Why the Monty Hall puzzle is categorically 1/2 and not 2/3rds

This is how I think about it:

The only outcome where it would be beneficial to you to not switch is when you guess correctly with your initial choice. As you know this happens 1/3 of the time. This would mean that the other 2/3 of the time you will get it correct if you switch.

In other words, every time you do not choose the correct door on your first try, switching is guaranteed to give you the correct door since you are currently on one wrong door and the other wrong door has been eliminated. You do not choose the correct door on your first try 2/3 of the time, therefor 2/3 of the time when you switch you will end up on the correct door.

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 Quote by TheIliaD This is how I think about it: The only outcome where it would be beneficial to you to not switch is when you guess correctly with your initial choice. As you know this happens 1/3 of the time. This would mean that the other 2/3 of the time you will get it correct if you switch. In other words, every time you do not choose the correct door on your first try, switching is guaranteed to give you the correct door since you are currently on one wrong door and the other wrong door has been eliminated. You do not choose the correct door on your first try 2/3 of the time, therefor 2/3 of the time when you switch you will end up on the correct door.
This is incorrect.
If you happen to be on a show with an IGNORANT Monty Hall, who happens to open a door with a goat, then it is irrelevant if you switch or not.
The critical issue is that a rule of the game is that you can rest assured that Monty Hall WILL open a door for the second chance (rather than revealing your first picked goat), and that he ALWAYS will reveal a goat after your first pick.
 Mentor Blog Entries: 9 You choose 1 door, so have a 1/3 chance of making the right pick. That means the odds of the car being behind one of the remaining 2 doors is 2/3. That should be clear and obvious, right? Why does the host opening one of those doors change the odds? The host shows you which one is NOT the car, there is a 2/3 chance of winning the car if you switch, 1/3 if you don't.
 Recognitions: Gold Member Homework Help Science Advisor Just following Integral: The reason why the probabilities DON'T change is that you know in beforehand that Monty will reveal at least goat happened to be present in the GROUP of doors you did not pick to begin with. Thus, the Group probability (of containing car) remains unchanged, relative to your initial information after Mobty opens the door. If you are blind, and are allowed to pick the opened door as well as the closed one, there is no reason to switch..

 Quote by arildno This is incorrect.
Why?
 If you happen to be on a show with an IGNORANT Monty Hall, who happens to open a door with a goat, then it is irrelevant if you switch or not.
I don't think anyone's suggested that Monty Hall has ever been ignorant of the contents.
 The critical issue is that a rule of the game is that you can rest assured that Monty Hall WILL open a door for the second chance (rather than revealing your first picked goat), and that he ALWAYS will reveal a goat after your first pick.
Precisely, so what TheIliaD said seems pretty correct to me. where does "revealing you first picked goat come" into it?

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 Quote by arildno This is incorrect.
I don't see anything wrong with what he said. He clearly said that the wrong door gets eliminated, not a random door.

 Quote by micromass I don't see anything wrong with what he said. He clearly said that the wrong door gets eliminated, not a random door.

This is one of the oldest and, usually, hottest mathematical discussion on the web. I remember it AT LEAST from 1987, when I

If mathematicians are involved one can usually expect they will understand more or less pretty quickly why the probability

really is 2/3, but if a greenhorn gets into the discussion....oh, my! In particular, greenhorns with an attitude and lots

of ignorance. Then not only peaceful agreement is NOT reached but things begin to rot pretty soon. Watch it!

DonAntonio

Hmm...

I still don't get it.

Does it depend if the rule changes or stays constant? (e.g. remove door C... left with two other doors. Meaning is the next part of the gameshow somehow unrelated to the first rule?)

honestly struggling to comprehend switching is 2/3rd chance.

 You choose 1 door, so have a 1/3 chance of making the right pick. That means the odds of the car being behind one of the remaining 2 doors is 2/3. That should be clear and obvious, right? Why does the host opening one of those doors change the odds? The host shows you which one is NOT the car, there is a 2/3 chance of winning the car if you switch, 1/3 if you don't.
Completely agree with the first couple of sentences... but the last sentence "The host shows you which one is not the car" not sure I understand the jump from 1 door being 2/3rds and the other 1/3. I'm not trolling I just don't get it.
 Recognitions: Homework Help All the talk of door opening just confuses things. The problem is equivalent to this. Suppose there are n>2 doors and one random one is it Choose one door Now guess if you were right. You are wrong more of the time as you picked one door and did not pick more than one. in particular your are right with probability 1/n and wrong with probability 1-1/n and 1/n<1-1/n since n >2 It is better to bet on being wrong.

 Quote by CaptainOrange Hmm... I still don't get it. Does it depend if the rule changes or stays constant? (e.g. remove door C... left with two other doors. Meaning is the next part of the gameshow somehow unrelated to the first rule?) honestly struggling to comprehend switching is 2/3rd chance. Completely agree with the first couple of sentences... but the last sentence "The host shows you which one is not the car" not sure I understand the jump from 1 door being 2/3rds and the other 1/3. I'm not trolling I just don't get it.

Think of this as follow: if you change your choice after the presenter opened a "bad" door (and yes, he always knows

what door contains the great prize), then:

1) From the beginning you chose first "bad" door, the presenter opens the second "bad" door, you switch then YOU WIN

2) From the beginning you chose second "bad" door, the present opens the first "bad" door, you switch then YOU WIN again

3) From the beginning you chose the "good" door, the present opens either "bad" door, you switch then YOU LOSE.

Be convinced the above exhaust all the possibilities, and thus by switching doors in two of them you win and in one

of them you lose, ergo the probability of winning by switching doors is 2/3. Q.E.D.

DonAntonio

 Quote by DonAntonio Think of this as follow: if you change your choice after the presenter opened a "bad" door (and yes, he always knows what door contains the great prize), then: 1) From the beginning you chose first "bad" door, the presenter opens the second "bad" door, you switch then YOU WIN 2) From the beginning you chose second "bad" door, the present opens the first "bad" door, you switch then YOU WIN again 3) From the beginning you chose the "good" door, the present opens either "bad" door, you switch then YOU LOSE. Be convinced the above exhaust all the possibilities, and thus by switching doors in two of them you win and in one of them you lose, ergo the probability of winning by switching doors is 2/3. Q.E.D. DonAntonio
Don, that was so eloquently explained! I don't know why I was stuck at the 50:50 stage (maybe I was imagining the problem differently). After reading that I confess I have eaten humble pie.

 Quote by CaptainOrange Don, that was so eloquently explained! I don't know why I was stuck at the 50:50 stage (maybe I was imagining the problem differently). After reading that I confess I have eaten humble pie.

I think we all did when we first meet with this problem. The most important lesson I learned back then: in mathematics, no matter

how good/old/veteran you think you are, always leave a little groove for doubt and humbleness to sit...and wait.

DonAntonio

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 Quote by DonAntonio Be convinced the above exhaust all the possibilities, ..
... and are equally likely ...

 Quote by haruspex ... and are equally likely ...

Of course, but since this is assumed to be a "choose in a random way something" out of several possibilites, this is

a rather straightforward assumption.

DonAntonio

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