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Quote:

Originally Posted by Slappin Fish (Post 19393337)

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.

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.

Quote:

Originally Posted by Dirty F (Post 19393341)

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

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

Quote:

Originally Posted by Slappin Fish (Post 19393343)

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

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

Quote:

Originally Posted by Dirty F (Post 19393577)

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.

Quote:

Originally Posted by woj (Post 19393782)

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, it does not.

Quote:

Originally Posted by Dirty F (Post 19393276)

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

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.

Quote:

Originally Posted by mikesouth (Post 19394039)

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.