From: warman@eagle.sangamon.edu (Stan Warman)
Subject: Re: Monty Hall/Shell Game..
X-Newsreader: TIN [version 1.1 PL6]
Organization: Sangamon State University
Date: Thu, 24 Dec 1992 05:44:53 GMT

Jerry M. Cox (sasjec@asimov.unx.sas.com) wrote:
> 
> Odds are that I could stand a review course in probability. But hear me 
> out anyway :-) If the contestant selects one door of three and that was 
> all there was to it then the odds of initially selecting the winning door 
> are one out of three. I have no problem with that, but I don't think 
> this describes the situation. If Monty is *always* going to reveal a 
> goat behind one of the doors, that door is removed from consideration. 
> The final choice then becomes a choice between one of two doors. If 
> Monty *always* removes one of the doors from consideration, then the 
> final choice (as well as the initial choice) is between a door with a 
> car behind it, and a door with the "other" (the one Monty doesn't show) 
> goat behind it. The odds of picking the car are 50/50 whether you switch 
> or not because Monty is *always* going to take away one of the losing 
> choices.
> 
> If he said "You can have the door you picked or the TWO doors that you 
> didn't pick", and didn't reveal anything about the two doors, then I 
> would agree that the best thing to do would be to switch. But, Monty 
> changes the odds when he shows you one of the goats behind one of the 
> two doors. He removes that door from consideration completely. Unless 
> you would like to take the goat Monty just showed you, you are left with 
> a one out of two choice. He equivalently says to you: "You can have the 
> car or the goat that I didn't show you", or "You can have the door you 
> initially chose or the door I didn't show you". That sounds like 50/50 
> odds to me.
> 
> To sum up the initial choice is not a one of three choice between 
> a car or goat #1 or goat #2. The initial (and final) choice is between a 
> car and one of the 2 goats. The other goat will be removed as a choice 
> by Monty. 

Cliff Heller writes:
>Monty equivalently says:  "You can have the door you chose, or BOTH doors
>that you did not chose" without opening anything.  Now would you switch?
>It is the same problem.

Of all the explanations I have read, this one (IMHO) puts it in the most
understandable way. It shows the importance of opening one of the losing

doors.

BTW, back in early '91, in Parade Magazine, Marilyn Vos Savant published
this puzzle in her column. When she stated that the odds of winning by
switching were 2 out of 3 she was swamped with letters(mine included I'm
sorry to say :-) taking her to task. Many of these letters were from
Mathematicians! When she defended her answer in a second column, I
decided to write a computer program to test her assertion. For those
who may be interested, the BASIC code follows:

------------------------------------------------------------------------
0 'MAKEDEAL.BAS STAN WARMAN 02-17-91
1 KEY OFF:CLS:PRINT"You are on Let's Make A Deal and you choose
 one of three doors trying to win a":PRINT"car (and avoid a goat).
  The host opens a door revealing a goat and asks if you"
2 PRINT"wish to switch to the remaining door. Marilyn Vos Savant
 of Parade Magazine":PRINT"says your chances are 2 out of 3 that
  switching will win you the car.":PRINT"Do you believe her or
   that the ratio is 1 out of 2? Answer the following"
3 PRINT"questions to test your belief in mathmatical logic."
:PRINT:PRINT
5 INPUT"How many trials";TRIALS:INPUT"Always keep 1st choice(K)
 or always switch(S)";CHOICE$:IF CHOICE$="K" OR CHOICE$="k" OR
  CHOICE$="S" OR CHOICE$="s" THEN 10 ELSE CLS:PRINT"Invalid
   response...Please try again":PRINT:GOTO 5
10 LOCATE 12,1:PRINT"Hits=";:RANDOMIZE TIMER:HITS=0:FOR I=
1 TO TRIALS:CHOICE1=INT(RND(1)2+1):PRIZE=INT(RND(1)3+1)
20 OPENDOOR=INT(RND(1)3+1):IF OPENDOOR=CHOICE1 OR OPENDOOR=PRIZE
 THEN 20
25 IF CHOICE$="K" OR CHOICE$="k" THEN CHOICE2=CHOICE1:GOTO 90
30 IF CHOICE1=1 AND OPENDOOR=2 THEN CHOICE2=3
40 IF CHOICE1=1 AND OPENDOOR=3 THEN CHOICE2=2
50 IF CHOICE1=2 AND OPENDOOR=1 THEN CHOICE2=3
60 IF CHOICE1=2 AND OPENDOOR=3 THEN CHOICE2=1
70 IF CHOICE1=3 AND OPENDOOR=1 THEN CHOICE2=2
80 IF CHOICE1=3 AND OPENDOOR=2 THEN CHOICE2=1
90 IF CHOICE2=PRIZE THEN HITS=HITS+1:LOCATE 12,6:PRINT HITS;
100 NEXT
110 LOCATE 12,12:PRINT"Out of";TRIALS;"trials.":PRINT HITS/TRIALS*100
;"%  Hit Rate.":INPUT"Play again";ANS$:IF LEFT$(ANS$,1)="Y" OR
 LEFT$(ANS$,1)="y" THEN 1 ELSE KEY ON:END
------------------------------------------------------------------------

Naturally, the more trials that are run, the closer to 67% you will
come(or to 33% if you always keep your initial choice).
This is truly a fascinating problem! N'est-ce Pas?

      Stan Warman: Microcomputer Specialist
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