I am assuming we are using varient one. Unless varient 2 means that
all the time he will get an empty door, then they are both the same.
Anyways,
Why do you still have a 1/3 chance of getting the prize? You have 2
doors left, one with a prize. The third door had nothing behind it,
and should just be ignored. Maybe a picture might help?

Door1  Door2  Door3

You pick door2
Host opens door1
Door1 is empty

[door1]  Door2  Door3
You now have 2 doors to pick from: Door2(your first choice) or door3
This means your first choice has a 1/2 chance of being correct. 

Why does the first door still count in the chances? 

Peter Greenwood


--- In [email protected], "Stefan Pochmann"
<[EMAIL PROTECTED]> wrote:
>
> Shelley already mentioned it, but I'd like to point it out more 
> clearly. I'll give you a new variant of the riddle that points out an 
> important detail that can easily be misunderstood, leading to 
> different results.
> 
> -----
> 
> There are three doors, one hides a car. You pick a door. The host 
> opens another and there's nothing. You choose between the two closed 
> doors and get what's behind it. Should you stay or switch?
> 
> -----
> 
> Variant 1: The host knows what's behind the doors and when he opens 
> one, he deliberately picks an empty one.
> 
> Variant 2: The host does *not* know what's behind the doors and when 
> he randomly opens one, it just happens to be empty.
> 
> -----
> 
> Variant 1 is the original riddle with 2/3 probability getting the car 
> if you switch. Variant 2 is the new one I propose. What's your answer 
> for it?
> 
> Cheers!
> Stefan
>






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