My Favorite Riddle/Statistical Question:

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Aug 26, 2002
You are on a game show. There are three doors, two of which contain a goat behind them and one that contains a car. If you choose the door with a car behindit you keep the car. The way the game works is like this: You choose a door and before that door is opened, the game show host opens one of the two remainingdoors. The game show host knows where the car is and will only open a door with a goat behind it. After the one door has been opened, do youchoose the original door you chose (which has not yet been opened), or do you switch to the other remaining door that is still closed.
Or does it not matter?
 
i think i would switch.

the original door seems like it would have a 1 out of 3 chance of holding the car. the other door has a 2 out of 3 chance of winning.
 
Originally Posted by Dirtylicious

statistically..
you switch

Wait, why? wouldn't you just choose the original door?

I'm confused now
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You switch you have a 50/50 shot if you switch, before he opened the door it was a 33.3% you would have picked the right door, after game show host reveals thegoat, the odds go up 50% that you choose the correct one.
 
Since there are 3 doors and two goats when you make your initial selection chances are you chose a goat. Since he
eliminates a door with the goat odds are the last door is the car. So you switch.

There is a wikipedia page about but I forgot the games show hosts' name which it was under.
 
i think a lot of people in here went and saw the movie 21
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Its an old question that was given to one of the smartest people in the world several years ago.
 
After eliminating one door it is still a 50% chance that you have the correct door. Statistically this doesn't make a damn bit of sense. Any door has a 33%chance. After one door is gone, all doors have a 50% chance. It doesn't matter if you change. Your percentages with either door are the same.
 
Originally Posted by overtime626

this is an old one, although it seems like a good idea to switch, statistically it doesnt matter.

actually, i think it does matter statistically. this is how i look at the problem:

these are the three doors, as well as what lies behind each:

[A] -- goat

-- goat

[C] -- car

there are three possible scenarios, each of which has a 33% chance of happening.

scenario #1
you choose door C. the host shows you either A or B (either of which has the goat). if you switch to either A or B, you lose. if you stay, obviously you winthe car.

scenario #2
you choose door A. the host shows you B, which has the goat (he or she can't show you C). if you switch to C, you win.

scenario #3
you choose door B. the host shows you A, which has a goat. if you switch to C, you win.

therefore, two out of the three above scenarios (which involve switching) could result in you winning.
btw, thanks for the wtaps vans
pimp.gif
 
Originally Posted by overtime626

this is an old one, although it seems like a good idea to switch, statistically it doesnt matter.
Actually it does matter. Statistically you would be better off to switch.
 
Originally Posted by roman736


You are on a game show. There are three doors, two of which contain a goat behind them and one that contains a car. If you choose the door with a car behind it you keep the car. The way the game works is like this: You choose a door and before that door is opened, the game show host opens one of the two remaining doors. The game show host knows where the car is and will only open a door with a goat behind it. After the one door has been opened, do you choose the original door you chose (which has not yet been opened), or do you switch to the other remaining door that is still closed.
Or does it not matter?

STOLEN FROM A SCENE IN:


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the answer is you change your mind, because your original pick of the door was based on a 33 1/3 chance. Now that there are only two left, your chance is a66 2/3, not just 1/2.
you sir, fail.
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