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Path: utzoo!watmath!watcgl!dmmartindale
From: dmmartindale@watcgl.UUCP (Dave Martindale)
Newsgroups: net.math
Subject: Re: Let's Make a Deal
Message-ID: <932@watcgl.UUCP>
Date: Thu, 6-Oct-83 15:42:08 EDT
Article-I.D.: watcgl.932
Posted: Thu Oct  6 15:42:08 1983
Date-Received: Fri, 7-Oct-83 10:57:41 EDT
References: <165@mi-cec.UUCP>
Organization: U of Waterloo, Ontario
Lines: 23

I disagree with Dan Klein.  His argument is essentially that knowing which
door is bad leaves the remaining doors with equal probability of being good.
Another way of thinking of this is that the door he chooses gives you no
new information, it just reduces the number of possible choices and the
remaining ones are equally likely.  But this isn't true.  You will pick the
good door 1/3 of the time; in that case Monty is showing you one of the two
bad doors but since both are bad his choice contains no useful information.
In this case, switching doors is guaranteed to give you a bad door.  But with
probability 2/3, you pick a bad door, and Monty shows you the other bad
door.  Here, he is TELLING YOU WHICH DOOR IS GOOD, which is useful
information, and switching doors is guaranteed to give you a good door.
Thus the strategy of switching doors gives you the good door 2/3 of the time,
while the strategy of keeping your original choice only gives you the good
door 1/3 of the time.

Another way of looking at it is that Monty is NOT picking one of the bad
doors at random, which is an assumption built into your argument.
He is picking at random 1/3 of the time, but 2/3 of the time he is picking
the ONLY remaining bad door.

	Dave Martindale
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