My favorite math problem

There are two arguments to this; one is math and one is logic:
1. with three doors, there is a 1/3 chance of any one door being correct.  Once one of the doors is open, then you now have a two thirds chance of being wrong; so switch.
2. Assuming that the game show host knows which door is correct, it is possible that he only asks to switch if the correct door is inially chosen; so stay...

I will never be on a game show...
1.This is the right line of thinking.... at least it is what I was looking for.

2. Interesting take. For simplicity, lets say the host always picks an empty door that doesn't have it first, no matter what.
 
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MC MUSTANG @ Mar. 06 2008 said:
1326739[/ATTACH] AM]
There are two arguments to this; one is math and one is logic:
1. with three doors, there is a 1/3 chance of any one door being correct.  Once one of the doors is open, then you now have a two thirds chance of being wrong; so switch.
2. Assuming that the game show host knows which door is correct, it is possible that he only asks to switch if the correct door is inially chosen; so stay...

I will never be on a game show...
Not correct.  Once one of the doors is open, your chances change from 1 in 3 to 1 in 2.  It makes no difference if you switch from A to C or not.  Whichever one you prefer, go for it!

--Wag--
You're first guess is 1/3 of being right. Correct? The host then removes an incorrect case.

So there is a 50% chance that you case A or C has the 50K. However, your selection was based on A,B and C. In order for you to take advantage of the new information, you must switch.

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I'd keep door A , I would kick myself more for giving it away then I would for not choosing it.
 
Simple, just ask the host to show you one more door and you will then make your selection.
 
I looked it up on wiki like rocket man said. Wow that is deep stuff man.
you could pick a worse information resource but you would be hard pressed... Ask wiki what the best tire for your bike is or what color you should paint your house...

Want even more fun? cite wikipedia on a research paper, most likely end up with an F...
 
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MC MUSTANG @ Mar. 06 2008 said:
1328019[/ATTACH] AM]
There are two arguments to this; one is math and one is logic:
1. with three doors, there is a 1/3 chance of any one door being correct. Once one of the doors is open, then you now have a two thirds chance of being wrong; so switch.
2. Assuming that the game show host knows which door is correct, it is possible that he only asks to switch if the correct door is inially chosen; so stay...

I will never be on a game show...
Not correct. Once one of the doors is open, your chances change from 1 in 3 to 1 in 2. It makes no difference if you switch from A to C or not. Whichever one you prefer, go for it!

--Wag--
You're first guess is 1/3 of being right. Correct? The host then removes an incorrect case.

So there is a 50% chance that you case A or C has the 50K. However, your selection was based on A,B and C. In order for you to take advantage of the new information, you must switch.
I disagree. In the first case, you do have a one in three chance of being right. However, once a choice has been removed, the original scenario no longer applies. In other words, your odds are now based on two choices not three choices because you no longer have three choices. You only have two. Therefore, the odds have also changed.

Capiche?
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--Wag--

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MC MUSTANG @ Mar. 06 2008 said:
1328026[/ATTACH] AM]
There are two arguments to this; one is math and one is logic:
1. with three doors, there is a 1/3 chance of any one door being correct.  Once one of the doors is open, then you now have a two thirds chance of being wrong; so switch.
2. Assuming that the game show host knows which door is correct, it is possible that he only asks to switch if the correct door is inially chosen; so stay...

I will never be on a game show...
Not correct.  Once one of the doors is open, your chances change from 1 in 3 to 1 in 2.  It makes no difference if you switch from A to C or not.  Whichever one you prefer, go for it!

--Wag--
You're first guess is 1/3 of being right. Correct? The host then removes an incorrect case.

So there is a 50% chance that you case A or C has the 50K. However, your selection was based on A,B and C. In order for you to take advantage of the new information, you must switch.
I disagree.  In the first case, you do have a one in three chance of being right.  However, once a choice has been removed, the original scenario no longer applies.  In other words, your odds are now based on two choices not three choices because you no longer have three choices.  You only have two.  Therefore, the odds have also changed.

Capiche?  
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--Wag--
Lou,
I agree with you in the sence that you now have a 50/50 shot, but the original question is whether or not to switch your initial answer, not to choose between a and c. The question is whether or not to second guess yourself. I agree that this is a matter of perspective; if you evaluate the second choice completely as an independant decision, then it is 50/50...

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MC MUSTANG @ Mar. 06 2008 said:
1328030[/ATTACH] AM]
There are two arguments to this; one is math and one is logic:
1. with three doors, there is a 1/3 chance of any one door being correct. Once one of the doors is open, then you now have a two thirds chance of being wrong; so switch.
2. Assuming that the game show host knows which door is correct, it is possible that he only asks to switch if the correct door is inially chosen; so stay...

I will never be on a game show...
Not correct. Once one of the doors is open, your chances change from 1 in 3 to 1 in 2. It makes no difference if you switch from A to C or not. Whichever one you prefer, go for it!

--Wag--
You're first guess is 1/3 of being right. Correct? The host then removes an incorrect case.

So there is a 50% chance that you case A or C has the 50K. However, your selection was based on A,B and C. In order for you to take advantage of the new information, you must switch.
I disagree. In the first case, you do have a one in three chance of being right. However, once a choice has been removed, the original scenario no longer applies. In other words, your odds are now based on two choices not three choices because you no longer have three choices. You only have two. Therefore, the odds have also changed.

Capiche?
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--Wag--
Lou,
I agree with you in the sence that you now have a 50/50 shot, but the original question is whether or not to switch your initial answer, not to choose between a and c. The question is whether or not to second guess yourself. I agree that this is a matter of perspective; if you evaluate the second choice completely as an independant decision, then it is 50/50...
Aha! Then it IS a question of psychology, not a math question!

Hence, the decision to NOT switch the answer. Non-mathematical odds are, the game show host knows the right answer, knows you picked it and is attempting to sway you to change. Otherwise had you initially picked an empty door, he would not show you yet another empty door, give you an option to change and risk the possibility that you WOULD change your answer.

The choice is ONLY given if you pick the right door to begin with.

Slick!

--Wag--

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MC MUSTANG @ Mar. 06 2008 said:
1330338[/ATTACH] AM]
There are two arguments to this; one is math and one is logic:
1. with three doors, there is a 1/3 chance of any one door being correct.  Once one of the doors is open, then you now have a two thirds chance of being wrong; so switch.
2. Assuming that the game show host knows which door is correct, it is possible that he only asks to switch if the correct door is inially chosen; so stay...

I will never be on a game show...
Not correct.  Once one of the doors is open, your chances change from 1 in 3 to 1 in 2.  It makes no difference if you switch from A to C or not.  Whichever one you prefer, go for it!

--Wag--
You're first guess is 1/3 of being right. Correct? The host then removes an incorrect case.

So there is a 50% chance that you case A or C has the 50K. However, your selection was based on A,B and C. In order for you to take advantage of the new information, you must switch.
I disagree.  In the first case, you do have a one in three chance of being right.  However, once a choice has been removed, the original scenario no longer applies.  In other words, your odds are now based on two choices not three choices because you no longer have three choices.  You only have two.  Therefore, the odds have also changed.

Capiche?  
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--Wag--
This is why I like this problem. Intuitively, it seems so wrong.

I agree there is a 50% chance of it being in A or C, but you picked between A, B and C, or a 1/3 chance.

The fact that the host shows you an empty door doesn't change your odds any. In order to change your odds you must switch. And they don't change unless you switch.

You can even try this with a friend with cups and a ball, and have you pick a cup, then have them show you a cup they know to be empty then ask you to switch.

If you don't switch, you will find the ball about 1/3 of the time. Better yet... play as the host, and bet on the outcome
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