Always switch when offered the option
 

There are 3 doors. Behind 1 one of them is a brand new car.
You can pick 1 door.
You randomly choose door 1. The quizmaster then opens door number 3 to show you the car is not behind that door.
He then asks you if you still will go for door 1 or if you want to switch to door 2.
Always switch to the other door because your chances will go from 1 in 3 to 2 in 3 instantly :thumbsup

need to smoke some joints to understand this

Monty Hall paradox.

Fascinating. You're like a savant or something. :)

I'm with the Morgan et al. 91 train of thought on this one. If the quizmaster opens door number three with nothing behind it you've already made a right choice, so your chances aren't 1/3 or 2/3 (those were your chance before any door was open) your chances are now 1/2.

If when all doors are closed the quizmaster says i'll open a door if you switch. then by switching your odds go to 2/3 but If you already picked door 1 and the host already opened door 3 odds are 1/2 .

Isn't there a movie in which this thing is explained? Can't remember which one...

Yes. That movie is called, "21".

:2 cents:

I don't get it - but I have heard it before.

In effect you now have 2 doors and one has the prize - It has to be 50/50 - Forget about what has happened before.

I'm still too drunk to even think about this. Will come back in 5hrs...

Then you will have a hang over (like me) and it will be even worse - bookmark it and come back tomorrow.

this is 50/50 case as EddyTheDog said.

because you know the 3rd is empty.



imho the car is behind 1st door.
the guy is trying to distract you.
i saw this type of disctracting in some indian movie

slumdog millionaire ..

no, you are assuming that they don't want to give away a car. game shows want people to win, not every time, or there'd be no suspense but without people winning big prizes and breaking down in tears there's no reason to watch a game show.

I thought the 'game show' was just a metaphor for a maths question - If the host in the game show knows where the car is then that changes everything.

Ah, of course, thanks.

I must've missed that part as i left the theather because the movie sucked so amazingly fucking bad!

No, you are wrong.

Switching results in a win 2/3 of the time. :)

If you have trouble understanding why it's not a 50/50 chance, replace "3 doors" with "10 doors" making the quizmaster open 8 doors. That should make it more clear.

hmmm, you have point.

ok, btw which car is it ? :winkwink:

A Fiat 500.

that still sounds like it turns into a 50/50 chance

No, obviously it doesn't.
The chance of picking the right door by switching is 9/10 in this case.

In that case I would ask if I could pick door #3.

but those 8 doors are eliminated after he already picked so that would, in turn, create the new odds after that would it not? Overall I see your 9/10 obviously yes.

I just Googled it and it and it is a trick question - The host does know where the car is and is trying to stop you winning the car.

That obviously changes everything.

explain why.

1. you pick a door, 1/3rd of the time it's a winner... 2/3rds it's not...
2a. IF the chosen door is a winner (1/3rd chance) and you switch = you lose
2b. IF the chosen door is not a winner (2/3rd chance) and you switch = you win

so by using the "switching" strategy, you win 2/3rds of the time...

The key to calculate the probability is continuance. Continuance in this case meaning that the winning door is decided before the game begins and is not changed during the game. If the winning door could be changed during the game, your chance of winning would be completely independent of probability and entirely up to the gamehost.

Due to the continuance, your first choice is actually completely irrelevant. The gamehost could easily skip it and just ask you: Would you like to open all but one door looking for the prize or would you like to open just one door?

Lack of continuance is - for instance - the reason that the silly roulette trick where you keep betting on one color and double your bet if you lose, does NOT work in the long run. One spin doesn't affect the other and hitting red 10 times in a row, won't increase the chance of black on the 11th spin. THAT is still 50/50.

He ofcourse knows that. He also knows that 9 out of 10 people won't switch. Resulting in way less cars to give away.

Best explanation I've ever seen of this "Monty Hall paradox".

Yeah, it's different :)

No, in this case they want to give you the idea that the chances of winning are equal but in reality they aren't. But because of the free choice most people won't realize this.

1+1=3


frankkkkkkkkkkkkkk

wtf - it makes no sense. there are 3 doors - one's been opened and has no car behind it.
now you have two doors left, you having picked one already is meaningless, each door has the same chance of having the car behind it so switching your choice from your first arbitrary one shouldn't increase your odds at all.

obviously i'm wrong but i still don't see it.

Yes, it doesn't make sense when you look at it that way, which is the logical way to look at it. It has been Mythbusted though and did hold true.

edit: here is another good explanation

I don't choosing the worst so ...

Always pick the one that you think it won't be, because Murphy's Law indicates that it will the the right one.

this is only true if the announcer does NOT know which door the car is behind

this guy in yt video is making it more complicated than it already is,

he says if you switch you have 66% chance of winning a car as he already knows that there is no car behind his 1st choice of door.

btw there are funny and good comments on this video on youtube .

I have yet to see a single Mythbusters that would actually hold up to a real scientific anaylasis, every experiment they have ever done has been full of holes.

The video you posted was good though.

Huh, no obviously he must know it or he might open the door with the car behind it.

1. What is the probability of winning the car by always switching?
2. What is the probability of winning the car given the player has picked door 1 and the host has opened door 3?

The answer to the first question is 2/3, as is correctly shown by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1/2. This is because Monty's preference for rightmost doors means he opens door 3 if the car is behind door 1 (which it is originally with probability 1/3) or if the car is behind door 2 (also originally with probability 1/3). For this variation, the two questions yield different answers. However as long as the initial probability the car is behind each door is 1/3, it is never to the contestant's disadvantage to switch, as the conditional probability of winning by switching is always at least 1/2. (Morgan et al. 1991)

There is disagreement regarding whether vos Savant's formulation of the problem, as presented is asking the first or second question.

source: wikipedia


you are answering question number one so yes probability is 2/3 , but the way Frank worded it with door 1 already picked and door 3 opened it's question number two. 1/2

http://www.bbc.co.uk/learningzone/cl...ned/11261.html Video explanation that's easy to understand :thumbsup

No, because there is a 2 in 3 chance it's behind door 2 or 3. So by picking the door that is left from those 2 (he eliminated one of them) you still have a 2 in 3 chance.

^--- notice how i explained it in 2 sentences? I could've done it in 1 actually.

:)

Your title is right but probability of winning by always switching is a distinct concept from the probability of winning by switching given the player has already picked door 1 and the host has already opened door 3. The title and the example you gave are two slightly distinct propositions. If you announce before a door has been opened you plan to switch you'll get 2/3 not if you already made your pick and host already opened door 3. Anyway yeah better to just always switch.