My Favorite Riddle/Statistical Question:

[4ever23]

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 a 66 2/3, not just 1/2.
you sir, fail.

No, you sir, fail. There are 2 doors, there is one car, 50/50. The chance of each having the car is the same. The 33.3 from the door that waseliminated is evenly distributed to the two doors left making them both 50%.

EDIT: I guess in theory both doors are 66.66% because the first door was a freebie allowing you to in a sense pick 2 doors the first time and not just one.

 

[whywesteppin]

lol @ people arguing the wrong side. Don't feel stupid, though. This question trips up a lot of other people too.

smh @ the movie using this example. Now a bunch of people are going to walk out of the movie thinking they're good at probability.

 

[bigtomgetsgwap]

damn i never saw 21

 

[cynical jordna]

saw 21 last week goin to atlantic city this weekend FTL lol

 

[roman736]

I never saw that movie. This is a statistical problem a lot older than 21...

 

[commodity05]

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?
thank you professor Rosa

 

[addictedtofreshkicks]

i wouldnt switch

 

[liljrdn]

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 win the 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.
Makes sense. I always thought that if you chose one of the wrong doors, the host would eliminate you automatically and not give you a chance tochoose another door. If the host can only show you a door with the goat behind it and you chose it, then he would probably tell you lost. That's why Ithought that if the host showed you the door with a goat, chances are that you picked the right door in your first try. But now seeing it your way, it doesmake sense to switch.

 

[Thats My Word]

Originally Posted by Shapeshiftah

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 win the 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

there are 2 scenarios in scenario #1. cant just bunch them together...

so really, there are 4 scenarios all together...

 

[shapeshiftah]

Originally Posted by derrty6232

Originally Posted by Shapeshiftah

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 win the 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

there are 2 scenarios in scenario #1. cant just bunch them together...

so really, there are 4 scenarios all together...

not really, because each scenario has one of two choices -- switch or stay. i just chose not to type out the other choices.
therefore, i could rewriteit as follows, for clarity:

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 with door C,obviously you win the 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. if you stay with door A, you lose.

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

 

[gettin it crackin]

Originally Posted by 4ever23

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 a 66 2/3, not just 1/2.
you sir, fail.

No, you sir, fail. There are 2 doors, there is one car, 50/50. The chance of each having the car is the same. The 33.3 from the door that was eliminated is evenly distributed to the two doors left making them both 50%.

EDIT: I guess in theory both doors are 66.66% because the first door was a freebie allowing you to in a sense pick 2 doors the first time and not just one.


4ever23 = Fail.

 

[roman736]

Alright look, this is the best way to explain it: When you make your first choice, there is a 2/3 chance that the car is in the other two doors, and a 1/3chance that it is behind your door. After the guy opens the other door, there is a 1/2 chance that the car is behind the door you did not choose because itMUST be in one of the remaining two unopened doors. HOWEVER, the 50% statistic does not apply to your choice because when you picked it it was only a 1/3chance that you were right, and the opening of one door statistically only becomes applicable to the door you did not choose. Thus, if you switch to the otherdoor you have a 1/2 chance of being right, as opposed to the 1/3 chance of the door you picked when all three were still closed.

 

[oidreez]

so basically you're saying switch doors in theory because the announcer not picking one of the doors means there's a good chance that that door mighthave the prize behind it. right?

 

[sneakerreaper]

Monty Hall Problem