Tony asks:

> Ok, you agree that door 2 has a 1 in 3 chance of being the right door, so
>  let us look at this closer.
>  Door 1 (my pick) has a 1 in 3 chance of winning, Door 2 (my buddy's pick)
>  has the same 1 in 3 chance and Door 3  has a 1 in 3 chance.
>
>  Lets say we eliminate 3 by opening it and finding it is a dud, my buddy,
who
>  picked 2, does not all of a sudden have twice as much chance as me get the
>  right one, as there were 2 out of 3 chances he would fail,  the same as
>  there were 2 out of 3 chance that I would fail.
>
>  The fallacy that creeps into everyone's thinking at this point is that
>  because I had a 2/3 chance against me, I would have a 2/3 chance of winning
>  if I gave up my choice and jumped over to the 2/3 chance side.
>
>  The problem with that is , Wouldn't my buddy also be able to get the same
>  2/3 chance if he switched to my side.??   Removing door 3 did not increase
>  my odds to 2/3 nor his.  What it did do is change the rules of the game to
>  give us both 1 out of 2 chance.  This is not a math problem, it is a logic
>  problem.

As I wrote several days ago, never let anyone tell you that properly
describing the event state space in probability problems is easy.

In this case, Tony subtly redefines the problem by having two people choose
two doors -- and then having Monty open the unchosen door. That isn't the
original problem. In the original problem, an intelligent agent (called a
"daemon" in physics) is in operation. In this case the agent/daemon is Monty
Hall, and he has god-like pre-knowledge of the situation.

With just one contestant, Monty has two doors to chose from -- and will
always have at least one door *without* a prize behind it, and that's the
door he'll show you. If he happens to have two doors without prizes, because
you have chosen the winning door, it doesn't matter which door he shows you.
Nonetheless, the odds remain the same. If you hold your original choice, you
have a 1/3th chance of winning. If you choose Monty's unopened door, you have
a 2/3th's chance of winning.

What happens when you have two contestants (you and your buddy), you
completely eliminate Monty's freedom to always select a non-winning door. He
is no longer a daemon; rather he has now become just a cog in the mechanism.
The unchosen door has a 1/3th chance of being the winning door. If that
proves to be true, you both lose. But if it's not, you and your buddy each
have a 1/2th chance of being the winner, just as you say.

But the problem is, you're playing a different game than was originally
described.

Wirt Atmar