After much thinking, I finally understand, the chances are better to
switch doors. 

Yoi have a 2/3 chance of not picking the car as your first choice.
Lets assume you don't pick the car, you pick an empty door. The host
only has 1 choice of doors to open. Now since you picked an empty
door, you would want to change! Now since you had a 2/3 chance of
being wrong at first, 2/3 times you would want to change.

Now if you did pick the car, the host would have 2 choices of doors,
and you would lose if you switched. But getting the car first shot is
1/3 chance, not much compared to 2/3 of being wrong. This shows the
chances are better to switch.

I hope this clears up other people's confusion (it sure helped me!)

Peter Greenwood


--- In [email protected], "Stefan Pochmann"
<[EMAIL PROTECTED]> wrote:
>
> --- In [email protected], pjgat09 <[EMAIL PROTECTED]> 
> wrote:
> >
> > This really is not making sense. How is opening an empty door
> > different than just having 2 from the start? In both cases you have
> > only 2 doors to choose from, one with the prize. Am I missing 
> something?
> > 
> > Peter Greenwood
> 
> You have not done the 20 experiment games I suggested, have you? Let 
> me also suggest you play the host while another family member playing 
> the player. And make sure you do 10 times one strategy and then 10 
> times the other strategy. I bet you'll notice something.
> 
> I just saw a hilarious variant: Suppose you are playing a seven door 
> version of the game. You choose three doors. Monty now opens three of 
> the remaining doors to show you that there is no prize behind it. He 
> then says, "Would you like to stick with the three doors you have 
> chosen, or would you prefer to swap them for the one other door I have 
> not opened?" What do you do?
> 
> Cheers!
> Stefan
>






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