A three-door game.

You are on a game show, and you are given a choice of three doors.  Behind exactly one of the doors, is a brand new car.  There is nothing behind the other two doors.  The game works according to certain rules.  First, you make a choice:  you are asked to PICK a door, without having it opened.  Second, soemthing is revealed to you:  the game show host opens one of the two doors which does NOT hide the car.  Third, you are allowed to rethink your choice.  You can stay with your original choice (if the host has not opened that door), or you can switch to one of the unopened doors.  Once you've made your final choice, the door you last picked is opened-and you either win a car or walk away with nothing.  The question is this: suppose that you were playing this game (or an equivalent simulation) over and over again.  Is any advantage gained by always picking any one door initially, and always switching your choice in the second phase?  Or do your odds remain the same if you always stay with whatever your first choice was?

Just to be absolutely clear:

The car, or prize, is always behind exactly one door throughout any one game, and does not 'shift' during a game.
The host KNOWS which door hides the car, and that the other doors are empty.  The host always opens a door with nothing behind it (obviously).  The host may open the door that you initially selected if the car is not behind it.

Phases of the game:

Selection:  you pick one of the three doors to be opened, without having it opened.

Revelation:  Exactly one of the doors not hiding the car is opened.

Re-selection:  You can switch to one of the unopened doors, or stay with your original choice if that door was not opened.  Your new choice here is the door that is opened.

Well there is a chance that the host will pick the selected door as the empty one. Therefore always switching, as opposed to always not switching, will be the better option.

(Isn't this supposed to have some weird complicated solution or something?)

Yay, the Monty Hall problem, one of the most trolled and beat-to-death math topics right after 0.9999...==1

Both logic and empirical studies prove that switching gives you better chances.

(in the original Monty Hall problem, the host will reveal one of the doors you HAVEN'T picked to be empty - is the OP different on purpose?)

My maths teacher used to give us this. It's pretty obvious unless you're dumb enough to believe that "it's either here or there, therefore 50/50"

You pick a door at random. There is a 1/3 chance that you have chosen the correct door. The host removes one of the doors from the equation, leaving your door with its 1/3 chance and the one remaining door which has a 2/3 chance of being the right one.

i'm dumb enough to believe that "it's either here or there, therefore 50/50" orz

Me and my high school friends used that logic with the lottery. We either win or don't win, so it's clearly a 50% chance. So to make sure we will win, we bought TWO TICKETS, because, hey, 50% + 50% = 100%.

We didn't win anything. OBVIOUSLY THE LOTTERY WAS RIGGED.

>>4

I'm phrasing the problem slightly differently as part of my effort to check out the whole thing.  Another analogue for the problem would be a situation where there are n doors, for natural n greater than or equal to three, and again, you pick a door without its being opened, and all but two doors in the entire set are opened.  Whether you stipulate that the door you initially pick cannot be opened in phase two or not is another side to it.

>>7
Damn politicians!

well, I was looking at a solution..

3 positions for the car x 3 choices for an opened door x 3 possible initial choices x 3 possible revised choices gives 81 'conceivabilities'.  (Don't correct me yet, I'm about to do it myself).  Clearly, we never open the door that the car is behind-this eliminates 27 'situations' right out.
3x2x3x3 instead.  If you make yourself a shorthand for possible games, and push everything through
step by step, you find that there's a class of arrangements where the final choice is the door that
was opened-which never reveals the car, since all those situations are also against the rules, and
were struck before.  There are 18 situations of this type, after removing the first illegal 27. 
81-27-18=36 legal, distinct, possible games.  Now, if you check your remaining 'games'
carefully, I claim you'll find that there's twelve games where the door you picked is the one that
the host opens, forcing you to switch in any event.  We can either check the game including these,
as they are allowed in my original version of the rules above, or we can omit them to check a
version of the game where opening the contestant's first choice is NOT allowed (24 such games). 
I will use a number shorthand to represent the 36 games in question.  the four numbers, in order,
are: the door the prize is behind, the door that is opened, the first choice, and the revised choice. 
Clearly, the first and second numbers cannot be equal, and the second and fourth numbers cannot be equal. 
The third and fourth numbers being the same represents staying with your original choice, and their
being different represents switching your choice.  The first and fourth number being the same
represents a won game, and their being different represents a loss.  The next post will have a scheme for
the 36 games.  W and L mean win and lose, C means 'switch' (3rd and 4th numbers are different), Y means
'stay' (same).  F means 'forced', meaning that the door you picked was opened, implying a switch for the
game to be legal.  The upshot: 6 games win and 6 games lose when you stay with the initial choice, and 12
games win and 12 games lose when you switch your choice (even if the 'switch' is forced).  To check the
original version of the problem, simply omit the 12 numbers with Fs: in this case, there are exactly six of
each type of possible game: win-stay, lose-stay, win-switch, and lose-switch.  Both versions of the game
seem on the surface to be 50/50 because of even distribution of scenarios, but I was under the impression
that the latter (original) version of the game was illegal precisely because you lose more when you never
switch.  So, I'm working on the assumption that my whole approach of bruteforcing scenarios is wrong. 
Someone please help me grasp this (and find errors in calculation if possible).  One other thing:  enen
though there's 36-24  possible games, even if you actively limit those games to 24 or 12 types by always
switching, odds should still be even as all versions of the game come up.... yes?  no?  maybe?

1211        W    Y   
1213    L    C
1221    W    C    F
1223    L    C    F
1231    W    C
1233    L    Y
1311        W    Y
1312    L    C
1321    W    C
1322    L    Y
1331    W    C    F
1332    L    C    F
2112        W    C    F
2113        L    C    F
2122    W    Y
2123    L    C
2132    W    C
2133    L    Y

2311        L    Y
2312    W    C
2321    L    C
2322    W    Y
2331    L    C    F
2332    W    C    F
3112        L    C    F
3113        W    C    F
3122    L    Y
3123    W    C
3132    L    C
3133    W    Y
3211        L    Y
3213    W    C
3221    L    C    F
3223    W    C    F
3231    L    C
3233    W    Y

and yes, I've been reading the wiki on it, so don't direct me to the internet or anything.  -.-

I understand that the 'host' is not a free agent.  But there's other stuff I'm fuzzy on.

The OP is different than the standard Monty Haul problem.  If the host can open the door you selected, I'm not sure about the chances and the advice of switching.  I'm going to think about it on my own.  Gotta keep that brain in shape over the summer.

I think the change in the OP doesn't change anything except making the game harder on the contestant.  There's a 1/3 the host will open your door, and if he does, you're down to a 1/2 chance of winning from the 2/3 from the standard problem.

bump lol