Monty on the Front Page

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Dave Alexander

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Jul 21, 1991, 11:58:18 PM7/21/91
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The Monty Hall Problem made page one of today's (21 July)
New York Times. Apparently, the Smartest Person in the Known
Universe(tm) ignited a firestorm by bringing up the MHP in her
column in Parade magazine. The casualties were staggering,
with the usual crop of math professors lamenting the *shameful*
innnumeracy of the American people and taking TSPitKU to task
because she *obviously* didn't think the problem through before
post^H^H^H^Hsubmitting it to Parade. [choke snort guffaw]

So are we doomed? Is this how it will end? Will we finally come
to Armageddon in a particularly acrimonious argument over whether
that beautiful new car is more likely to be behind Door #1 or
Door #2? Or whether Deckard was a replicant? Will alien
archaeologists someday come to dig our remains out from under a
pile of postcards and speculate about the devotions that we paid
to a god named "Craig Shergold"?

Or will they understand?


-- Dave Alexander

--
"But 2 weeks of blue balls? I can hardly fault the guy."
-- Kevin Darcy

Dave Seaman

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Jul 22, 1991, 3:10:12 AM7/22/91
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In article <1991Jul22.0...@dartvax.dartmouth.edu> da...@eleazar.dartmouth.edu (Dave Alexander) writes:
>
>The Monty Hall Problem made page one of today's (21 July)
>New York Times. Apparently, the Smartest Person in the Known
>Universe(tm) ignited a firestorm by bringing up the MHP in her
>column in Parade magazine.

The column is "Ask Marilyn" by Marilyn vos Savant, and the episode was reported
in the "Skeptical Inquirer" (Summer, 1991, pp. 342-345) in an article entitled
"Nation's Mathematicians Guilty of 'Innumeracy'". She did not "bring up" the
problem. It was submitted by a reader.

--
Dave Seaman
a...@seaman.cc.purdue.edu

Theodore A. Kaldis

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Jul 22, 1991, 10:02:26 AM7/22/91
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> The Monty Hall Problem made page one of today's (21 July) New York
> Times.

Did it, now?

> Apparently, the Smartest Person in the Known Universe(tm) ignited a
> firestorm by bringing up the MHP in her column in Parade magazine.
> The casualties were staggering, with the usual crop of math professors
> lamenting the *shameful* innnumeracy of the American people and taking
> TSPitKU to task because she *obviously* didn't think the problem

> through before submitting it to Parade. [choke snort guffaw]

Well, I seem to recall the column in question (it was perhaps as much
as a year or more ago). The "math professors" in question would do
well to resign their positions, because as I recall "the Smartest
Person in the Known Universe" (whose name escapes me now) had it
straight.

To review (and perhaps to reignite the firestorm), the question
concerns a feature on the old game show, _Let's Make a Deal_. The
contestant would pick one of three doors, at which point the MC (Monty
Hall) would show what was behind one of the unpicked doors -- and then
give the contestant the opportunity to switch doors. The question is,
does the contestant have a greater probability of winning by switching
or by not switching -- or is the probablitity the same in either case?
It seems that many -- particularly those unschooled in statistics and
probability -- intuitively believe that the probability is the same
either way.

To demonstrate the solution of this problem, I have prepared some
tables. In the first table, I have listed every possible combination
of the three events: which door the big prize is behind, which door
the contestant picks, and which door the MC shows. Note, however,
that certain combinations will never occur. The MC will never open
the door with the big prize, nor will he reveal what's behind the door
that was picked. The outcome charted here is the result of one
switching doors when given the opportunity to after the contents of
one of the unpicked doors is revealed.

Prize
is Door MC
behind: picked: shows: outcome:

1 1 1 ---
1 1 2 LOSE
1 1 3 LOSE
1 2 1 ---
1 2 2 ---
1 2 3 WIN!
1 3 1 ---
1 3 2 WIN!
1 3 3 ---
2 1 1 ---
2 1 2 ---
2 1 3 WIN!
2 2 1 LOSE
2 2 2 ---
2 2 3 LOSE
2 3 1 WIN!
2 3 2 ---
2 3 3 ---
3 1 1 ---
3 1 2 WIN!
3 1 3 ---
3 2 1 WIN!
3 2 2 ---
3 2 3 ---
3 3 1 LOSE
3 3 2 LOSE
3 3 3 ---

This table is pretty long, and (unless you have a very long window
open) it won't fit on one screen, so lets knock out all the
combinations which will never occur.


Prize
is Door MC
behind: picked: shows: outcome:

1 1 2 LOSE
1 1 3 LOSE
1 2 3 WIN!
1 3 2 WIN!
2 1 3 WIN!
2 2 1 LOSE
2 2 3 LOSE
2 3 1 WIN!
3 1 2 WIN!
3 2 1 WIN!
3 3 1 LOSE
3 3 2 LOSE

Okay. Now this table seems to indicate that, regardless of whether
one switches or not, the odds of winning are the same. But wait!
There's a little problem here! Let's say the contestant picked a
losing door. Then the MC will always show the other losing door. But
when the winning door is picked, in half the cases the MC will show
one of the losing doors, and in the other half he will show the other!
So let's rearrange the table somewhat to reflect this anamoly.


Prize
is Door MC
behind: picked: shows: outcome:

1 1 2 or 3 LOSE
1 2 3 WIN!
1 3 2 WIN!
2 1 3 WIN!
2 2 1 or 3 LOSE
2 3 1 WIN!
3 1 2 WIN!
3 2 1 WIN!
3 3 1 or 2 LOSE

Voila! Here it is, and this should put the argument to rest and send
all those errant ersatz "math professors" packing for the hills. If
the contestant switches doors, he stands a 2-1 advantage of winning.
This is exactly the answer published in Parade, and the "Smartest
Person in the Known Universe" stands vindicated by none other than
Theodore A. Kaldis.

[I should expect the folks in sci.math and rec.puzzles to know this.
Therefore followups are directed to alt.flame.]
--
Theodore A. Kaldis
{...}!rutgers!remus.rutgers.edu!kaldis
kal...@remus.rutgers.edu

Michael Greenwald

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Jul 22, 1991, 1:34:18 PM7/22/91
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>> The Monty Hall Problem made page one of today's (21 July) New York
>> Times.

There were two or three pleasant surprises in/about this article.
First, they "got it right", basically. It was actually a coherent
explanation aimed at the general readership. Second, they interviewed
Monty Hall about the problem, and he apparently understood the
problem, and the "correct" answer in theory. He didn't seem as
brainless as I expected (the problem, of course, was in my
expectations).

He introduced another variant to the problem (how much would it take
to "bribe" you >not< to choose another door: i.e. how much would
"Monty Hall" have to offer you before the immediate payoff (if you
choose to accept it) was more than your expected value from choosing
another door.)

Finally, Monty Hall also stated that the problem wasn't an accurate
reflection of "Monty Hall"'s power in the show: the assumption that
"Monty Hall" behaves uniformly no matter where the goats and cars lie
is a bad one. There was no requirement that he show you another door,
there wasn't even a requirement that he offer you a choice. The issue
on the show was mainly psychological. To prove his point he
consistently forced the interviewer to choose the goat in a number of
successive trials. His closing line was, (paraphrased) "If you can
get me to offer you $5000 rather than choose, just take the money and
sit down".

Russell Turpin

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Jul 22, 1991, 3:03:25 PM7/22/91
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-----
In article <Jul.22.10.02....@remus.rutgers.edu> kal...@remus.rutgers.edu (Theodore A. Kaldis) writes:
> ... The contestant would pick one of three doors, at which
> point the MC (Monty Hall) would show what was behind one of
> the unpicked doors -- and then give the contestant the
> opportunity to switch doors. The question is, does the
> contestant have a greater probability of winning by switching
> or by not switching -- or is the probablitity the same in
> either case? ...

This, unfortunately, resembles the way the problem is usually
stated, and as such, it is not well formed. In proceeding to
analyze the problem, the above writer notes:

> ... Note, however, that certain combinations will never occur.

> The MC will never open the door with the big prize, nor will

> he reveal what's behind the door that was picked. ...

In order to get the "correct" solution, that by switching one
gets a 2/3 chance at the prize, the problem must be stated in a
fashion that includes the above assumptions. Unfortunately, it
is often stated sloppily, so that the "right" answer is no longer
determined from the problem statement.

The newspaper article was very good. It pointed out the
importance of these assumptions. Indeed, Monty Hall himself
noted that one should never switch if the MC makes it a rule
to always reveals the first chosen door whenever it has a
goat behind it, since any offer to switch is then but a ruse.

Russell

Mark T. Lakata

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Jul 23, 1991, 1:47:38 AM7/23/91
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da...@eleazar.dartmouth.edu (Dave Alexander) writes:

>The Monty Hall Problem made page one of today's (21 July)
>New York Times.

I'm surprised more people didn't actually try out the game to
verify the theory and that so many people had so much faith
in their reasoning. The two probabilities involved were 50% (wrong) or
66.7% (right) , a difference of 16.7%. If we assume some type of reasonable
distribution, (like with error bars of 1/sqrt(N)) then a good numbers of trials
is N = (1/16%)^2 about 40. It took me and my dad a few minutes to try out
the game and I remember the ratios were something like
when switching: 13 wins 6 losses
when not switching: 7 wins 14 losses
I have't any real good statistics class, but I think this is a case of
95% reliability (at least). I realize that the answer may seem
trivial to some and that this experimentation was stupid, but
I needed to prove it to myself. I was convinced.
Mark

Doc O'Leary

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Jul 23, 1991, 4:05:24 AM7/23/91
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Believe me, Mark, it is FAR more intelligent to get the right answer by hand
than it is make an ass out of yourself on the front of the New York Times!
Anybody want to post the names of those great mathematical minds that couldn't
grasp the Monty problem (it'll help to eliminate a few choices for grad
school :-)?

--------- Doc


********************** Signature Block : Version 2.7 *********************
* | *
* "Was it love, or was it the idea | Time flies whenever it damn *
* of being in love?" -- PF | well pleases *
* (BTW, which one *is* Pink?) | *
* | --->ole...@ux.acs.umn.edu<--- *
****************** Copyright (c) 1991 by Doc O'Leary ********************

Marc Roussel

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Jul 23, 1991, 12:09:06 PM7/23/91
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I have seen a lot of bashing of the "eminent professors" who messed
up the Monty problem in this thread. One poster went to far as to
suggest that he would use the identities of the offenders to eliminate
grad schools from his consideration. Now what I'd be curious to know is
whether any of these professors were statisticians. If not, what are
you all so upset about? The Monty problem is very subtle and unless
someone has explained it to you properly, you're not likely to get it
right and indeed are quite likely to make an ass of yourself trying to
defend the wrong answer. How does understanding the Monty problem
affect one's ability to solve problems in topology? Not at all.
Unless you can tell me with a straight face that you have never and
will never make a horrendous mistake in an area where you're not an
expert, let's leave these guys alone. They'll suffer enough in their
own staff rooms where their run-in with Mr. Hall will provide ammunition
to the departmental jokers for years to come. If these guys were all
statisticians, I withdraw my comments.

Marc R. Roussel
mrou...@alchemy.chem.utoronto.ca

Mark Johnson

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Jul 23, 1991, 1:58:53 PM7/23/91
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mrou...@alchemy.chem.utoronto.ca (Marc Roussel) writes:
> I have seen a lot of bashing of the "eminent professors" who messed
>up the Monty problem in this thread. One poster went to far as to
>suggest that he would use the identities of the offenders to eliminate
>grad schools from his consideration. Now what I'd be curious to know is
>whether any of these professors were statisticians. If not, what are
>you all so upset about? The Monty problem is very subtle and unless
>....

I don't know about the people who posted, but for myself, it was the
professors who bashed Marilyn with their credentials that bugged me.
They were so *arrogant* (I read the ones she printed). I understand
_anyone_ getting the problem wrong, including professors. But when
someone says, "I'm a MATH PHD and I ABHOR the mathematical illiteracy
you are propagating, blah, blah, blah..." then it's a different ballgame.

It reminds me of something that happened to a friend of mine when he
moved out east. He mentioned something about Iowa to an easterner,
and they (rather condescendingly) corrected him: "Oh... around here,
we pronounce it ``Ohio.''"

just my two cents....
Mark Johnson / mjoh...@math.wisc.edu

Eric Postpischil (Always mount a scratch monkey.)

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Jul 23, 1991, 2:22:30 PM7/23/91
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In article <1991Jul23.1...@alchemy.chem.utoronto.ca>,
mrou...@alchemy.chem.utoronto.ca (Marc Roussel) writes:

> Now what I'd be curious to know is whether any of these
> professors were statisticians. If not, what are you all so

> upset about? . . . How does understanding the Monty problem


>affect one's ability to solve problems in topology? Not at all.

It is more than "not at all". Mathematics has an essence of proof -- of
certainty of knowledge -- that is difference from any other field.
Deduction and rigor are the hallmarks of disproving another's statements
in mathematics, not calumny and authority. A competent topologist
should not write, about a probability problem, "Our math department had
a good, self-righteous laugh at your expense". Further, consider the
mathematician who wrote "You blew it! . . . . As a professional
mathematician, I'm very concerned with the general public's lack of
mathematical skills. Please help by confessing your error and, in the
future, being more careful.". This mathematician has violated their own
advice, by failing to be careful.

Even if probablity can be tricky, these people professed to authority
erroneously and when they should have used deduction to challenge vos
Savant's statements. Some of them continued even after several more
columns appeared. Another thinks perhaps we should vote on correct
answers: "You are utterly incorrect . . . How many irate mathematicians
are needed to get you to change your mind?". No field of mathematics,
topology or otherwise, should settle its issues by counting irate
mathematicians.


-- edp (Eric Postpischil)
"Always mount a scratch monkey."
e...@jareth.enet.dec.com

Jeff Kenton OSG/UEG

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Jul 23, 1991, 3:23:08 PM7/23/91
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In article <24...@shlump.lkg.dec.com>, e...@jareth.enet.dec.com (Eric

Postpischil (Always mount a scratch monkey.)) writes:

|>
|> "You are utterly incorrect . . . How many irate mathematicians
|> are needed to get you to change your mind?". No field of mathematics,
|> topology or otherwise, should settle its issues by counting irate
|> mathematicians.
|>

You mean you've never heard of proof by intimidation?

-----------------------------------------------------------------------------
== jeff kenton Consulting at ken...@decvax.dec.com ==
== (617) 894-4508 (603) 881-2742 ==
-----------------------------------------------------------------------------

Mark T. Lakata

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Jul 23, 1991, 10:31:43 PM7/23/91
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ole...@ux.acs.umn.edu (Doc O'Leary) writes:

>Anybody want to post the names of those great mathematical minds that couldn't
>grasp the Monty problem (it'll help to eliminate a few choices for grad
>school :-)?

> --------- Doc
I don't have the NY times article but I have the feb 17,1991
"Ask Marilyn" article which was the third one on the subject. These are the
people she mentions, as well as their quotes.

Frank Rose, PhD, U of Michigan
"You are in error -- and you have ignored good counsel-- but Albert
Einstein earned a dearer place in the hearts of the people after he
admitted his errors."

James Rauff, PhD, Millikin University
"I have been a faithful reader of your column and have not, until now,
had any reason to doubt you. However, in this matter, in which I do
have expertise, your answer is clearly at odds with the truth."

Charles Reid, PhD, University of Florida
"May I suggest that you obtain and refer to a standard textbook on
probability before you try to answer a question of this type again?"

W. Robert Smith, PhD, Georgia State University
"Your logic is in error, and I am sure you will receive many letters
on this topic from high school and college students. Perhaps you
should keep a few addresses for help with future columns."


E. Ray Bobo, PhD, Georgetown University
"You are utterly incorrect about the game-show question, and I hope
this controversy will call some public attention to the serious
national crisis in mathematical education. If you can admit your error,
you will have contributed constructively toward the solution of a
deplorable situation. How many irate mathematicians are needed to get

you to change your mind?"

Kent Ford, Dickinson State University
"I am in shock that after being corrected by at least three mathematicians,
you still do not see your mistake."

Don Edwards, Sunriver, Ore.
"Maybe women look at math problems differently than men."

Glenn Calkins, Western State College
"You are the goat!"
[This doesn't make much sense until you realize that behind the
two doors with no car are goats.-Mark]

Everett Harman, PhD, U.S. Army Research Institute
"You're wrong, but look at the positive side. If all those Ph.D.s were wrong,
the country would be in very serious trouble.

I hope that the above individuals are not offened if they happen to read
news, but that's what you get when you criticize (incorrectly)
people publicly.

Mark
p.s. Marilyn posted^H^H^H^H^H^Hprinted one letter from

Seth Kalson, PhD, MIT
"You are indeed correct. My collegues at work had a ball with this
problem, and I dare say that most of them--including me at first--thought
you were wrong."
[I thought she was wrong too!-Mark]

Jan Willem Nienhuys

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Jul 24, 1991, 6:12:18 AM7/24/91
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In article <1991Jul24.0...@nntp-server.caltech.edu> lak...@nntp-server.caltech.edu (Mark T. Lakata) writes:
Keywords:

In article <1991Jul24.0...@nntp-server.caltech.edu> lak...@nntp-server.caltech.edu (Mark T. Lakata) writes:


I have looked in the World Directory of Mathematicians (1990), and in the
Combined Membership List of the American Mathematical Society, the
Mathematical Association of America, and the SIAM, for 1990/91
and I have found (Note: ??= not found anywhere)


>
>Frank Rose, PhD, U of Michigan
>"You are in error -- and you have ignored good counsel-- but Albert
>Einstein earned a dearer place in the hearts of the people after he
>admitted his errors."

??; also not listed as professor in The World of Learning.


>
>James Rauff, PhD, Millikin University
>"I have been a faithful reader of your column and have not, until now,
>had any reason to doubt you. However, in this matter, in which I do
>have expertise, your answer is clearly at odds with the truth."

There is a James Vernon Rauff, member MAA (basically a teacher's association,
the AMS is more oriented towards research mathematicians, I believe),
in the College of Lake County, Ill.


>
>Charles Reid, PhD, University of Florida
>"May I suggest that you obtain and refer to a standard textbook on
>probability before you try to answer a question of this type again?"

??


>
>W. Robert Smith, PhD, Georgia State University
>"Your logic is in error, and I am sure you will receive many letters
>on this topic from high school and college students. Perhaps you
>should keep a few addresses for help with future columns."

??


>
>
>E. Ray Bobo, PhD, Georgetown University
>"You are utterly incorrect about the game-show question, and I hope
>this controversy will call some public attention to the serious
>national crisis in mathematical education. If you can admit your error,
>you will have contributed constructively toward the solution of a
>deplorable situation. How many irate mathematicians are needed to get
>you to change your mind?"

MAA, Associate Professor


>
>Kent Ford, Dickinson State University
>"I am in shock that after being corrected by at least three mathematicians,
>you still do not see your mistake."

??


>
>Don Edwards, Sunriver, Ore.
>"Maybe women look at math problems differently than men."

According to WDM, there is a Don Edwards at Un. of South Carolina
there is also a Don A. Edwards, member MAA, Houston Community College.

>
>Glenn Calkins, Western State College
>"You are the goat!"
>[This doesn't make much sense until you realize that behind the
>two doors with no car are goats.-Mark]

MAA


>
>Everett Harman, PhD, U.S. Army Research Institute
>"You're wrong, but look at the positive side. If all those Ph.D.s were wrong,
>the country would be in very serious trouble.
>

??


>
>
>I hope that the above individuals are not offened if they happen to read
>news, but that's what you get when you criticize (incorrectly)
>people publicly.
>
>Mark
>

>Seth Kalson, PhD, MIT
>"You are indeed correct. My collegues at work had a ball with this
>problem, and I dare say that most of them--including me at first--thought
>you were wrong."
>[I thought she was wrong too!-Mark]


J.W. Nienhuys,
Research Group Discrete Mathematics
Dept. of Mathematics and Computing Science
Eindhoven University of Technology
P.O. BOX 513, 5600 MB Eindhoven
The Netherlands

e-mail: wsa...@urc.tue.nl

Ron Evans

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Jul 24, 1991, 10:13:50 PM7/24/91
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The New York Times front page article (7/21/91) states:

"A few mathematicians were familiar with the puzzle long before vos
Savant's column. They called it the Monty Hall problem - the title
of an analysis in the journal American Statistician in 1976 -..."

The problem appeared in this journal in 1975(sic) under the title
"A Problem in Probability". (See pp. 67 and 134
of the Feb. and Aug. issues, resp.)

The problem was unfortunately stated ambiguously in American
Statistician as well as in columns of vos Savant. I don't recall
seeing a succinct, unambiguous statement of the problem in the
NYT article, but the NYT did reproduce the following flawed
version:

"Suppose you're on a game show, and you're given the choice of
three doors. Behind one door is a car; behind the others, goats.
You pick a door, say No. 1, and the host, who knows what's behind
the other doors, opens another door, say No. 3, which has a goat.
He then says to you, 'Do you want to pick door No. 2?' Is it to
your advantage to take the switch?"


Here is a way to state the problem without the ambiguity:


Suppose you're on a game show, and you're given the choice of
three doors. Behind one door is a car; behind the others, goats.
You select a door (which stays unopened) and the host,
who knows what lies behind the other doors,
opens another door intentionally revealing a goat. He then says,
"You may switch your selection to the other unopened door."
Is it to your advantage to take the switch?


The major change I made above is insertion of the word "intentionally".
To assign probabilities, we need a *repeatable* experiment,
but in the first formulation, it is not clear that Monty *always*
reveals a goat. It leaves room for the possible interpretation
that instead Monty always opens one of the two unselected doors at
random, with the given condition (so conditional probability will
apply) that on THIS occasion it happens that he reveals a goat rather
than the car. Indeed, with this interpretation, one cannot conclude
that the odds in favor of switching are two out of three.

Another change I made is the removal of references to specific
door numbers. One could of course NAME the selected door "No.1",
and NAME the opened door "No.3". As I remember the show
"Let's Make A Deal", the doors are named No.1, No.2, and No.3
from left to right (from the vantage point of the audience),
and it is desirable to eliminate any confusion about which doors
are selected or opened. Let's agree that the doors are
numbered #1,#2,#3 from left to right. Now try my variant
of the Monty Hall problem below.


THE MONTY HALL PROBLEM WITH A TWIST

Suppose you're on a game show, and you're given the choice of
three doors. Behind one door is a car; behind the others, goats.
You select a door (which stays unopened) and the host,
who knows what lies behind the other doors,
opens another door intentionally revealing a goat. He then says,
"You may switch your selection to the other unopened door."
ON THIS OCCASION YOU HAPPEN TO SELECT DOOR #1 AFTER WHICH MONTY
HAPPENS TO OPEN DOOR #3. Which of the following is true?
(A) It is to your advantage to take the switch on this occasion.
(B) There is no advantage to switching on this occasion.
(C) There is not enough information to decide between (A) and (B).
(D) It is not possible to decide between (A),(B), or (C).

Please think about this before looking at my solution below.

SOLUTION
If Monty has a choice of TWO doors to open after your selection,
how does he decide which one to open? Well, it makes no
difference in the original Monty Hall problem (where switching
improves your odds), but it does make a difference here.
SCENARIO 1: Monty's strategy
(announced to the world) is to always open the leftmost door
when he has a choice. (The leftmost door is the door with
smaller number.) In Scenario 1, the answer would be (A), since the
car is necessarily behind door #2.
SCENARIO 2: Monty's strategy
is to always open the rightmost door when he has a choice. In
Scenario 2, the answer would be (B), since the car is equally likely
to be behind door #1, door #2. Since Monty's strategy isn't known,
the correct answer to the puzzle is (C).


--
Ron Evans, Department of Mathematics, UCSD (rev...@math.ucsd.edu)

Jim Propp

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Jul 24, 1991, 2:30:06 PM7/24/91
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It's one thing to quote (with attribution) the remarks of those who
over-hastily dismissed Marilyn vos Savant's explanation of the Monty
Hall paradox. It's quite another to start trying to track down the
individuals, as at least one poster to sci.math seems to be doing.

We should think long and hard before embarking on anything that might
remotely resemble an intellectual witch-hunt -- especially since it's
by no means clear that the individuals in question ever intended
their remarks to be given such wide publicity. (Even knowing nothing
about D. E., I am sure that he in particular would have been a lot
more circumspect if he'd known that his words would someday appear
in the Times!) And remarks taken out of context sometime undergo a
subtle but significant shift in meaning.

Also, in taking a smug attitude toward the "vos Savant Nine" as some
posters to this thread have done, there is the danger that we may
come to resemble those we deride.

And who knows? Maybe we'll all wake up tomorrow in a universe where
these other guys are _right_. I myself don't believe it, but it's good
epistemological hygiene to keep such possibilities in mind.


Jim "I'm probably over-reacting, but..." Propp


P.S. Speaking of game shows: were there ever any mathematicians (famous
or otherwise) on those old game-shows where you had to guess someone's
profession, or sort out the real "X" from the two fake ones?

==================================================================
|| ||
|| A mathematician is someone who works hard at being lazy. ||
|| ||
==================================================================

Jan Willem Nienhuys

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Jul 25, 1991, 8:04:54 AM7/25/91
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In article <1991Jul24.1...@galois.mit.edu> pr...@carleman.mit.edu (Jim Propp) writes:
>It's one thing to quote (with attribution) the remarks of those who
>over-hastily dismissed Marilyn vos Savant's explanation of the Monty
>Hall paradox. It's quite another to start trying to track down the
>individuals, as at least one poster to sci.math seems to be doing.

Maybe you mean me. But I'm not interested in tracking down, but in
finding out whether all these over-hasty guys can be called mathematicians
at all. 5 out of 9 are not in readily accessible lists of U.S. mathematicians,
so some doubt is retrospectively justified. One name hailing from Oregon,
(according to v.S.) is found only in Houston and in South Carolina.

>
>We should think long and hard before embarking on anything that might
>remotely resemble an intellectual witch-hunt -- especially since it's
>by no means clear that the individuals in question ever intended
>their remarks to be given such wide publicity.

>Also, in taking a smug attitude toward the "vos Savant Nine" as some


>posters to this thread have done, there is the danger that we may
>come to resemble those we deride.

In a column last December Vos Savant mentioned (primary) reactions from
three more. I won't mention their names, but one IS listed in lists of
mathematicians, even with his email-address.
It is not clear what exactly they had been writing to Vos Savant.
The solution of the puzzle depends on the assumption that M.H.
(1) always will make the offer to switch (2) that he chooses only between
goat doors (3) that his choice between possibilities (if any) is random.
Ms v. S. didn't make this perfectly clear, maybe some of the people
writing her made a different assumption on points 1-3, or explained that
she ought to have been more clear, or maybe said that Monty Hall actually
did not operate that way, or whatever. We have only one or two sentences
excerpted from their letters by Ms. v. S. to go by.

>
>Jim "I'm probably over-reacting, but..." Propp
>

One more thing: I recall reading in a post somewhere that Ms. v.S. is
supposed to have an IQ of 230. Technically this would mean that she is,
in a normally distributed population, 130/15 standard deviations above
average, so by definition only 1 in 10^16 or so people would be so good
in solving IQ-test puzzles. Allow me to be a bit skeptical about the claim
that she is so bright. I find it hard to imagine that this bragging
about IQ is done without her approval.

martin.brilliant

unread,
Jul 25, 1991, 10:53:39 AM7/25/91
to
From article <21...@sdcc6.ucsd.edu>, by rev...@euclid.ucsd.edu (Ron Evans):
> .....

> The problem was unfortunately stated ambiguously in American
> Statistician as well as in columns of vos Savant.....

> Here is a way to state the problem without the ambiguity:
>
> Suppose you're on a game show, and you're given the choice of
> three doors. Behind one door is a car; behind the others, goats.
> You select a door (which stays unopened) and the host,
> who knows what lies behind the other doors,
> opens another door intentionally revealing a goat. He then says,
> "You may switch your selection to the other unopened door."
> Is it to your advantage to take the switch?

Unfortunately, this statement of the problem leaves unresolved the
ambiguity mentioned in the New York Times. That is, we don't know
whether the host has the option of not opening a door at all. They
point out that in the real game show, the host does not give you a
second chance if you pick a goat on the first try. If you can rely
on that strategy, you should never switch if given a chance to.

There is probably a minimax solution, but I don't offhand know how to
derive it. That is, the host knows you know he has a choice of
strategies. So he chooses a strategy randomly to minimize his
expected loss under the assumption that you use your optimum
randomized strategy, which you choose to maximize your expected gain
under the assumption that the host uses his optimized random strategy.

Marty
ma...@hoqax.att.com hoqax!marty
Martin B. Brilliant (Winnertech Corporation)

Philip Weissman

unread,
Jul 25, 1991, 4:41:15 PM7/25/91
to

>NYT article ... the NYT did reproduce the following flawed
>version:
>

> "Suppose you're on a game show, and you're given the choice of
> three doors. Behind one door is a car; behind the others, goats.
> You pick a door, say No. 1, and the host, who knows what's behind
> the other doors, opens another door, say No. 3, which has a goat.
> He then says to you, 'Do you want to pick door No. 2?' Is it to

> your advantage to take the switch?"
>
>
>Here is a way to state the problem without the ambiguity:
>
>
> Suppose you're on a game show, and you're given the choice of
> three doors. Behind one door is a car; behind the others, goats.
> You select a door (which stays unopened) and the host,
> who knows what lies behind the other doors,
> opens another door intentionally revealing a goat. He then says,
> "You may switch your selection to the other unopened door."
> Is it to your advantage to take the switch?
>
>
>The major change I made above is insertion of the word "intentionally".
>To assign probabilities, we need a *repeatable* experiment,
>but in the first formulation, it is not clear that Monty *always*
>reveals a goat. It leaves room for the possible interpretation
>that instead Monty always opens one of the two unselected doors at
>random, with the given condition (so conditional probability will
>apply) that on THIS occasion it happens that he reveals a goat rather
>than the car. Indeed, with this interpretation, one cannot conclude
>that the odds in favor of switching are two out of three.
>
>--
>Ron Evans, Department of Mathematics, UCSD (rev...@math.ucsd.edu)

You are correct in saying that it is impossible to assign
probabilities with the problem as stated by the New York Times. Your
restatement, however, does not alter this condition. The reason being
that Monty was not forced to offer the switch by the rules of the
game. If Monty's strategy were to only offer the switch if you
correctly picked the winning door, then switching would offer a zero
probability of winning. If he offered the switch only if you picked a
losing door then the probability of winning by switching would be one.
Other strategies that Monty might employ could make the probability of
winning by switching anything between zero and one.

You cannot analyze one particular instance where Monty offered the
switch, as in the problem, since Monty is free to make the offer or
not at his own discretion. This point was made in the NY Times
article. Thus it is impossible to answer the question on purely
mathematical grounds.

Since the situation, as described by the NYT, is the result of Monty's
discretion it is impossible to duplicate. In Let's Make a Deal, Monty
does have the option to offer the switch or not to offer the switch.
Therefore, the only way to do an experiment is to watch Let's Make a
Deal and keep a count.

If you're still not convinced try this. Imagine that you're on stage
with Monty. He just offered you the option of switching to the other
unopened door. Now say to yourself "Monty will allow me to switch
doors." "WHY DID HE GIVE ME THIS OPTION?" "Is he trying to get me to
switch away from the winning door?" You must also consider the money
Monty has put in your hand. Psychological factors should affect your
decision at least as much as the mathematical ones. There are people
who, by watching Monty's actions, would rarely make a wrong decision.
Of course Monty wouldn't always give them a decision to make in the
first place.

In order to asses probabilities, it MUST be stated that Monty will
ALWAYS allow you to switch after he reveals the identity of a loosing
door. This subtle point is far from trivial. The problem as stated
by the NYT falls under the discipline of game theory which combines
probability with the psychological element present in this game, poker
and many other more common aspects of life such as finding parking
spaces. :~) Other formulations of the problem do state that Monty will
ALWAYS allow you to switch after he reveals the identity of a loosing
door. That type of formulation can be analyzed mathematically to
determine that the probability of winning by switching is 2/3.

Philip Weissman

Leads Network News

unread,
Jul 26, 1991, 11:31:06 AM7/26/91
to
>If you're still not convinced try this. Imagine that you're on stage
>with Monty. He just offered you the option of switching to the other
>unopened door. Now say to yourself "Monty will allow me to switch
>doors." "WHY DID HE GIVE ME THIS OPTION?" "Is he trying to get me to
>switch away from the winning door?" You must also consider the money
>Monty has put in your hand. Psychological factors should affect your
>decision at least as much as the mathematical ones. There are people
>who, by watching Monty's actions, would rarely make a wrong decision.
>Of course Monty wouldn't always give them a decision to make in the
>first place.

AARRGHH!!! I thought only my wife thought like this! When I laid
out the game for her and asked her if she would switch, she said "I
don't know, it depends on whether I thought he was trying to trick me".

>In order to asses probabilities, it MUST be stated that Monty will
>ALWAYS allow you to switch after he reveals the identity of a loosing
>door.

It was clear to me that such was the case in this problem, i.e., the
way the game was to be played was pre-determined. Of course on the *REAL*
show that's not true. But we weren't talking about the real show, were we?
In any case, you're right - if that point is not made clear (which I
don't think it was in vos Savant's column), then the problem is different.
=============================================================================
Bob Marshall ^ \\
Lockheed Missiles & Space Co. |L| \\ Marshall's Theorem :
Sunnyvale, CA |M| \\ "2 + 2 approximately equals 5 for
(408)756-5737 |S| \\ large values of 2"
mars...@force.decnet.lockheed.com |C| \\
/ \ \\
=============================================================================

Alton Harkcom

unread,
Jul 23, 1991, 8:24:25 AM7/23/91
to

If you weren't so arrogant about your ignorance, I wouldn't even have
bothered correcting you...

In article <Jul.22.10.02....@remus.rutgers.edu>
kal...@remus.rutgers.edu (Theodore A. Kaldis) writes:

Your logic has enough holes to fill the Albert Hall...

=}_Let's Make a Deal_. The
=}contestant would pick one of three doors, at which point the MC (Monty
=}Hall) would show what was behind one of the unpicked doors -- and then
=}give the contestant the opportunity to switch doors. The question is,
=}does the contestant have a greater probability of winning by switching
=}or by not switching -- or is the probablitity the same in either case?

=}The MC will never open
=}the door with the big prize, nor will he reveal what's behind the door
=}that was picked.

=} switching implies
=}
=} behind: picked: shows: outcome:
=} 1 1 2 LOSE
=} 1 1 3 LOSE
=} 1 2 3 WIN!
=} 1 3 2 WIN!
=} [same data for other 2 doors]
=}
=}Okay. Now this table seems to indicate that, regardless of whether
=}one switches or not, the odds of winning are the same.

You must mention that not switching will give the exact opposite
series of answers (WIN WIN LOSE LOSE) to be able to make that claim...

=}There's a little problem here!

Yes, you attempted to solve a problem without the proper tools...

=} Let's say the contestant picked a
=}losing door. Then the MC will always show the other losing door. But
=}when the winning door is picked, in half the cases the MC will show
=}one of the losing doors, and in the other half he will show the other!
=}So let's rearrange the table somewhat to reflect this anamoly.
=}
=} behind: picked: shows: outcome:
=} 1 1 2 or 3 LOSE
=} 1 2 3 WIN!
=} 1 3 2 WIN!
=} [same data for other 2 doors]
=}
=}Voila!

Another "Mr. Kaldis" display of ignorance... Nowhere do you mention
probability... But giving your arguments a slight resemblence to
mathematics...

----------------------------------------------------------------------------
The prize is behind door A (A can be 1, 2, or 3). The player chooses
door B (B can be 1, 2, or 3), and the MC reveals door C (C can be 1, 2,
or 3). The rules are as stated above...

With C always being a losing door, the following states are possible...

behind picked shows with switch no switch
A B C A==B LOSE WIN
where C is one of the losing doors
A B C A==B LOSE WIN
where C is the other losing door
A B C A!=B WIN LOSE
where B is one of the losing doors
A B C A!=B WIN LOSE
where B is the other losing door

Which collapses into...

behind picked shows with switch no switch
A B C A==B LOSE WIN
A B C A!=B WIN LOSE
where B is one of the losing doors
A B C A!=B WIN LOSE
where B is the other losing door

So obviously, the chance of winning by switching is 2:1 compared to
the chance of winning without switching... Conclusion is that the chance
of winning is greater if one switches.
----------------------------------------------------------------------------

Where the logic fails is in that tricky little part which groups the
two losing doors together in one situation, but seperates them in another.
But "Mr. Kaldis", being the king of double standards that he is, can't
recognize this as improper behaviour... Carrying his arguments one step
further will give...

behind picked shows with switch no switch
A B C A==B C!=B LOSE WIN
A B C A!=B C!=B WIN LOSE

No change for switching... But what is the PROBABILITY of winning in
either case????

There are three doors. We'll call them W, L1, and L2. The probability
that the player will choose W, L1 or L2 is .3_ (where _ means continue to
infinity). The MC will reveal either L1 or L2. Either the player chose
W (.3_) or the remaining L door (.3_). If the player switches, then he
merely trades his door's .3_ probability of being the winner for the
other door's .3_ probability of being the winner...

=}This is exactly the answer published in Parade, and the "Smartest
=}Person in the Known Universe" stands vindicated by none other than
=}Theodore A. Kaldis.

Parade? Well at least we now know where you get all of that great
misinformation which you post. Well part of it anyway, the Enquirer
must give you a few hints...

=}[I should expect the folks in sci.math and rec.puzzles to know this.
=}Therefore followups are directed to alt.flame.]

I should expect the folks in sci.math and rec.puzzles would know
enough to find the correct answer and recognize your ignorance. I am
replying from alt.flame to rationalize my flaming of you. My apologies
to those in the non alt.flame groups for the heat...

Al

martin.brilliant

unread,
Jul 29, 1991, 12:53:31 PM7/29/91
to
From article <17...@leadsv.UUCP>, by net...@leadsv.UUCP (Leads Network News):
> .....

> It was clear to me that such was the case in this problem, i.e., the
> way the game was to be played was pre-determined. Of course on the *REAL*
> show that's not true. But we weren't talking about the real show, were we?

Why weren't we talking about the real show? Is this a pure math, not
applied math, forum? Maybe I should unsubscribe.

Frederick W. Chapman

unread,
Jul 29, 1991, 12:27:19 PM7/29/91
to
>From: har...@spinach.pa.yokogawa.co.jp (Alton Harkcom)
>Date: 23 Jul 91 12:24:25 GMT
>Organization: Yokogawa Electric Corporation, Tokyo, Japan
>In-reply-to: kal...@remus.rutgers.edu's message of 22 Jul 91 14:02:26 GMT


[NOTE: The following is a comment from Alton Harkcom regarding a post
by Theodore Kaldis.]

> If you weren't so arrogant about your ignorance, I wouldn't even have
>bothered correcting you...
>
>In article <Jul.22.10.02....@remus.rutgers.edu>
> kal...@remus.rutgers.edu (Theodore A. Kaldis) writes:
>


[Mr. Kaldis's original analysis and Mr. Harkcom's flame deleted.]


Whereas Mr. Kaldis's analysis may not have been the most *concise* that
I have seen, I found no arrogance whatsoever in the manner of his
presentation. I am left wondering what on earth precipitated this
personal attack by Mr. Harkcom. I have seen no evidence of provocation
on sci.math.


[NOTE: The following analysis is by Mr. Harkcom.]

> There are three doors. We'll call them W, L1, and L2. The probability
>that the player will choose W, L1 or L2 is .3_ (where _ means continue to
>infinity). The MC will reveal either L1 or L2. Either the player chose
>W (.3_) or the remaining L door (.3_). If the player switches, then he
>merely trades his door's .3_ probability of being the winner for the
>other door's .3_ probability of being the winner...

This analysis is incorrect, the main flaw being that the argument fails
to take into account the new information provided by the opening of one
of the (loosing) doors. The probability that the original choice was
the winning choice is 1/3; this fact remains constant throughout the
problem. Since the prize lies behind either the originally chosen door
or the one remaining door (new information), the probability of winning
by changing to the remaining door is 1 - 1/3 = 2/3.

Whereas many people (myself included) originally believed that there was
no advantage to switching choices, most of these people (myself
included) now realize that our first analysis was in error. The answer
printed in Parade Magazine *is* correct. For another concise, correct,
and even more general analysis of this problem, I recommend Trent
Tobler's July 26 post to sci.math on the topic "Re: Monty Hall Problem".

If you are not convinced that there is an advantage to switching doors,
why not run a computer simulation? I have a simulation written in Maple
which I will gladly give to anyone who wants a copy. The simulation
makes no assumptions beyond those required to define the problem, and
confirms the Parade answer. Someone else recently posted some C
programs for doing simulations of variations of the problem.


[More Kaldis commentary and Harkcom flame deleted.]


[NOTE: The following comment is from Mr. Harcom.]

> I should expect the folks in sci.math and rec.puzzles would know
>enough to find the correct answer and recognize your ignorance. I am
>replying from alt.flame to rationalize my flaming of you. My apologies
>to those in the non alt.flame groups for the heat...
>
>Al

The folks in sci.math and rec.puzzles *do* know enough to recognize the
correct answer, and we recognize ignorance when we see it, unfortunately
for you. I find your utterly uncalled for personal attack on Mr.
Kaldis to be all the more distasteful in light of your own public
display of ignorance and arrogance.

Such behavior is completely inappropriate on any USENET news group. A
public apology to Mr. Kaldis is certainly in order.


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Frederick W. Chapman (N3IJC) Campus Phone: (215) 758-3218
User Services Group Network Server UserID: FC03
Computing Center Internet: FC...@NS.CC.LEHIGH.EDU
Lehigh University Bitnet: FC03@LEHIGH

Theodore A. Kaldis

unread,
Jul 29, 1991, 2:11:17 PM7/29/91
to
In article <HARKCOM.91...@spinach.pa.yokogawa.co.jp> har...@spinach.pa.yokogawa.co.jp (Alton Harkcom) writes:

[in response to me]

> If you weren't so arrogant about your ignorance, I wouldn't even have
> bothered correcting you...

> Your logic has enough holes to fill the Albert Hall...

> [...]

> There are three doors. We'll call them W, L1, and L2. The probability
> that the player will choose W, L1 or L2 is .3_ (where _ means continue
> to infinity). The MC will reveal either L1 or L2. Either the player
> chose W (.3_) or the remaining L door (.3_). If the player switches,
> then he merely trades his door's .3_ probability of being the winner
> for the other door's .3_ probability of being the winner...

Alton Harkcom still doesn't get it. Quite simply: The player picks
one of three doors, of which only one is the potential winner. The
probability of the player picking the winning door is 1/3. The MC
will then reveal a losing door, and subsequently give the player an
opportunity to switch to the remaining unpicked door. In 1 out of
three cases, the player would be switching from the winning door to a
losing door, and in 2 out of three cases, the player would be
switching from a losing door to the winning door.

Scott Dixon

unread,
Jul 29, 1991, 8:50:17 PM7/29/91
to
I have been following the "Monty on the Front Page" thread with great interest.
When I first read the Marilyn Vos Savant Article in the "Parade" section of
the San Jose Mercury News, I thought "She's wrong!. She sure made a fool of
herself." In trying to prove her wrong, I kept proving her right. I even
simulated the problem with Verilog (a hardware description language) to win
a free lunch from a colleague. However, the problem continued to bother me
because I couldn't come up with a simple way to describe the 2-to-1 advantage
of switching versus non-switching. I eventually developed the following
resoning, which seems to make the choice of switching very obvious.

If you don't switch doors, you have a 1-in-3 chance of winning the prize.

If you always switch doors (after Monty always shows you an empty door), you
will get the prize if it was behind either of the two doors that you didn't
pick originally. Always switching is the same as picking two doors - while
never switching is the same as picking one door. Would you rather pick two
doors or one door? Obviously, you would prefer two doors, and your advantage
is 2-to-1.

When I have explained this to others, they have asked, "How do I pick two
doors"? I tell them to mentally pick two doors and then choose the third
door. When Monty asks if you want to switch, say yes. You will get the
prize if it was behind either of the other two doors.

If you agree with my explanation, then perhaps it should be added to the
decision/switch.p FAQL.

By the way, the Monty article made the front page of the San Jose Mercury
News.

Bruce Leban

unread,
Jul 29, 1991, 5:49:51 PM7/29/91
to

> If you weren't so arrogant about your ignorance, I wouldn't even have
> bothered correcting you...

I'm sick of this problem. If you want to argue about who's more arrogant do
it on e-mail. Look -- if I go on Let's Make a Deal regardless of which door
I pick I still leave with more than I came so I'm a winner. Unless of
course, the chicken suit cost more than I can sell the goat for.

--- Bruce
Le...@cs.umass.edu @amherst.mass.usa.earth

ObPuzzle: Here's a new puzzle *never before seen on this newsgroup*:

THE PROBLEM OF THE UNEXPECTED PLUMBER:

On Monday, you call a plumber. He promises that he will come
fix your broken pipes before the end of the week (Friday). He
also tells you that he will show up on a day that you don't
expect him. How is this possible?

BEFORE YOU FLAME ME, READ THE SPOILER:

The plumber will come on Monday of the week after.

Patrick J. Fleury

unread,
Jul 30, 1991, 12:18:40 AM7/30/91
to
In article <1991Jul24.1...@galois.mit.edu> pr...@carleman.mit.edu (Jim Propp) writes:
>It's one thing to quote (with attribution) the remarks of those who
>over-hastily dismissed Marilyn vos Savant's explanation of the Monty
>Hall paradox. It's quite another to start trying to track down the
>individuals, as at least one poster to sci.math seems to be doing.
>
>We should think long and hard before embarking on anything that might
>remotely resemble an intellectual witch-hunt -- especially since it's
>by no means clear that the individuals in question ever intended
>their remarks to be given such wide publicity. (Even knowing nothing
>about D. E., I am sure that he in particular would have been a lot
>more circumspect if he'd known that his words would someday appear
>in the Times!) And remarks taken out of context sometime undergo a
>subtle but significant shift in meaning.
>
I am afraid I must disagree here. I do not think this is the beginning of a
"witch hunt" against the "Marilyn Nine." I am sure they are all very, very
embarassed at being pilloried on the front pages of the and have been
punished enough.

However, the fact that some of these people do not belong to the main
professional organizations in their supposed fields of expertise does say
something about the way that the media - oh, that horrible hydra headed
media monster - does treat mathematics. The Times enthusiastically
published the remarks of "experts" without bothering to check on their
credentials. The attitude of the media in general and the Times in
particular toward mathematics has been slipshod for years.

Do you remember when they endorsed Walter Mondale for President? They asked
us to vote for him in spite of the fact that he had "all the pizzazz of a
trigonometry teacher." ( I think that's a fairly accurate quote.) I voted
for him in spite of the NYT's goofy remark.

Do you remember their front page article on the way finite difference
methods were going to replace calculus? Or what about the article on
catastrophe theory and the way it would revolutionize mathematics?

The Times has such a bad record on mathematics that the appearance of the
Monty problem on the front page may be the best thing that could ever happen
to sci.math. We may never hear of it again.

The rest of the media does not seem any better. Do you remember Arnold
Arnold - I think that's his name - and his proof of Fermat's Last Theorem? ??m
That got a pretty big play on National Public Radio about 8 years ago.

I am not really angry about this, but I am terribly disappointed. When it
comes to mathematics, the media just does not do its homework. Years ago,
I'll bet the reporters and editors who publish this stuff probably did not
do their's. That is probably the reason these people became reporters rather
than mathematicians or physicists or engineers.

Yet, some time shortly in the future, we shall read an article or an
editorial fulminating about the shortage of scientists and engineers or
demanding accountability from teachers of mathematics. AAnd the worst part of
it is that the Times has influence in the halls of Congress, but the AMS
does not.

Just though I'd mention it.

Diane MacMartin

unread,
Jul 30, 1991, 11:38:53 AM7/30/91
to
I like to give the following example:

I ask you to select the Ace of Spades from a deck of cards. I show you
the back of a deck, and ask you to select a card at random.

I now LOOK at the front of the cards, and discard all but one card,
keeping the Ace of Spades if it is in the remainder of the deck. I now
hold one card, you now hold one card. Are your chances better to find
the Ace of Spades by sticking with your random selection or switching to
my card?

Obviously, you have a 1 in 52 chance of holding the Ace of Spades. I
have a 51 in 52 chance, so you are better off switching.

This is the same problem as the Monty problem.

--
Diane MacMartin Bell-Northern Research, Ltd Ottawa, Canada
m...@bnr.ca Any opinions are my own and do not represent BNR.
=========>> If this is printed on paper, please recycle it <<==========

Dave Alexander

unread,
Jul 31, 1991, 3:41:23 AM7/31/91
to

In article <1991Jul29.2...@jfwhome.FUNHOUSE.COM>
j...@jfwhome.FUNHOUSE.COM (John F. Woods) writes:

> You have no idea how much it PAINS me to point out that T*d
> is correct, though his presentation could use a great deal
> of work.


Does it hurt you as much as this hurts me?


"the probability of the door that Monty has shown
causes several hundred math books to get thrown
at Marilyn who's sitting intelligently alone
but the tears on her cheeks are from laughter

"I wish I could give William Buckley his great thrill
I'd put him in chains on Capitol Hill
and call for T*d Kaldis and maybe George Will
he could die happily ever after"


-- Dave Alexander

--
"But 2 weeks of blue balls? I can hardly fault the guy."
-- Kevin Darcy

Mark Prysant

unread,
Jul 30, 1991, 11:48:17 AM7/30/91
to

I don't have access to alt.flame . If some one feels this is
worthy, could you post it there?

> If you weren't so arrogant about your ignorance, I wouldn't even have
>bothered correcting you...
>
>In article <Jul.22.10.02....@remus.rutgers.edu>
> kal...@remus.rutgers.edu (Theodore A. Kaldis) writes:
>
>Your logic has enough holes to fill the Albert Hall...
>
> =}_Let's Make a Deal_. The
> =}contestant would pick one of three doors, at which point the MC (Monty
> =}Hall) would show what was behind one of the unpicked doors -- and then
> =}give the contestant the opportunity to switch doors. The question is,
> =}does the contestant have a greater probability of winning by switching
> =}or by not switching -- or is the probablitity the same in either case?
>

[He then derives a chance of 1/3]

>
> Another "Mr. Kaldis" display of ignorance... Nowhere do you mention
>probability... But giving your arguments a slight resemblence to
>mathematics...

>


> There are three doors. We'll call them W, L1, and L2. The probability
>that the player will choose W, L1 or L2 is .3_ (where _ means continue to
>infinity). The MC will reveal either L1 or L2. Either the player chose
>W (.3_) or the remaining L door (.3_). If the player switches, then he
>merely trades his door's .3_ probability of being the winner for the
>other door's .3_ probability of being the winner...
>

Didn't your mother teach you that probabilities should sum to 1?
(Actually, mutually exclusive exhaustive event probabilities should
sum to one, and that is what we are dealing with here.)

Tsk. Tsk. If you are going to flame someone it would be a good idea to
know that the original poster is wrong and that you are right,
neither of which applies in this case.

First, we have to make the assumption that Monty will always reveal a
door, it will be a loser, and the loser is chosen randomly among
the losers which you did not choose (this is the tacit assumption
almost everyone seems to make when seeing this puzzle).

Call the three doors W, L1, and L2. The probability that the player
will choose W, L1, or L2 is 1/3 (where did you get this .3_ notation
idea?). The MC will reveal eitheer L1 or L2. If the player chose
W (prob = 1/3) and switches, it will be a loss. Suppose the player
chose L1. Then Monty opens L2, and the player would win by
switching (1/3). Suppose the player chose L2. Then Monty would
open L1, and the player would win by switching (1/3). So the
probability of winning with a switch is 2/3.

If you don't believe this, run some tests.

> =}This is exactly the answer published in Parade, and the "Smartest
> =}Person in the Known Universe" stands vindicated by none other than
> =}Theodore A. Kaldis.
>
> Parade? Well at least we now know where you get all of that great
>misinformation which you post. Well part of it anyway, the Enquirer
>must give you a few hints...
>
> =}[I should expect the folks in sci.math and rec.puzzles to know this.
> =}Therefore followups are directed to alt.flame.]
>
> I should expect the folks in sci.math and rec.puzzles would know
>enough to find the correct answer and recognize your ignorance. I am
>replying from alt.flame to rationalize my flaming of you. My apologies
>to those in the non alt.flame groups for the heat...
>
>Al

Apology not accepted. It's difficult to justify an incorrect flame.

mark 'no signature' prysant

Keith Kushner

unread,
Aug 2, 1991, 7:19:58 AM8/2/91
to
kal...@remus.rutgers.edu (Theodore A. Kaldis) writes:

> Alton Harkcom still doesn't get it. Quite simply: The player picks
> one of three doors, of which only one is the potential winner. The
> probability of the player picking the winning door is 1/3. The MC
> will then reveal a losing door, and subsequently give the player an
> opportunity to switch to the remaining unpicked door. In 1 out of
> three cases, the player would be switching from the winning door to a
> losing door, and in 2 out of three cases, the player would be
> switching from a losing door to the winning door.
> --

Ah, but can you find support for your explanation of what the "MC will
then" do in the problem as given? I doubt you can: the problem merely
stated that the MC reveals a booby prize behind door three. It is
only your assumption that the MC will *always* reveal a booby prize.

There seem to be three different views of the problem:

1) That the MC will always open, of the two unpicked doors, one with
a booby prize behind it. If so, then your analysis is correct, and
the player should switch. You would be hard-put, however, to
prove, from the problem as stated, that this is the case.

2) That the MC will always open door three if door three is not
picked by the player. Or that the MC flips a coin - heads he picks
the higher-numbered unpicked door to open, tails the lower - to decide.
Or anything roughly analogous to having the door opened at random.
Without the filtering effect of the assumptions in the first view, the
odds would therefore be 50/50. The problem as stated, however, does
not support this interpretation either.

3) That the problem as stated is seriously flawed; that any answer to it
is based on the answerer's preconceptions as to the MC's motives in
opening the door; that asking for an assessment of probability given
*one* trial is ridiculous; that one can impute to the MC any of a host
(pun intended) of motives and rationales, each with its own set of odds
varying from nil to certainty that the winning prize is behind the picked
door. (i.e., why should your assumption that the MC will always reveal
a booby prize behind an unpicked door carry any more weight than the
assumption that the MC will always reveal whatever prize is behind the
highest-numbered unpicked door? Or that the MC will *only* offer the
player a chance to switch if he's picked the winning door?)

You might guess which view of the problem I hold. Still, if you
believe you can demonstrate from the problem as stated that your
assumptions about it are indeed correct, please do so.


==================================================================
* myc...@dorsai.com * Keith Kushner * dorsai!myc...@uu.psi.com *
==================================================================
* The heart has its reasons which reason does not know. - Pascal *
==================================================================

Ron Evans

unread,
Aug 3, 1991, 8:37:15 PM8/3/91
to
In article <qmZ46...@dorsai.com> myc...@dorsai.com (Keith Kushner) writes:

>m...@ccrwest.UUCP (Mark Prysant) writes:
>> First, we have to make the assumption that Monty will always reveal a
>> door, it will be a loser, and the loser is chosen randomly among
>> the losers which you did not choose (this is the tacit assumption
>> almost everyone seems to make when seeing this puzzle).

>Well put! Of course, you'll be hard-put to _justify_ this assumption
>from the problem as stated, but since "almost everyone seems" to make


In a very recent post entitled something like
"The Monty Hall Problem: Description Wanted",
Mr. Turpin gave an unambiguous description of the problem.
One does NOT need to know that Monty opens randomly
when he has a choice of two doors to open.
Mr. Turpin is correct. Mr. Kushner is wrong.

Keith Kushner

unread,
Aug 3, 1991, 11:02:25 AM8/3/91
to
m...@ccrwest.UUCP (Mark Prysant) writes:

>
> First, we have to make the assumption that Monty will always reveal a
> door, it will be a loser, and the loser is chosen randomly among
> the losers which you did not choose (this is the tacit assumption
> almost everyone seems to make when seeing this puzzle).
>

Well put! Of course, you'll be hard-put to _justify_ this assumption
from the problem as stated, but since "almost everyone seems" to make
the identical assumption (Do they really? Those that insist it's a
50/50 chance if the player switches certainly don't.) it must, of
necessity in a democratic society, be true. You have come up with the
way to solve _any_ difficult problem in the hard sciences: Vote
on it. How Politically Correct of you...

One might as well make the "tacit assumption that Monty will" only
give the player the chance to switch if the player has already picked
the winning door as to make the assumption you have. Why - aside from
your unproven assertion that "almost everyone seems to" think so -
should your assumption have any more validity than that one, or any
other?

Leads Network News

unread,
Aug 3, 1991, 1:38:56 PM8/3/91
to
In article <yNu26...@dorsai.com>, myc...@dorsai.com (Keith Kushner) writes:
>kal...@remus.rutgers.edu (Theodore A. Kaldis) writes:
>
>> Alton Harkcom still doesn't get it. Quite simply: The player picks
>> one of three doors, of which only one is the potential winner. The
>> probability of the player picking the winning door is 1/3. The MC
>> will then reveal a losing door, and subsequently give the player an
>> opportunity to switch to the remaining unpicked door. In 1 out of
>> three cases, the player would be switching from the winning door to a
>> losing door, and in 2 out of three cases, the player would be
>> switching from a losing door to the winning door.
>> --
>
>Ah, but can you find support for your explanation of what the "MC will
>then" do in the problem as given? I doubt you can: the problem merely
>stated that the MC reveals a booby prize behind door three. It is
>only your assumption that the MC will *always* reveal a booby prize.

Come on, let's put this thing to rest. How much clearer can Mr. Kaldis
say it : "The MC will then reveal a losing door". This is an a priori
ground rule for the game, not a recap of what happened during one particular
trial of the game. So, yes, I think there is plenty of support for making
the assumption that the MC will *always* reveal a losing door. True, if
you were a real contestant on a real show, you would not be told the MC's
strategy in advance, and none of the analysis applies. However, THIS IS
ONLY A HYPOTHETICAL QUESTION!. You *DO* know the MC's strategy, and the
analysis takes off from that point.

Do we have to get bogged down in semantics on every probability problem that
comes up in this group? Most of the problems are adequately stated, and for
the ones that are not, the intended statement can usually be discerned. It
seems that some people make a conscious effort to devise some convoluted
interpretation, knowing good and well what the intended premises of a given
problem are. There are few problems I have run across that have been
truly ambiguous.

Charlie Geyer

unread,
Aug 3, 1991, 7:21:46 PM8/3/91
to
In article <17...@leadsv.UUCP> mars...@force.decnet.lockheed.com writes:

> Do we have to get bogged down in semantics on every probability problem that
> comes up in this group? Most of the problems are adequately stated, and for
> the ones that are not, the intended statement can usually be discerned. It
> seems that some people make a conscious effort to devise some convoluted
> interpretation, knowing good and well what the intended premises of a given
> problem are. There are few problems I have run across that have been
> truly ambiguous.

People don't like probability. It is counterintuitive. Thus they want to
find something "wrong" with the problem.

Monty and the three doors is a counterexample to an ingrained heuristic
for human decision making. You make a "fearless forcast." Some time goes
by, new data comes in, and none of it contradicts your original guess.
Do you change your mind now? No, not even if your original forecast was a
pure guess. The fact that it has held up so far tells you that you must
have made a good guess. Right? Wrong! The Monty problem shows why.

So nobody likes the conclusion that you should switch even though the new
evidence doesn't contradict your original guess. This means that a lot of
other "fearless forecasts" that people are so fond of should be continuously
second guessed, even before they are shot down by data. But who can think
that way?

So we keep sticking to our original guesses. Our decision making remains
suboptimal. But we don't like to think of ourselves as irrational.

--
Charles Geyer visiting Seattle
Department of Statistics Department of Statistics
University of Chicago University of Washington
ge...@galton.uchicago.edu ge...@stat.washington.edu

David M.V. Utidjian

unread,
Aug 5, 1991, 3:11:41 AM8/5/91
to
In article <52966...@dorsai.com> myc...@dorsai.com (Keith Kushner) writes:
>
>I see... because you "think there is plenty of support for making the
>assumption that the MC will *always* reveal a losing door," it
>becomes fact.
>

It is not neccessary to make the assumption that the MC will
*always* reveal a losing door in order to solve the problem as stated.

>
>Shall we examine the question again?
>
> "Suppose you're on a game show, and you're given a choice


>of three doors. Behind one door is a car; behind the others,

>goats. You pick a door - say, No.1 - and the host, who knows
>what's behind the doors, opens another door - say, No.3 -


>which has a goat. He then says to you, 'Do you want to pick

>door No.2?' Is it to your advantage to switch your choice?"
>--- Parade Magazine, 17 February 1991, page 12.
>

If I was on a game show and everything transpired as stated
above, my best choice would be to switch. That answers the question
correctly regardless of what the MC does at other times.

>
>The article then goes on to say "(Try to forget any
>particular television show)" so you have no recourse
>to a priori assumptions.
>

Regardless of the game show, it makes no difference to the
problem as stated in PM.

>Now, from the question, as given, can you justify your assumption that
>the MC will *always* reveal a losing door?
>

No. It is only important that the game show host reveal a
losing door at the time I play the game, otherwise it is a different
problem entirely and not the one that was stated.

From watching "Let's Make a Deal", the host would have
variations on the theme. Sometimes there were desirable prizes
behind each door, sometimes there were more than three doors,
sometimes there was only one undesirable prize....etc
It still makes no difference to the problem as stated in
PM, or the correct answer.... Choice1 = 1/3, Choice2 = 2/3.

That is that.

-Dave-
Utidjian at your service...

Ron Evans

unread,
Aug 4, 1991, 11:55:53 PM8/4/91
to
In article <g9066...@dorsai.com> myc...@dorsai.com (Keith Kushner) writes:

>However, since Mr. Turpin's unambiguous description was of a problem
>that varied in some points from the problem that appeared in print,
>it's not a description of THE problem, which, of course, is
>ambiguous. Indeed, my point is that the question, as written, is
>ambiguous.

And my point is that if you are going to support the specific
insertion of an additional hypothesis, then you should state
the precise version of the problem to which *you* are referring
rather than adding to the confusion.

Keith Kushner

unread,
Aug 4, 1991, 4:21:15 PM8/4/91
to
net...@leadsv.UUCP (Leads Network News) writes:

> In article <yNu26...@dorsai.com>, myc...@dorsai.com (Keith Kushner) write
> >

> >Ah, but can you find support for your explanation of what the "MC will
> >then" do in the problem as given? I doubt you can: the problem merely
> >stated that the MC reveals a booby prize behind door three. It is
> >only your assumption that the MC will *always* reveal a booby prize.
>
> Come on, let's put this thing to rest. How much clearer can Mr. Kaldis
> say it : "The MC will then reveal a losing door". This is an a priori
> ground rule for the game, not a recap of what happened during one particular
> trial of the game. So, yes, I think there is plenty of support for making
> the assumption that the MC will *always* reveal a losing door. True, if
> you were a real contestant on a real show, you would not be told the MC's
> strategy in advance, and none of the analysis applies. However, THIS IS
> ONLY A HYPOTHETICAL QUESTION!. You *DO* know the MC's strategy, and the
> analysis takes off from that point.
>
> Do we have to get bogged down in semantics on every probability problem that
> comes up in this group? Most of the problems are adequately stated, and for
> the ones that are not, the intended statement can usually be discerned. It
> seems that some people make a conscious effort to devise some convoluted
> interpretation, knowing good and well what the intended premises of a given
> problem are. There are few problems I have run across that have been
> truly ambiguous.

I see... because you "think there is plenty of support for making the
assumption that the MC will *always* reveal a losing door," it
becomes fact.


Shall we examine the question again?

"Suppose you're on a game show, and you're given a choice
of three doors. Behind one door is a car; behind the others,
goats. You pick a door - say, No.1 - and the host, who knows
what's behind the doors, opens another door - say, No.3 -
which has a goat. He then says to you, 'Do you want to pick
door No.2?' Is it to your advantage to switch your choice?"
--- Parade Magazine, 17 February 1991, page 12.

The article then goes on to say "(Try to forget any
particular television show)" so you have no recourse
to a priori assumptions.

Now, from the question, as given, can you justify your assumption that
the MC will *always* reveal a losing door?

Keith Kushner

unread,
Aug 4, 1991, 4:46:39 PM8/4/91
to
rev...@euclid.ucsd.edu (Ron Evans) writes:

> In a very recent post entitled something like
> "The Monty Hall Problem: Description Wanted",
> Mr. Turpin gave an unambiguous description of the problem.
> One does NOT need to know that Monty opens randomly
> when he has a choice of two doors to open.
> Mr. Turpin is correct. Mr. Kushner is wrong.
>
> --
> Ron Evans, Department of Mathematics, UCSD (rev...@math.ucsd.edu)

However, since Mr. Turpin's unambiguous description was of a problem


that varied in some points from the problem that appeared in print,
it's not a description of THE problem, which, of course, is
ambiguous. Indeed, my point is that the question, as written, is
ambiguous.

Timothy Alan Mohler

unread,
Aug 4, 1991, 9:50:41 PM8/4/91
to
Article <52966...@dorsai.com>, myc...@dorsai.com (Keith Kushner):

>Now, from the question, as given, can you justify your assumption that
>the MC will *always* reveal a losing door?

Common sense? Because it would be pretty pointless for the MC to reveal
which door had the prize, and then asked you if you wanted to switch?

Tim
----
I've shredded my copy of the bible to build a paper mache image of a bull,
which I've painted gold, and use as a centerpiece for Satanic Rituals to
restore Ted Kaldis's manhood.
-Owen Rowley

Keith Kushner

unread,
Aug 5, 1991, 4:21:48 AM8/5/91
to
rev...@euclid.ucsd.edu (Ron Evans) writes:

Actually, I could take exception to that statement, seeing as how it's
not qualified by "in my opinion" or something similar, and request that
you point out where "Mr. Kushner is wrong" in anything I've posted here
that predates that response. All I've done is point out the various
points-of-view of the problem and request that other posters justify,
from the problem as stated, their assumptions that Old Monty will always
reveal a goat behind one of the unpicked doors. And you will note that
most of those posters stated that this *was* an assumption; this shows
that they, too, were responding to the "ambiguous" form of the question
rather than an unambiguous one, else they'd not have called it such.

"First we have to make the assumption...." "This is the tacit
assumption almost everyone seems to make..." This seems to point
out that it was the Parade version of the problem that everyone
save you were considering, a version that is, indeed, ambiguous, but
which is the version that this thread is about.

MICHAEL MASCHLER

unread,
Aug 4, 1991, 4:48:57 AM8/4/91
to
In article <1991Jul30.0...@cadence.com>, sco...@cadence.com (Scott Dixon) writes...

>I have been following the "Monty on the Front Page" thread with great interest.
>When I first read the Marilyn Vos Savant Article in the "Parade" section of
>the San Jose Mercury News, I thought "She's wrong!. She sure made a fool of
>herself." In trying to prove her wrong, I kept proving her right. I even
>simulated the problem with Verilog (a hardware description language) to win
>a free lunch from a colleague. However, the problem continued to bother me
>because I couldn't come up with a simple way to describe the 2-to-1 advantage
>of switching versus non-switching. I eventually developed the following
>resoning, which seems to make the choice of switching very obvious.
..
How about forming a Usenet group dedicated solely to Monty problem?...

Michael Maschler One should observe two rules
Department of Mathematics in order to succeed in one's career:
The Hebrew University
Jerusalem, Israel 1. Never reveal to others everything
you know.

Barry Schwartz

unread,
Aug 5, 1991, 7:10:30 PM8/5/91
to
In article <52966...@dorsai.com>
myc...@dorsai.com (Keith Kushner) writes:

]Shall we examine the question again?


]
] "Suppose you're on a game show, and you're given a choice
]of three doors. Behind one door is a car; behind the others,
]goats. You pick a door - say, No.1 - and the host, who knows

^^^^^^^^^
]what's behind the doors, opens another door - say, No.3 -
^^^^^^^^^^^^^^^^^^^^^^^
]which has a goat. He then says to you, 'Do you want to pick


]door No.2?' Is it to your advantage to switch your choice?"
]--- Parade Magazine, 17 February 1991, page 12.
]
]
]The article then goes on to say "(Try to forget any
]particular television show)" so you have no recourse
]to a priori assumptions.

Whatever you say, Himey.

]Now, from the question, as given, can you justify your assumption that


]the MC will *always* reveal a losing door?

"Electronics are in me."


--
Vote YES for KEVIN DARCY to sit beside Theodore A. Kaldis
on the COUNCIL OF NET.IDIOTS. Send your vote to
kee...@crdgw1.crd.ge.com and help Kevin win!

Daniel Lance Herrick

unread,
Aug 5, 1991, 9:48:04 PM8/5/91
to
In article <52966...@dorsai.com> myc...@dorsai.com (Keith Kushner) writes:
>Shall we examine the question again?
>
> "Suppose you're on a game show, and you're given a choice
>of three doors. Behind one door is a car; behind the others,
>goats. You pick a door - say, No.1 - and the host, who knows
>what's behind the doors, opens another door - say, No.3 -
>which has a goat. He then says to you, 'Do you want to pick
>door No.2?' Is it to your advantage to switch your choice?"
>--- Parade Magazine, 17 February 1991, page 12.
>
>
>The article then goes on to say "(Try to forget any
>particular television show)" so you have no recourse
>to a priori assumptions.
>
>Now, from the question, as given, can you justify your assumption that
>the MC will *always* reveal a losing door?

Your quote from Parade asks for a strategy to apply to one play
of a game in which events were as described. We only get to play
once, but we want to base our bet on probabilities that apply to
the situation facing us.

Hence, we assume that our situation was drawn from a hat containing
several million game events identical in all relevant details to
what we know about the one event that faces us. One of the details
assumed identical in all game events in the hat is that the MC does
what he did in the event facing us.

So, we use language like, "assume we have a large collection of events
all identical in the features we know about the one interesting event
and having their unknown features drawn from a uniform distribution"

Except we don't use language like that, we abbreviate it to the
language that is causing so much confusion here.

dan herrick d...@NCoast.org

Barry Schwartz

unread,
Aug 5, 1991, 3:21:10 PM8/5/91
to
In article <yNu26...@dorsai.com>
myc...@dorsai.com (Keith Kushner) writes:

]kal...@remus.rutgers.edu (Theodore A. Kaldis) writes:
]
]> Alton Harkcom still doesn't get it. Quite simply: The player picks
]> one of three doors, of which only one is the potential winner. The
]> probability of the player picking the winning door is 1/3. The MC
]> will then reveal a losing door, and subsequently give the player an
]> opportunity to switch to the remaining unpicked door. In 1 out of
]> three cases, the player would be switching from the winning door to a
]> losing door, and in 2 out of three cases, the player would be
]> switching from a losing door to the winning door.
]> --
]
]Ah, but can you find support for your explanation of what the "MC will
]then" do in the problem as given? I doubt you can: the problem merely
]stated that the MC reveals a booby prize behind door three. It is
]only your assumption that the MC will *always* reveal a booby prize.

...

]You might guess which view of the problem I hold. Still, if you


]believe you can demonstrate from the problem as stated that your
]assumptions about it are indeed correct, please do so.

BZZZZZZZZZZT!!!!

Go directly to Jail. Do not pass Go. Do not collect $200.


[The purpose of computing is insight, not numbers. -- Hamming]

Andrew Clayton

unread,
Aug 6, 1991, 8:39:11 AM8/6/91
to
In article <1991Aug6.0...@NCoast.ORG>, Daniel Lance Herrick writes:

> In article <52966...@dorsai.com> myc...@dorsai.com (Keith Kushner) writes:
> >Shall we examine the question again?
> >
> > "Suppose you're on a game show, and you're given a choice
> >of three doors. Behind one door is a car; behind the others,


Suppose you are in a newsgroup, and you are presented with the
following people to flame;

Kevin Darcy.
Kent Paul Dolan.
Patricia O Tuama.
Barry Schwartz.

The favourite target it ol' "bloatobix for breakfast, again!" Trish,
but you have to wait for the prizes first;

Behind the stars are the following

o One tattered combined dictionary/thesaurus with 'Kevin is a
Wonker' inscribed on the front cover.

o A hit of Thorazine

o Two dozen giant economy size packs of Tampax

o Three oldish books on relativity and a slightly masticated fish

Who would YOU chose to flame?

[gur nafjre, bs pbhefr, vf gbgnyyl veeryrinag.]

Dac
--

James D Dolan

unread,
Aug 6, 1991, 10:45:02 AM8/6/91
to
In article <Icb_vli00...@andrew.cmu.edu>, tm...@andrew.cmu.edu
(Timothy Alan Mohler) writes...

>Article <52966...@dorsai.com>, myc...@dorsai.com (Keith Kushner):
>
>>Now, from the question, as given, can you justify your assumption that
>>the MC will *always* reveal a losing door?
>
>Common sense? Because it would be pretty pointless for the MC to reveal
>which door had the prize, and then asked you if you wanted to switch?


no, your common sense is dead wrong. the mc's policy might be to pick the door
to reveal at random, but then to offer you the opportunity to switch, only if
the revealed door did not have the prize. this would be no more senseless than
any other aspect of this whole stupid problem, and it would give the exact same
"filtering" effect (as described quite clearly by kushner) as before, on those
outcomes where the revealed door does not have the prize.

in general, the effect of a filter is unaffected by the manner in which you
treat those inputs that are already guaranteed not to pass the filter.

james dolan v088...@ubvms.cc.buffalo.edu

J. A. Durieux

unread,
Aug 7, 1991, 7:04:54 AM8/7/91
to
In article <b9p3d...@wang.com> masc...@vms.huji.ac.il writes:
>How about forming a Usenet group dedicated solely to Monty problem?...

Well, a newsgroup rec.puzzles.probabilistic might not be such a bad idea.
There are lots of counterintuitive probabilistic puzzles (Monty Hall,
switching envelopes, other child a girl, Petersburg Paradox, ...),
the errors made by half of the poster are mostly the same all the time,
as is the stubbornness with which people stick to their erroneous viewpoints,
they are (generally) boring for those who understand probability, and they
tend to evoke *long* discussions.

Jan Willem Nienhuys

unread,
Aug 7, 1991, 9:15:15 AM8/7/91
to
In article <10...@star.cs.vu.nl> xe...@cs.vu.nl (J. A. Durieux) writes:
>Well, a newsgroup rec.puzzles.probabilistic might not be such a bad idea.
>There are lots of counterintuitive probabilistic puzzles (Monty Hall,
>switching envelopes, other child a girl, Petersburg Paradox, ...),

Here is something I found. I wrote a popular piece about the Benveniste
flap. I explained the Poisson distribution. Example: take some rice, and
mix it with a small portion of rice that you have colored black with ink,
say. Take a spoonful out of the mixture. Count the black grains. Repeat.
You will find that the variance about equals the mean.

I have been severely berated for maintaining such stupid ideas, and for
not knowing that in general there is no relation between mean and
variance. Needless to say, the critics never took the trouble to do
the experiment (which I had done).

And of course one of the top immunologists of France, even after being
explained this by Randi, Maddox and Stewart, kept saying that this was
mere theory and that in his laboratory the variances were much smaller,
because counting was done much more careful ....

I'm afraid that the people who ought to read rec.puzzles.probabilistic,
won't.

JWN

Andrew Ofiesh

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Aug 9, 1991, 10:28:12 AM8/9/91
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I picked up this thread late so please excuse me if I have repeated anything.

The problem is very simple if you look at it correctly. Suppose you choose #1 and Monty says "You may now choose the other 2 doors instead, and by the way, as you know at least one of them is a goat. Also if you don't mind, I will show you what one of the goats looks like. Whatever goat(s) you win, you may leave here." Then Monty opens up door #3 to reveal one of the two goats. The fact that Monty tells you something you already know and shows you a goat should not dissuade you from picking the other tw

o doors instead of your original one because you then have two chances to win. The fact that you now know which of the other two doors is certain to have a goat behind it is irrelevant.

However, if you know that he may open up the door with car behind it things change. In that case, if he opens door #3 and you don't see the car, you know your chances are no longer 2 to 1 if you choose the other two doors. Also, if you know that when there are two goats behind the other two doors Monty will tend to choose #3, then Monty is telling you something you need know when he reveals the goat. This because when he opens #2 it is more likely that he didn't open #3 because there was car there and

not a goat, and when he opens number #3 it is less likely that he was forced to pick #3 because there was a car in #2. The statements above can be proven with Bayes Rule.

So for this problem to work you have to assume that Monty will never show the contestant a car, and he will always pick randomly when given a choice of goats to display. This is a common theme in Bridge play,called restricted choice, and more discussion of this topic can be found on occasion in rec.games.bridge. Also, it is interesting that this problem is equivalent to the problem, "If you flip two coins and one of them comes up heads, what are the odds that the other one comes up tails?.

Andrew Ofiesh "All tautologies are tautologies"

Alton Harkcom

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Aug 6, 1991, 7:31:14 PM8/6/91
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In article <4...@ccrwest.UUCP> m...@ccrwest.UUCP (Mark Prysant) writes:

=}In article <HARKCOM.91...@spinach.pa.yokogawa.co.jp>
=} har...@spinach.pa.yokogawa.co.jp (Alton Harkcom) writes:
=}
=} > There are three doors. We'll call them W, L1, and L2. The probability
=} >that the player will choose W, L1 or L2 is .3_ (where _ means continue to
=} >infinity). The MC will reveal either L1 or L2. Either the player chose
=} >W (.3_) or the remaining L door (.3_). If the player switches, then he
=} >merely trades his door's .3_ probability of being the winner for the
=} >other door's .3_ probability of being the winner...
=}
=}Didn't your mother teach you that probabilities should sum to 1?

No, I read it in a text on probability. Some of us come from
backgrounds where our parents weren't given the educational
opportunity to be able to teach us such things...

=}Tsk. Tsk. If you are going to flame someone it would be a good idea to
=}know that the original poster is wrong and that you are right,
=}neither of which applies in this case.

Yes, my solution was wrong. A sample set of data gave the number of
wins when switching to be twice that of the number of wins when not
switching. Another look at my solution shows that my assumption that
the probability of choosing A losing door was .3_, it is actually .6_

=}(where did you get this .3_ notation idea?)

I don't use standard notations... A bad scientist indeed...

=}Apology not accepted.

Who cares if you accept it or not...

=} It's difficult to justify an incorrect flame.

Is there such a thing as a 'just flame'? If you don't want that
apology then try the following one...

I apologize to the readers of sci.math and rec.puzzles for posting
an incorrect solution in your groups (especially since I don't read
them). I also apologize for posting on a topic which apparently isn't
welcome in this group.

Al

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