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From: chongo@nsc.uucp (Landon Noll)
Newsgroups: net.math
Subject: i have the deal - generalized door answer - SPOILER
Message-ID: <465@nsc.uucp>
Date: Fri, 14-Oct-83 21:28:49 EDT
Article-I.D.: nsc.465
Posted: Fri Oct 14 21:28:49 1983
Date-Received: Thu, 20-Oct-83 03:04:13 EDT
Organization: Proper UNIX, San Leandro, CA
Lines: 46

to prove that you have a 2/3 chance to win by reselecting to a friend
i setup a 3 door experiment (using the Knuth 3 Vol. set as doors) where
i used a die to place a large ball behind one of the three vols.
"is it behind Sorting and Searching, or Semi-numerical Algor. ... ?"
repeated trials showed that the answer was near the value 2/3.

my friend then wrote a simulation prog. which also showed the answer
was near 2/3.  so from this we sat down and came up with the
generalised case:

the chance of winning by reselecting in an  N  door game is:

			N - 1
		       -------
			 2
			N -2*N

that is:  you select a door, Monty shows you a bad prize door and offers
to allow you to select a new door if you like.  you will win by
this reselection (n-1)/(n^2-2*n) of the time.

the value for n=3 yields:   2/3

here is how this value comes up:

you will  ONLY  win by reselecting  IF  your first choice is  BAD.  since
your first choice will be good 1/n of the time, your first choice
will be bad  1-(1/n) = (n-1)/n of the time.

when you reselect you will not select the first choice door and you will
not select the door which Monty showed you.  so out of n doors, you will
have n-2 to reselect from.  the chance of selecting a single door out of
n-2 is  1/(n-2).

now the prize being one of the n-2 doors only occurs (n-1)/n of the time.
i.e., when you didnt select the good prize in the first place.  since
the 1/(n-2) chance depends of the first (n-1)/n chance we multiply
them and get:  (n-1)/(n^2-2*n).

Note that for finite N: (N-1)/(N^2-2*N)  >  1/N

1/N is the chance you will win by keeping your first choice.

THUS:  FOR A FINITE NUMBER OF DOORS, IT IS ALWAYS BEST TO RESELECT!

chongo /\../\