Ok, I know believe that Switching is the strategy.

Here is the reason :

1. You choose one door (p = 1/3)
2. An empty door is opened.
And here is the interesting fact :
If you don't switch : the fact that the other door is opened doesn't
change any thing and you still a p = 1/3.
Now, if you switch : there are 2 cases :
- If the car is indeed behind your door, the fact that the other is
opened doesn't change anything and you still have a p = 1/3
- If the car is not behind your door, this means that the other door
was opened because
there was no other possibility to open an empty door. So this means
the p now equals 2/3.

This is not very clear yet in my head, but I believe that Switching is
the dominant strategy.

2005/12/25, Pedro <[EMAIL PROTECTED]>:
> ??
>
>   Pedro
>
> "Chris Sz..." <[EMAIL PROTECTED]> escreveu:
>   In case you still don't believe us - try it for yourself!
> http://tinyurl.com/9rmy2
>
> --Chris
> Merry X-mas!
>
> --- In [email protected], Pedro <[EMAIL PROTECTED]>
> wrote:
> >
> > Yeah, most of you are right. The chance of winning is 2/3 if you
> switch the door. It looks strange at first, but it's not. See:
> >
> >   On your firts choice, you have 1/3 of chance of winning, and 2/3
> of losing. The host *just* can open an empty door. So, suppose you
> choose door A or 1. Each door has 1/3 of chance of containing the
> prize. So, your door has 1/3, and the other 2, together, have 2/3 of
> chance. So, if the host opens an empty door, the 2/3 probability is
> now on the other door (the one you didn't choose.) So, you'll win more
> times if you switch the door. Pay atention, you will *NOT ALWAYS WIN*.
> You'll just win more times than if you continue with your first choice.
> >
> >   Pedro
> >
>
>
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