🚀 go-pugleaf

I can't believe we are going through this again, but I suppose it is
somewhat educational.

In <X382VB1w164w@perseus.spk.wa.us> don@perseus.spk.wa.us (Donald Byrne) writes:
>     He was asked to change only after the elimation of one of the other 
>doors. Now we know that there are two non-win doors in the game, and the 
>object of Monty Hall is to elimanite one by the rules. If our player has 
>any smarts he knows that he will either have a winner or a looser, and 
>that Monty Hall will asure him of that. The reason is that of the three 
>doors he will only really deal with two. Monty Hall will elimaniate the 
>other. So unless I'm not seeing the game properly it seems to me the 
>final results should be 50%.
>     Now explain why all of you say 66% of a chanch if he changes to the 
>other one of the remaining two??

Think about it this way:  If the person guessed correctly initially,
then switching would not benefit him.  Since he selected one door from
three, the odds of having a correct initial guess are 1/3.  There should be
no dispute on this point or a review course in probability is necessary.

Now, this isn't quite as easy to see, but if the person guessed
INCORRECTLY initially, then he automatically wins by switching.  Why?
Because Monty always reveals a non-winning door.  He equivalently says to
you: you can have the door that you picked, or the TWO doors that you
didn't pick.  Since He always reveals a goat, if the initial guess was
incorrect, then the remaining door contains the prize.  The odds of the
initial guess being incorrect are 2/3 by the same reasoning as above.

If you want an empirical demonstration, we can set up an experiment and I
will win all of your money without cheating.
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