Okay, I will now step in on Tim's behalf. The answer to your question is the
third person will have no chance. The eliminated cards are known to not be the
Ace. We are starting to go away from the point Tim was trying to make (BTW it
was an excellent point Tim). Monty Hall is eliminating a door that he knows FOR
SURE is not the prize. When you pick 1 out of 3, you have 1/3 of a chance to be
correct. The other two doors represent a 2/3's chance. Just because Monty
knowingly eliminates one door that is not the prize, that doesn't improve your
chance. The other door now represents both doors with the same 2 out of 3
chance.

Hope this finally helps,

Garo K. Akcelik
System Administrator
Regence BlueCross BlueShield of Oregon / Regence HMO Oregon
==========================================================
The opinions expressed, here, are mine and mine alone.

---------------------- Forwarded by Garo K Akcelik/BCBSO/TBG on 03/02/2000 03:39
PM ---------------------------

From: Phil Anthony <[log in to unmask]> on 03/02/2000 03:09 PM

Please respond to Phil Anthony <[log in to unmask]>


To:   [log in to unmask]
cc:    (bcc: Garo K Akcelik/BCBSO/TBG)

Subject:  Re: OT: Probability question - Wanna Play?




Tim wrote:

>Wrong.  The odds have not changed!

>With my original card I had a 1 in 52 chance and you had a 51
>in 52 chance - those odds don't change just because you've
>turned over 50 of the cards you have!

--------------------------------

Okay, I swore I wasn't gonna get into this thread, but here I am (sorry,
James).  Tim, sure looks to me like the odds changed to 50/50.  Let me ask
you this:  If a third person came along and picked one of the cards that we
eliminated, would they also have a 1 in 52 chance?

I'm so confused . . .