http://www.ukpsychics.com/3doors_answer.html

this one has always intrigued me. i remember the original "Ask Marilyn" article. I still can't bring myself to agree with her conclusion...

I dont get it

Gig 'Em

makes sense to me how the contest is set up.

You have doors A, B, C

You pick A,
B is revealed not to have the prize.
You must chose between A and C
Go with C.

very confused... what is the question

I dont understand how the not chosen door/box/whatever changes to 2/3. Why is it not still 1/3?

It doesnt seem like the revealed door should effect the probability.

Gig 'Em

The odds of B or C is 2/3 because B = 1/3 and C = 1/3. I don't buy that removing B changes the odds of C to 1/3, but then again I'm not a mathematician so I can't refute it

I'm not trying to support or refute this hypothesis, but I wonder if this is a case where semantics are getting in the way of the mathematics?

ok this part is on a previous page

quote:

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The three doors dilemma
So you think you understand gambling odds, eh? Well, try this...
Imagine you're on a game show, and you're given the choice of three doors. Behind one door is a car; behind each of the other two, a goat.
You pick a door, say No. 1, and the host opens one of the remaining two doors, say No. 3, to reveal a goat. He says to you, "Do you want to stick with the door you've chosen, or do you want to change your mind and go for door No. 2 instead?"
Question: Is it to your advantage to switch your choice of doors?

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When the two remaining doors (B and C) are whittled down to one remaining door (just B), the odds are in favor of the remaining door (B).

You get to hold two doors up to the fire, and when one burns, the other has a better chance of being "the one." You get to "test" the two doors that you think are wrong, while your door hasn't been tested at all. The door that is tested and survives is better than your un-tested door.

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If you can read this, thank a teacher.
If you can read this in English, thank a Soldier.

so if there are only 2 doors, if you switch your choice you have a 100% chance?

Makes sense to me.

quote:

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The odds of B or C is 2/3 because B = 1/3 and C = 1/3. I don't buy that removing B changes the odds of C to 1/3, but then again I'm not a mathematician so I can't refute it

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If I choose Door A, there is a 2/3 chance that the car is behind B or C. If I reveal that the car is NOT behind Door B, there is STILL a 2/3 chance that the car is behind B or C. The odds didn't change. But since we now know the car is not behind B, that puts the entire 2/3 odds squarely on Door C.

Theres 3 doors and one goes away leaving 2 doors. Read the damn link

DING DING DING

quote:

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that puts the entire 2/3 odds squarely on Door C

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We have a WINNER! Way to go Racer, you may claim your car at the Texags Staff office.

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If you can read this, thank a teacher.
If you can read this in English, thank a Soldier.

that is just stupid and wrong.

quote:

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that is just stupid and wrong.

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You'll find that many things in statistics are counterintuitive, but they are nonetheless correct.

I dont think so. Once a door has been eliminated, there are two doors. one has a car, one has a goat. your chance is 50/50. why would you still be working in "thirds " when you only have two doors left.

based on what is says if you pick door 1. Host has door 3 opened to show you the prize isnt in there. So the prize is behind door 1 or door 2. You have a 50/50 shot that 1 is correct and a 50/50 shot that 2 is correct. Saying that there is a 66% chance is crap because 1 choice has been eliminated, hence no longer a choice.

The odds of it being behind door 2 do not increase unless it is not truely random.

howdydamnit04: Please re-read:

You get to "test" the two doors that you think are wrong, while your door hasn't been tested at all. The door that is tested and survives is better than your un-tested door.


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If you can read this, thank a teacher.
If you can read this in English, thank a Soldier.

Did you read the article? They have two simulators that you can run that will prove to you that it is correct. You can even do the same thing with 3 playing cards so you know it isn't fixed.

The logic makes my head hurty.

No I didnt read the entire article because I read this part

quote:

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The three doors dilemma
So you think you understand gambling odds, eh? Well, try this...
Imagine you're on a game show, and you're given the choice of three doors. Behind one door is a car; behind each of the other two, a goat.
You pick a door, say No. 1, and the host opens one of the remaining two doors, say No. 3, to reveal a goat. He says to you, "Do you want to stick with the door you've chosen, or do you want to change your mind and go for door No. 2 instead?"
Question: Is it to your advantage to switch your choice of doors?

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that is the "question" that this link is the "answer" to. I started to read their answer but it was wrong so I stopped.

so basically I was purely answering the original question. If you want to put stipulations on the question, you do not do that in the answer section.

ok now i have read their explanation and they are still wrong. Sorry, just wrong. I dont care if there are stats to back it up. If your pick is truely random and the prize is located behind a truely random door then the odds will not change.

[This message has been edited by howdydamnit04 (edited 10/18/2005 2:59p).]

There are no stipulations. It's the same question and answer.

http://planettom.home.mindspring.com/vossavant.htm

Try it out.

quote:

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ok now i have read their explanation and they are still wrong. Sorry, just wrong. I dont care if there are stats to back it up. If the host is picking a truely random door and your pick is truely random and the prize is located behind a truely random door then the odds will not change.

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You're right. That's exactly the point. The odds DO NOT CHANGE even after you reveal that the prize is not behind one of the doors. The odds are still 1/3 that you picked the right door and 2/3 that you picked the wrong one. That's the point!

Ok imagine this. There are two doors. A prize behind one and a goat behind another. You get to pick a door. Exact same thing.

The first round is BS. It is the second round you make the real choice as one of the 2 "goat" doors is eliminated.

Howdydamnit04, come to Bonfire Cut on the 29th and I'll explain in person. I'll be the guy with the white spring break t-shirt and a black pot with "Opie03" on the back.

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If you can read this, thank a teacher.
If you can read this in English, thank a Soldier.

It isn't the exact same thing, though. They are entirely different problems.

Just try it with 3 playing cards. Take 2 number cards and a face card. Have a friend shuffle them and lay them face down, noting where the face card is. Then you put your finger on one. Then have your friend reveal the one of the other two cards that is NOT the face card. Then you either put your finger on the other card or keep it where it is. Keep track of how many times you "win" doing each. Do it about 25 times each way. You'll win twice as often when you switch.

The first round IS important. You are chosing which doors will represent the 2/3 odds that you are wrong. When one door is removed from those two, the remaining one STILL represents the 2/3 odds that you are wrong.

I'll go with 2/3 over 1/3 any day.

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If you can read this in English, thank a Soldier.

I dont need it explained. If you read the post right above yours, you will see my position. This is basic logic that is getting confused with math.

quote:

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This is basic logic that is getting confused with math.

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If you play the card game I suggested, you will see that your logic is flawed.

Sorry guys I know this is pissing you off if I get the choice between two doors and one has a prize behind it and the other definately does not... that is a 50/50 chance. I cant explain it any other way. Anything that happens before that does not matter, short of thems saying "We are going to eliminate door 3" and whispering in my ear "it isnt behind 2".