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on ur first try you ahve 1/3 chance to get it right. On the second try you have a 1/2 chance. So by not switching you have given urself only a 33% chance of getting it correct.
Incorrect....
There's no point to switching.
If he picks door 1 and reveals door 3 all that happens is either Door 3 is the prize and it doesnt matter if he switches or door 3 is a chicken and he's 50/50 to pick the right door (and his initial pick is just as viable)
Unless he can switch to door 3 after revealing it, there's no point.
Incorrect..
I'll explain it blow, scroll down if your ready for the answer..
With your first choice, you have (A) (B) (C) .. 1/3 chance that it's behind each door...
(A) = 1/3
(B) = 1/3
(C) = 1/3
So you have a 1/3 chance it's the door you pick, and a 2/3 chance that it's behind one of the other two doors.
(A) = 1/3
(B) or (C) = 2/3
You pick (A)
The game show host opens one of the remaining doors to show (and this is the key) WHICH of the 2 remaining doors (with a 2/3 chance it's behind one of them) is the incorrect one.
(A) = 1/2
(B) or (C) = 2/3
Shown it's not (B) leaves the 2/3 chance behind door (C)
Another way to look at it is this:
If you always change, you can only lose if you start with the winning door (and we all agree you have a 1/3 chance of starting with the loseing door)
for example:
you pick door (A) and it's behind door (A) then the host opens door (B) or door (C) (either one) and you change to the other, you lose.
You pick door (B) and its behind door (A), he opens (C), you change to (A), you win
You pick door(C) and its behind door (A), he opens (B), you change to (A), you win.
I wrote a program that actually runs the problem if you want to see:
http://www.startingdreams.com/doors.php
If you dont believe that, take 3 cups, put a coin in one, have a friend mix them up, you pick one, have him show you an empty cup from the remaining two, and then write down how often you win if you stay, and how often you win if you change...
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