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.

Still then. Because there is still a 2/3 chance it's behind one of the 2 other doors. No matter what.

Edit: i now understand you. You mean he already opened the door. That doesn't make sense. I mean that's a completely different situation.
Then there simply are 2 doors. Not 3.

Well no because he ruled out a door. 1 pick left. 2 doors left. 1/2

Edit : saw your edit yeah that's what I mean. Your title is correct but this example for some reason often used is actually a slightly different question

Then you know the other door that you didn't pick originally has a 2 in 3 chance of having a car behind it.

Slappin Fish is right, the solution changes if by "door #1" and "door #3" you mean a specific door, not just a label to illustrate the problem... :2 cents:

no he isn't - watch the video link I posted a couple posts back.

No, it does not.

thats the point if he always knows and always pics one of the two doors that are NOT the right one the odds if you change are 2/3

if he doesnt know and randomly pics a door (and 33% of the time it has the car) then the odds are also 66%

there are variations

lets say he doesnt always off the choice, since he knows where the car is he can manipulate things....say you guess correctly and that is the only time he offers the choice or he only offers it if you guess wrong.

but those strategies would be easy to figure out

NOW it can be shown that if he always opens a door when the contestant is right and half the time when he's wrong--a perfectly rational approach--over the long haul the odds of the prize being behind Door #1 versus Door #2 are 50-50.

http://www.bbc.co.uk/learningzone/cl...ned/11261.html

If you pick a door, before there are only 2 doors left, there is never a 50/50 chance at any point. Forget the fact there are only 2 doors LEFT, the fact is you are picking when there are more than 2 doors, so the odds are never 50/50, regardless if you whittle unopened doors from 3 (or a hundred, to grasp it) doors to 2 doors.

I posted the clip again because it's obvious no-one watched it.