car, even though there was a 33% chance he could have, should not change our original plan of switching doors. Even though it has occurred by chance, and not by deliberate manipulation on Monty's part, our original scenario remains the same.
Apparently, I am wrong on this.
You seem to comprehend so please be so kind to say why this makes sense to you.
does anyone care?
I think I just figured it out.
The reason that it does not matter which door you choose if Monty does not know which door the car is behind, is that his choosing door three does not quarantee the door is behind door two the way it would if, and only if, he had previous knowledge that door three was empty in, and only in, the scenario that door one was empty.
Refer to must first answer for completion of my logic.
Now, does anyone agree or disagree with what I just said?
Does anyone understand what I just said?
Does anyone give a fu*k?
I think I just figured it out.
The reason that it does not matter which door you choose if Monty does not know which door the car is behind, is that his choosing door three does not quarantee the door is behind door two the way it would if, and only if, he had previous knowledge that door three was empty in, and only in, the scenario that door one was empty.
Refer to must first answer for completion of my logic.
Now, does anyone agree or disagree with what I just said?
Does anyone understand what I just said?
Does anyone give a fu*k?
This is statistical smoke and mirrors. Take the third door out of the puzzle and you have 2 doors. The car is behind one of them. You have a 50% chance of being right and a 50% chance of being wrong. Switching still give you the same odds.
The car is either there, or it is not. 2 choices, 2 chances. Switching does nothing to change your odds.
The 3rd door is a confounding variable that has been eliminated from the equation after it is opened.
The car is either there, or it is not. 2 choices, 2 chances. Switching does nothing to change your odds.
The 3rd door is a confounding variable that has been eliminated from the equation after it is opened.
DistantVoyeur
You are not learning.
I heard this puzzle on the movie "21" but they did not explain any of the logic so I posted this puzzle this morning hoping to be enlightened and I was.
It turns out this is a very famous puzzle because it defies our common sense, which is the problem you are having.
It does not matter what door you choose, either 1 or 2, ONLY if Monty DID NOT KNOW that door 3 was empty before he opened it. In such a case the odds are 50/50 on either door.
If Monty knew door 3 was empty before he opened it then you increase your odds to 66.6% by choosing door 2.
I have tried to articulate the reasoning as best I can in past threadlings and will not do so again as you can go back and read them.
If you still have questions I will try to answer them.
It took many PHD's some time before they accepted this answer but now the debate has subsided.
You are not learning.
I heard this puzzle on the movie "21" but they did not explain any of the logic so I posted this puzzle this morning hoping to be enlightened and I was.
It turns out this is a very famous puzzle because it defies our common sense, which is the problem you are having.
It does not matter what door you choose, either 1 or 2, ONLY if Monty DID NOT KNOW that door 3 was empty before he opened it. In such a case the odds are 50/50 on either door.
If Monty knew door 3 was empty before he opened it then you increase your odds to 66.6% by choosing door 2.
I have tried to articulate the reasoning as best I can in past threadlings and will not do so again as you can go back and read them.
If you still have questions I will try to answer them.
It took many PHD's some time before they accepted this answer but now the debate has subsided.
Take the question to an extreme to see how it works.DistantVoyeur said:
Say you have 100 doors, behind one of which is a car. You select one door, and before you open it, Monty opens up 98 of the other 99 doors. He knows which door the car is behind, and deliberately chooses which doors to reveal so that the car is still hidden.
So, now you have two doors unopened still -- one that you originally chose, and the other that Monty chose to leave unopened.
Which do you think you should now choose to open to try to find the car?
I would switch to the door Monty chose.
Why?
The first door I chose had a 1/100 chance of having the car.
The other set of doors had a 99/100 chance of having the car. What Monty did was simply narrow that set of doors down to one for me. His hand is forced a bit, as he has to choose empty doors.
So, the door Monty chose has a 99/100 chance of having the car. Choose it.
Past behaviour still has no influence on the outcome of the event. The argument that Monty has to go for empty doors only works until you get down to 2 doors. At that point in time he is obligated to ask you to switch whether you have the right door or not.
If you use your analogy of 100 doors then the cumulative addition of odds means that you have a 98% chance if you switch. Make it a 1000 doors and you have a 99.9% chance.
If you are given only 2 doors, and only start with 2 doors, you would say it is 50% absolutely without question. At that point in time, you only have 2 doors. The hosts question has no influence at that point in time. He has to ask you the question whether you have the right door or not.
If you use your analogy of 100 doors then the cumulative addition of odds means that you have a 98% chance if you switch. Make it a 1000 doors and you have a 99.9% chance.
If you are given only 2 doors, and only start with 2 doors, you would say it is 50% absolutely without question. At that point in time, you only have 2 doors. The hosts question has no influence at that point in time. He has to ask you the question whether you have the right door or not.
