Let's say you are on a game show. There are three doors, Door 1, Door 2, and Door 3. Behind two of the doors is a goat. Behind one of the doors is a car.

You pick one door. Before your door is revealed, the host decides to make the game more interesting. He opens one of the other doors to reveal a goat. He asks if you'd like to switch to the other remaining door.

Would you say:

A. Your odds of winning are greater by sticking with the door you picked
B. Your odds of winning are greater by switching to the other remaining door
C. It makes no difference which door you pick



Think about it.




I'm guessing all of you picked C. After all, with two doors, the odds are 50/50.


This is incorrect. There is actually a 66.66% chance that you would win if you switched doors.


The average probability of winning the car by switching can be illustrated using Venn diagrams. After choosing Door 1, for example, the player has a 1/3 chance of having selected the door with the car, leaving a 2/3 chance between the other two doors, as shown below. Note that there is a 100% chance of finding a goat behind at least one of the two unchosen doors, because there is only one car.




The host now opens Door 3. Under the conditions of the problem statement the host is equally likely to open either unchosen door, so opening this door does not affect the average probability of winning the car by staying with the original choice which remains 1/3. The car is not behind Door 3, so the entire 2/3 probability of the two unchosen doors is now carried only by Door 2, as shown below. Another way to state this is that if the car is behind either door 2 or 3, by opening Door 3 the host has revealed it must be behind Door 2.



[Edited on April 11, 2008 at 11:12 AM. Reason : ]

4/11/2008 11:11:45 AM

MY MOUTH WAS A BROKEN JPEG

4/11/2008 11:12:10 AM

i'm actually okay with getting the goat

4/11/2008 11:12:25 AM

srsly, I already have a car

4/11/2008 11:14:47 AM

yes

slashdot controls many a viral thought

4/11/2008 11:15:47 AM

i hate statistics for this reason since one door is open there s a 50% chance of it being behind either door you can't just add the chance it's behind one of the doors if i remember this correctly

4/11/2008 11:21:27 AM

^^ I don't read slashdot, faggot.

4/11/2008 11:23:34 AM

still 50/50 in my mind. forget about the 3rd one once it's revealed.

4/11/2008 11:29:54 AM

God must have just watched the movie 21. They explained the same thing in it.

4/11/2008 11:33:33 AM

I don't watch shitty movies like that.

4/11/2008 11:36:01 AM

ohh now i've though some more i get it - when you picked door 1 it had 33% chance of the car being there - so 2&3 = 66% when door 3 is closed since that section is grouped it becomes door 2 is 66% chance of having car - door one is the one constant one - now if door 3 was opened before you decided then it would be 50 - 50

4/11/2008 11:38:37 AM

Quote :

"I'm guessing all of you picked C. After all, with two doors, the odds are 50/50. This is incorrect. There is actually a 66.66% chance that you would win if you switched doors."

From the way you explained the problem, YOU are the one that is incorrect here.

You need to add some information to your original problem for your answer to be true. Namely, you need to explain the fact that the host both
a) knows the location of the car beforehand
and
b) was required beforehand to open a door with a goat behind it after you make your selection. (i.e. he's not "conditionally" opening a door based on if you picked the car or not)

Assuming those two conditions are true, then yes, you'd be better off switching every time.

The way you explained it, the answer C is correct, it doesn't really matter which one you choose.

4/11/2008 11:48:11 AM

I dunno, I still have to see the gameshow myself

4/11/2008 11:49:05 AM

Someone saw 21

4/11/2008 11:50:33 AM

i remember making this thread like 4 years ago

4/11/2008 11:52:49 AM

^^ I HAVENT SEEN THAT SHITTY MOVIE JESUS CHRIST

4/11/2008 11:54:31 AM

OMG, I UNDERSTAND SOMETHING FOR ONCE IN MY LIFE

GIVE ME A FUCKING PRIZE

4/11/2008 11:54:37 AM

You can't recreate the airplane on a treadmill phenomenon if you give all/accurate info and descriptions on the first page!!

4/11/2008 11:54:55 AM

how can i say this....




































THIS THREAD DOES NOT DELIVER!!!!

We talked about this stuff in CSC 200 (Go Dana!!!) and it was still boring back then

4/11/2008 11:57:15 AM

this is actually one of my favorite problems

but it's obvious that god is posting about it because of generated hype due to a movie

he's a geek tagalong

the worst kind

4/11/2008 11:59:01 AM

^^^well as I've already explained, he didn't give all/accurate info and descriptions on the first page.

I'd say it's more like, you can't recreate the airplane on a treadmill phenomenon if you use a lame example from day one of a statistics 101 course taken straight from the #1 movie for three weeks straight.

but yeah, this thread has like 10 different kinds of [fail] oozing out of every pore.

^heh, a geek tagalong that is apparently unaware of what exactly he's tagging along to.

4/11/2008 11:59:51 AM

neat-o simulator:
http://math.ucsd.edu/~crypto/Monty/monty.html



and dude, don't give away the answer on the first fucking post. otherwise we can't have any fun with this thread.

[Edited on April 11, 2008 at 12:28 PM. Reason : ]

4/11/2008 12:24:39 PM

^^^ I actually read about this years ago as well.

I just never made a thread about it until now.

So shut the fuck up.

4/11/2008 1:41:38 PM

you're funny

OMG STOP MAKING THE FUNS AT ME!!!!!!!!!!!!!!!!!!!

4/11/2008 1:45:39 PM

although this "theory" appears to be true, i will hopefully someday disprove it

but the better question is why do we always think its 50/50?
why are our instincts wrong in this case?

4/11/2008 3:24:43 PM

IIRC it will still be 50/50 if the host doesn't know which door has the car

4/11/2008 3:31:15 PM

^ STATISTICALLY SPEAKING..ITS NOT

you disappoint me nerdchick



and IRL..... those are the numbers behind the reasoning. Its not 50/50 and will never be 50/50 given the situation.

[Edited on April 11, 2008 at 3:35 PM. Reason : x]

4/11/2008 3:32:52 PM

The only thing I dislike about this little situation is that you have discredit your first instinct nearly every time.

[Edited on April 11, 2008 at 3:33 PM. Reason : or at least 2 out of 3 times, anyway.]

^I think she's right. But then again, if the host doesn't know, he might open the "car".

[Edited on April 11, 2008 at 3:34 PM. Reason : +]

4/11/2008 3:33:38 PM

I'm right, I double checked on wikipedia



http://en.wikipedia.org/wiki/Monty_hall_problem#Other_host_behaviors

4/11/2008 3:37:21 PM

^ She's not right and neither are you. Its not 50/50.


^^




the only way it would be 50/50 is if the door with the goat had been revealed before chosing the door. Statistically speaking, your chances are better if you stick with the door you went with. If you did an experiment like this, you would see that the numbers support the reasoning.




^ referencing Wikepia for a math problem just lost you any credibility




[Edited on April 11, 2008 at 3:39 PM. Reason : x]

[Edited on April 11, 2008 at 3:40 PM. Reason : x]

4/11/2008 3:37:35 PM

Congratulations! You're a winner!

Recap: You originally picked door 1 and then stayed with that door.
Here is a summary of how previous contestants have fared.
# of Players Winners Percent Winners
Switched 1020 700 68.6
Didn't Switch 584 196 33.6

[Edited on April 11, 2008 at 3:39 PM. Reason : http://math.ucsd.edu/~crypto/Monty/monty.html]

4/11/2008 3:39:22 PM

Quote :

"^ referencing Wikepia for a math problem just lost you any credibility "

yeah, it's not like God copy-pasted the entire thing from Wikipedia

4/11/2008 3:41:37 PM

Two players

In this variant, two players are each allowed to choose a different door. The game host eliminates a player who has chosen a door hiding a goat; if either player has chosen the car the other is eliminated, otherwise one of the players is eliminated at random. The host then opens the eliminated player's door, and offers the remaining player a chance to switch to the originally unchosen door. Should the remaining player switch?

The answer is no. Switching in this game leads to a win if and only if both players originally picked goats; the likelihood of this is only 1/3. By sticking with the original choice, the remaining player wins in the remaining 2/3 of the cases. So stickers will win twice as often as switchers.

There are three possible scenarios, all with probability 1/3:

* Player 1 picks the door with the car. The host must eliminate player 2. Switching loses.
* Player 2 picks the door with the car. The host must eliminate player 1. Switching loses.
* Neither player picks the car. The host eliminates one of the players randomly. Switching wins.

4/11/2008 3:42:49 PM

Quote :

"Someone saw 21"

4/11/2008 3:43:06 PM

^^^ I did the last two paragraphs.

Jesus christ. You people.

4/11/2008 3:53:01 PM

you mean "people" who don't like uppity little bitch idiots spewing obscenities at them?

4/11/2008 4:00:05 PM

EAT MY FUCK

4/11/2008 4:00:29 PM

i mean... seriously

4/11/2008 4:01:24 PM

haha, my friend lost $50 because of the Monty Hall Problem during a poker game

4/11/2008 4:12:14 PM

The statistics after I played...

# of Players Winners Percent Winners
Switched 1068 735 68.8
Didn't Switch 641 218 34.0

102.8%??

from the ncsu link.

[Edited on April 11, 2008 at 4:30 PM. Reason : .]

# of Players Winners Percent Winners
Switched 1076 742 69.0
Didn't Switch 650 223 34.3

[Edited on April 11, 2008 at 4:31 PM. Reason : 103.3%]

4/11/2008 4:29:07 PM

^ that's 102.8%.

i have a serious question for you, God;
why did you make this thread?

4/11/2008 4:31:09 PM

those percentiles are % won of the people who did that action

the numbers will converge to 66.6 and 33.3 with enough trials

[Edited on April 11, 2008 at 4:33 PM. Reason : .]

4/11/2008 4:32:51 PM

^^^That's just saying when they switched, they won 68.8% of the time, and when they didn't switch they only won 34% of the time. You don't add the numbers as they don't really mean anything together.

^yeah, that.

4/11/2008 4:33:22 PM

^^^ It would be interesting to people who hadn't heard about it yet.

I took a long time explaining it to my coworkers today.

[Edited on April 11, 2008 at 4:33 PM. Reason : FUCKING THREE ]

4/11/2008 4:33:25 PM

^^^That makes more sense. Thanks.

[Edited on April 11, 2008 at 4:33 PM. Reason : ^]

4/11/2008 4:33:35 PM

haha...my statistics teacher told us this a couple weeks ago

4/11/2008 4:37:46 PM

Probably after watching that 21 movie.

4/11/2008 4:38:45 PM

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4/11/2008 4:43:40 PM

The only problem with the movie (well, there are many problems with that movie) is this is a basic principle of statistics so why's kevin spacenuts talkin about it in a nonlinear algebra class?

4/11/2008 4:45:32 PM

i was suprised to see this in 21, i thought everyone knew about it, and if they are wiz kids then they should've known about it too


besides this and the airplane/treadmill problem what else is out there?

4/11/2008 4:51:29 PM