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- From: warman@eagle.sangamon.edu (Stan Warman)
- Subject: Re: Monty Hall/Shell Game..
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- References: <BzqJx8.3Fv@unx.sas.com>
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- Organization: Sangamon State University
- Date: Thu, 24 Dec 1992 05:44:53 GMT
- Lines: 90
-
- 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
- Office Hours: M-F 8:30-5:00
- Phone Number: (78)6-6549
-