On 28 Nov 95 03:03pm, ROBERT J  PEASE wrote to ALL:

 RJ> People are missing the MAIN POINT.
 RJ> From her comments, she obviously didn't UNDERSTAND the
 RJ> difference between square feet and feet square.

That makes me wonder if she really is the world's smartest
person. <g>

Jud McCranie  jud.mccra...@camcat.com

 * Silver Xpress V4.02B03P SW20178

In article <8B5F3D6.02AE000434.uu...@swsbbs.com>,

As I understand it, Marilyn vos Savant's claim to fame is that she has the
highest IQ test score submitted to the Guinness Book of World Records.
A person's intelligence, however, isn't a singly-quantifiable attribute as,
say, height or weight.  I think it'd be as difficult to determine the world's
smartest person as to determine, for example, the world's most moral person.

Perhaps it's best to think of her as an advice columnist with a penchant for
puzzles.

Top
--
W. Top Changwatchai
w...@cs.utexas.edu

In article <8B5F3D6.02AE000434.uu...@swsbbs.com>,

Jesus, give it a rest!  Everyone makes mistakes, no matter how
much raw intelligence they have.  Besides, it's possible her
editor *changed* it to square feet.

You all idolize this woman's intelligence as if it's a gift
from God.  Some magical power that you can never posess.
I've got an idea.  Instead of desperately trying to salvage
your last few shreds of self-esteem by looking for proof that
brings her down to your level, why don't you spend your time
improving yourself?  Then you would actually have something
to be happy about.

-TomM

On 1 Dec 95 06:43pm, TOM MACIUKENAS wrote to ALL:

 TM> You all idolize this woman's intelligence as if it's a gift
 TM> from God.

Actually I don't think she's nearly as smart as she thinks she
is.  She scored very high on some imperfect IQ test.  But she
hasn't shown much intelligence otherwise.

Jud McCranie

 * Silver Xpress V4.02B03P SW20178

Thank you for debunking this quack at a much higher level than I
could. I simply did not like the way she put down the
mathematicians who tried to explain the "Monty Hall Puzzle" to
her. She doesn't understand the problem or the solutions
presented to her. And she refuses to even consider that she might
be mistaken.

--
- Don                                       dch...@relay.nswc.navy.mil

|>In article <9512040109.H...@AIC.NRL.Navy.Mil> h...@aic.nrl.navy.mil (Dan Hoey) writes:
|>>> Perhaps it's best to think of her as an advice columnist with a penchant for
|>>> puzzles.
|>>
|>> ... who fancies herself an expert in mathematics.
|>> ... who takes pride in her opposition to the "mathematical establishment".
|>> ... who was obviously out of her depth in writing a book that tried
|>>         to explain Andrew Wiles's proof of Fermat's Last Theorem.
|>> ... who nonetheless made up a spurious theory that Wiles's proof was not
|>>         applicable to FLT because of its use of "hyperbolic geometry".
|>> ... who has never explained how she came to the mistaken belief that
|>>         there was any use of hyperbolic geometry in that proof.
|>>
|>>Perhaps if she goes on making grade-school level mathematical errors,
|>>the public may come to understand how incompetent she is to critique
|>>serious mathematics.  If ever a fraud deserved deflation, she's it.
|>>
|>
|>Thank you for debunking this quack at a much higher level than I
|>could. I simply did not like the way she put down the
|>mathematicians who tried to explain the "Monty Hall Puzzle" to
|>her. She doesn't understand the problem or the solutions
|>presented to her. And she refuses to even consider that she might
|>be mistaken.

Could you elaborate on the "Monty Hall Puzzle" incident? If it's what I'm
thinking of, she and Gardner were correct, and all the mathematicians who told
her she was wrong were incorrect.

--

Chris Volpe                             Phone: (518) 387-7766 (Dial Comm 8*833
GE Corporate R&D                    Fax:   (518) 387-6560
PO Box 8, Schenectady, NY 12301         Email: volp...@crd.ge.com

Failing to be unsuccessful at not being unclear,

Alas, MVS is guilty of spreading misinformation about subjects other
than mathematics.

As a frequent lurker in alt.folklore.urban, I can assure you that she
has been a vector for the spread of a number of blatantly false urban
legends (ULs), the most notorious being the old "glass flows"
chestnut.

For those unfamiliar with this, it is the (mistaken) belief that glass
is a liquid, and actually flows over time.  Unevenness and varying
thickness of extremely old glass in windowpanes is often cited as
"proof".  However, such irregularities in thickness were due to the
manufacturing processes in use at the times.  The AFU archives at
cathouse.org have more detailed info, if anyone is interested....

In any case, MVS may have the title in the Guinness Book of Shameless
Self-Promotion, but she is far from being the "World's Smartest"
person....

 - Rick "Motto!" Dickinson

w...@cs.utexas.edu (Wichaya Top Changwatchai) writes:

I've heard the test that distinguishes her score from those of other
expert test-takers was one she had a hand in developing or
administering, at least in an earlier version.  It's amusing to
speculate whether this experience gave her an advantage in taking the
test.

 ... who fancies herself an expert in mathematics.
 ... who takes pride in her opposition to the "mathematical establishment".
 ... who was obviously out of her depth in writing a book that tried
         to explain Andrew Wiles's proof of Fermat's Last Theorem.
 ... who nonetheless made up a spurious theory that Wiles's proof was not
         applicable to FLT because of its use of "hyperbolic geometry".
 ... who has never explained how she came to the mistaken belief that
         there was any use of hyperbolic geometry in that proof.

Perhaps if she goes on making grade-school level mathematical errors,
the public may come to understand how incompetent she is to critique
serious mathematics.  If ever a fraud deserved deflation, she's it.

Dan Hoey
H...@AIC.NRL.Navy.Mil

Uh.

That may not be the best example you could have chosen!
She was, in fact, right about that problem, and quite a few serious
mathematicians were wrong.

The sad thing is that this episode apparently convinced her that she
didn't have to listen to serious mathematicians anymore, and that
conclusion in turn led her to write some embarassing things about the
Wiles proof.

-Jamie

I think she was. She may not have stated it explicitly enough, but she
gave the correct answer to the intended problem.

(Classic Monty Hall problem description, deleted)

Ok, you're right, this problem is underspecified.

Depends what you mean by 'agree with these analyses.'
In the first place, the relevance of a maximin solution is unclear here.
In the second place, the two options have exactly the same minimum (a
goat), so either choice maximizes the minimum payoff.

That is the solution to the Classic problem, of course.

I don't understand this (uncommon) usage of 'begs the question'.

Fair enough. (Except for the guy who actually admitted that he was
wrong--I'm sure I've seen excerpts from that letter, possibly in the NY
Times.)

-Jamie

I take it that readers of this newsgroup are aware that IQ tests are considered
biased and a bad indication of ones intelligence. For one they are thought of
as biased against cultures or race (a person scoring high in a Chinese IQ test
may not score highly in one for Australians). Anyway, isn't it EQ (Emotional
Quotient) that's the in thing these days - if I remember correctly it covers
things like empathy amongst other things.

Cheers,

      _                                 "If a picture paints
     | |                                 a thousand words,
     | | ___ _ __ ___   ___ _ __         then doodles don't
 _   | |/ _ \ '_// _ \ / _ \ '_  \       amount to much."
| |__| |  __/ | | (_) |  __/ | | |             - HEADLESS CHICKENS
 \____/ \___|_|  \___/ \___|_| |_|

E-mail : jer...@gp.co.nz

I'm not sure if this is "classic game-theoretic analysis," but the
host, by varying the conditional probabilities with which he takes
various actions (revealing a goat, allowing the contestant to switch,
and so forth) can alter the expected values of "switch" and "not
switch."  Suppose we rephrase the situation as a game played between
the host and the contestant as follows.  At the beginning of the game,
before any choices are made, the host and the contestant each secretly
write out their strategies in enough detail so that the rest of the
game can be played out by a computer (using some suitable random
number as input to make any probabilistic choices).  Then we have a
two-player game in which each player makes one move at the same time
as the other, without knowledge of the other player's move.

Now any pure (non-probabilistic) strategy the host might take
falls into one of the following four categories:

  C:  Offer the switch if the contestant chose the car, but not if
      she chose the goat.
  G:  Offer the switch if the contestant chose the goat, but not if
      she chose the car.
  A:  Always offer the switch regardless of the contestant's choice.
  N:  Never offer the switch.

Most generally, the host's (possibly mixed) strategy will be
equivalent to some probabilistic combination of the above strategies.

The contestant's strategy meanwhile is some probabilistic mix of the
following:

  Sw:  Switch if the switch is offered.
  St:  "Stay" (don't switch) if the switch is offered.

Now fill in a payoff matrix for the contestant using the *expected*
number of cars that the contestant will win during the game.  (The two
players' moves do not themselves determine whether the contestant will get
the car; there is one more variable, the 1/3 chance that the contestant
will initially choose the door that hides the car.)  The matrix then is:

    Sw  St

C   0   1/3

G   1   1/3

A  2/3  1/3

N  1/3  1/3

Clearly the contestant's best worst case is in the "Stay" column
(value 1/3).  The host's best worst case is in rows C and N, i.e. the
host loses at most 1/3 expected value as long as he never offers the
switch when the contestant initially chooses a goat.

Where this differs from my understanding of a "classic" game theory
approach is that I have taken two of the choices of the host and
contestant, namely where the car is hidden and which door to choose
first, and eliminated them from the payoff matrix, replacing them with
the assumption that the contestant guess right the first time with
probability 1/3.  I did this because, intuitively, if both the host's
and contestant's "pure" strategies include a specific choice of door
(e.g., "put the car behind door 1," "choose door 1 and then switch if
offered"), the player's worst case payoff for any pure strategy is 0
(e.g., when the host's strategy puts the car behind one of the other
two doors and then never offers a switch), but the player can still
get 1/3 payoff from the mixed strategy of "choose door N and don't
switch" where N=1, 2, or 3 with probability each 1/3.  (And no, this
isn't the complete argument, but I hope you can see where it's going.)

-- David A. Karr (k...@cs.cornell.edu)
-- <URL:http://www.cs.cornell.edu/Info/People/karr/home.html>

Marilyn vos Savant, the self-styled world's most intelligent woman
(why not "person"?) gets wrong the majority of physics questions
that she tackles.

Her record is so appalling that I don't think it is due to chance
or ignorance.  I think she is deliberately misphrasing these questions
such that specialists will call her on it, and whip up interest in
her columns.   Most non-specialists will not follow the debate, and
(here's the best part) *she* gets to judge the outcome of the debate
by printing selected extracts in her column!

She got wrong the old "falling bullets" question a few years ago.
She claimed they "returned harmlessly to earth".  I wrote to her
enclosing newsreports of people in the Bay Area who had been killed
by bullets falling on them (there are one or two such tragedies
each year here, at New Years Eve).

She never replied, but in a later column, she said that her example
explained what happens "in theory" and was never intended to apply
to the real world.   Of course, if you abstract away the friction
caused by the bullet rubbing against the air as it falls, so that
you have a theoretical model, her wrong statement becomes even more
acute.   So she couldn't even get her cover-up excuse right.

She did do something that I admire: she was one of the few columnists
to come out and say in print that of course OJ Simpson should have
been found guilty, and here's why.   Very few other opinion and
advice columnists had the guts to give it straight from the shoulder
like that.

ObPuzzle: When someone says "His name is mud" is there any connection
      to the Dr Mudd who fixed the leg of Lincoln's assassin?

--
Peter van der Linden       lin...@Eng.sun.com        Fun in California.        

 We changed our minds: now we're saying "the computer is the network", OK?

pl436...@brownvm.brown.edu (Jamie Dreier) writes, failing to cite

Uh indeed.

That was indeed not the best example (and one I chose not to raise),
but no, she was _not_ "right about that problem".  The explanation is,
I'm afraid, somewhat complicated, but since the question has been
raised, here it is.

There is a "classic" Monty Hall problem, that goes something like
this:

    "There is a game where there are three doors, one of which
    conceals a prize).  The game always proceeds in three steps:
        1. The player chooses a door,
        2. A second door is opened to reveal the lack of a prize, and
        3. The player must choose between the first door and the
           third (remaining) door.
    How should the player choose in step 3 to maximize the probability
    of taking the prize?"

which is easily confused with the "common" Monty Hall problem, which
was posed as follows (from a 1990 "Ask Marilyn" column):

    "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?"

The two problems are not the same, because in the second there is no
assertion that the host's always offers "you" a choice to switch.  In
fact, we might imagine that had "you" chosen a different door at
first, you might have been sent home with your first choice.  The
distinction is not immediately obvious (and leads to a problem
symmetry that may be a reason for the common mistakes with the classic
solution), but the mathematics is clear:
    In the "classic" problem, it is always to the player's advantage
        to switch, by straightforward Bayesian probability, but
    In the "common" problem, the player can maximize his minimum
        payoff by not switching, by classical game-theoretic analysis.
If you don't agree with these analyses, perhaps you should read the
FAQ or send me e-mail; I've got dozens of explanations filed away.

But Ms. vos Savant's response to the problem was:
    "Yes, you should switch.  The first door has a 1/3 chance of
    winning, but the second door has a 2/3 chance.  Here's a good way
    to visualize what happened.  Suppose there are a million doors,
    and you pick door #1.  Then the host, who knows what's behind the
    doors and will always avoid the one with the prize, opens them all
    except door #777,777.  You'd switch to that door pretty fast,
    wouldn't you?"
an analogy that begs the correct question to the classic problem, but
begs the wrong question to the common problem that was posed.  She
received a number of letters, some from mathematicians, telling her
she was mistaken, and printed excerpts from nine of them.  She also
excerpted a tenth letter telling her she was correct.  There are also
reports of an eleventh letter from a professor who said she was wrong,
then later changed his mind.  But the excerpts never included
discussion of the problem, just a few phrases that (perhaps out of
countext) sound insulting or abusive or arrogant.  There is no way to
tell whether she was being incorrectly chastised or was being
corrected on the distinction between the problem that was posed and
the problem that was solved.

As far as I know, she has never addressed the difference between the
two problems.  And I do not know how many of her correspondents tried
to bring that difference to her attention and were rewarded with
misquotation or nonresponsiveness.  But the fact remains that her
answer was neither correct nor relevant to the problem that was posed.

But, as you say, this was not the best example to bring up.  Not
because she was right in her answer--she was not.  But because
whenever this topic is brought up, a number of people (like you) will
simply assert she got it right, ignoring (as she did) the difference
between the two problems, and injecting confusion into the discussion.
That is not particularly helpful when less complicated examples of her
escapades into misinformation, illogic, and dishonesty abound.

Dan Hoey                        posted and e-mailed
H...@AIC.NRL.Navy.Mil

In article <pl436000-0612951017150...@bootp-124.college.brown.edu>, pl436...@brownvm.brown.edu (Jamie Dreier) writes:
|>
|>> which is easily confused with the "common" Monty Hall problem, which
|>> was posed as follows (from a 1990 "Ask Marilyn" column):
|>>
|>>     "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?"
|>
|>Ok, you're right, this problem is underspecified.
|>
|>> The
|>> distinction is not immediately obvious (and leads to a problem
|>> symmetry that may be a reason for the common mistakes with the classic
|>> solution), but the mathematics is clear:
|>>     In the "classic" problem, it is always to the player's advantage
|>>         to switch, by straightforward Bayesian probability, but
|>>     In the "common" problem, the player can maximize his minimum
|>>         payoff by not switching, by classical game-theoretic analysis.
|>> If you don't agree with these analyses, perhaps you should read the
|>> FAQ or send me e-mail; I've got dozens of explanations filed away.

I couldn't obtain the FAQ. It was expired at my site (except for the
unhelpful weekly version), even on news.answers. And rtfm.mit.edu wouldn't
let me connect. So forgive me for asking the following question.

What's the difference? Are you saying it matters whether Monty "knew" which
door to show you, or "intended" to show you a goat? I don't think so. When you
picked the first door, that door had a 1/3 chance of being the correct one.
Unless Monty shows you the car behind another door, the door you've chosen
still has a 1/3 chance of being right. There's a 2/3 chance that one of the
other doors is correct. Since you already know  that one of those two
remaining doors (the one Monty opened) is an incorrect door, the whole 2/3
probability of the car is "bundled up" into the third door. So it is always
advantageous to  switch.

If not, why not?

--

Chris Volpe                             Phone: (518) 387-7766 (Dial Comm 8*833
GE Corporate R&D                    Fax:   (518) 387-6560
PO Box 8, Schenectady, NY 12301         Email: volp...@crd.ge.com

|>In article <4a5itb$...@rdsunx.crd.ge.com> vo...@ausable.crd.ge.com writes:
|>>
|>>What's the difference? Are you saying it matters whether Monty "knew" which
|>>door to show you, or "intended" to show you a goat? I don't think so. [...]
|>
|>Technically, you're right.  What really matters is that *you* know for
|>a certainty that Monty was going to open a door revealing a goat and
|>give you a chance to switch.  As long as you know that, and didn't
|>yourself know were the car was before you made your first choice, then
|>based on your knowledge there is a 2/3 chance, *given* that you picked
|>door #1 and Monty opened door #3, that the car is behind door #2.

So far so good.

|>If Monty behaves differently, the odds can change considerably.
|>Suppose we know for a certainty that Monty *never* gives the second
|>chance to people whose first choice was a goat.  In that case, what
|>should you compute as the probability that the car is beind door #2,
|>given that you picked door #1 and Monty opened door #3 (and gave you a
|>chance to switch)?  Obviously, it's ZERO, because if the car were
|>behind door #2 Monty would never have given you the chance to switch.

Agreed. If events occur that allow you to logically rule out certain
possibilities, then the relative probabilities change.

|>
|>>Unless Monty shows you the car behind another door, the door you've chosen
|>>still has a 1/3 chance of being right.
|>
|>A simple application of Bayes' Theorem will show that this statement
|>is incorrect for a large number of variations of the problem.  In the
|>extreme example I described above, Monty's action makes your first
|>choice have probability 1 of being right.  More generally, the
|>probability that the car is behind the door you first chose is
|>
|>   p/(p + 2q)
|>
|>where p is the probability that Monty gives the second chance to a
|>contestant whose first choice was a car, and q is the probability when
|>the first choice was a goat.
|>
|>For the "classic" version of the problem, p = q = 1, and the correct
|>answer is indeed 1/3 for the door you first picked (and therefore
|>2/3 for the remaining door).  But if p and q are allowed to take on
|>values other than 1, the "1/3" probability can end up being anything
|>in the range 0 to 1.

Can you give a realistic description of the problem in which the chance are
equal (50-50) whether or not the person chooses to swap?

--

Chris Volpe                             Phone: (518) 387-7766 (Dial Comm 8*833
GE Corporate R&D                    Fax:   (518) 387-6560
PO Box 8, Schenectady, NY 12301         Email: volp...@crd.ge.com

Certainly.  Since in the real game show Monty was not *obliged* to
give a second chance, but *was* allowed to offer various "deals" to
the player rather than immediately revealing the player's first
choice, I think it is "realistic" to take any p from 0 to 1 inclusive,
and any q from 0 to 1 inclusive.

Perhaps you meant "easy for Monty to implement."  One way for Monty to
make the outcome 50-50 would be to assign "heads" to the first door
not chosen by the player, and "tails" to the other, and then flip a
coin to tell himself which door to open.  (Or have an off-camera
assistant flip the coin and give him a signal.)  If the coin tells him
to reveal the car, he would then simply try one of his other stunts
instead, something completely different.  This gives p = 1, q = 1/2.

But in fact Monty could rather easily make p and q any values that he
can generate by any random process before the show.  All he has to do
is run the first process, producing a "Yes" or "No" which he then
interprets as the answer to the question, "If the contestant chooses
the door with the car today, will I offer the chance to switch?"
He then uses the other process to answer the same question but with
"goat" instead of "car."

-- David A. Karr (k...@cs.cornell.edu)

In article <4a5itb$...@rdsunx.crd.ge.com>,
Christopher R. Volpe <vo...@ausable.crd.ge.com> wrote:

If Monty disregards the location of the car and simply opens an
unchosen door at random, then the probabilities are:

        contestant wins by switching            1/3
        contestant loses by switching           1/3
        host reveals car; game ends             1/3.

In 1/3 of the cases, the contestant begins by choosing the car.  Since
the remaining two doors both have goats, it doesn't matter which one
Monty opens.  It is guaranteed that Monty will not reveal the car, and
the contestant always loses these games after switching.

In 2/3 of the cases, the contestant begins by choosing a goat.  Given a
cooperative host, the contestant would always wind up switching to the
car after Monty reveals a goat.  If Monty chooses at random, he spoils
half the games that the contestant would have won.

The conditional probability of winning by switching, given that Monty
happens to choose a goat door at random, is 1/2.

Dave Seaman

Technically, you're right.  What really matters is that *you* know for
a certainty that Monty was going to open a door revealing a goat and
give you a chance to switch.  As long as you know that, and didn't
yourself know were the car was before you made your first choice, then
based on your knowledge there is a 2/3 chance, *given* that you picked
door #1 and Monty opened door #3, that the car is behind door #2.

What Monty "knows" or "intends" is really irrelevant except that it's
a mechanism we use to explain why we're sure Monty will always give
the second chance.

If Monty behaves differently, the odds can change considerably.
Suppose we know for a certainty that Monty *never* gives the second
chance to people whose first choice was a goat.  In that case, what
should you compute as the probability that the car is beind door #2,
given that you picked door #1 and Monty opened door #3 (and gave you a
chance to switch)?  Obviously, it's ZERO, because if the car were
behind door #2 Monty would never have given you the chance to switch.

A simple application of Bayes' Theorem will show that this statement
is incorrect for a large number of variations of the problem.  In the
extreme example I described above, Monty's action makes your first
choice have probability 1 of being right.  More generally, the
probability that the car is behind the door you first chose is

   p/(p + 2q)

where p is the probability that Monty gives the second chance to a
contestant whose first choice was a car, and q is the probability when
the first choice was a goat.

For the "classic" version of the problem, p = q = 1, and the correct
answer is indeed 1/3 for the door you first picked (and therefore
2/3 for the remaining door).  But if p and q are allowed to take on
values other than 1, the "1/3" probability can end up being anything
in the range 0 to 1.

-- David A. Karr (k...@cs.cornell.edu)
-- <URL:http://www.cs.cornell.edu/Info/People/karr/home.html>

In fact I don't see any a priori reason to take maximin either.
But it does seem to happen that this approach turns out to work
for this one game.

Informally, here's how I see it: Make this as close to a
game-theoretic situation as you can (rational players, completely
secret decision-making, etc.).  Suppose the host chooses a strategy
that sometimes offers a switch away from a goat.  Either this doesn't
make any difference at all---because contestants never switch in any
case, or they never pick door #1 which is the one that the host is
willing to offer the switch from, or any other complicated condition
you care to name---or this will lose some games to contestants who do
(perhaps by chance) satisfy the conditions for a successful switch.
Meanwhile offering the switch has no effect on contestants who don't
meet the conditions for a switch (choose the wrong door, are unwilling
to switch, etc.).  So the host can always improve such a strategy
(make fewer contestants win) by eliminating the chance that he'll
offer a switch away from a goat, and this is what a rational host
will do if he wants to minimize losses.

A rational player, knowing the above, will realize that any switch
must be away from a car, so any strategy that sometimes switches will
just lose the car in those cases where that rule applies.  Therefore
the player will never switch.

If you try to run this analysis the other way, however---start with
the contestant's decision, then figure out the host's---I don't think
it works, so I suppose it's not really maximin, it just happens to
match the maximin decisions.

I did have such things in mind, but trying to cover all of them
formally would take a lot of effort that I don't think is worth it.

In any case, my analysis above depends on two important assumptions,
one of which I am sure is not true and the other of which I suspect is
not true for the real-life game show:

(1) The contestant has no way of predicting the host's actions except
    by assuming he's a rational player.  In fact in real life the
    contestant would most likely have seen the show before and been
    able to form a statistical estimate of the host's likely behavior
    which could be used to modify or replace the above analysis.

(2) The host plays the game strictly to minimize each contestant's
    chance of winning the car.  I suspect that if the host offered the
    switch only away from the car and never away from a goat, this
    tactic would soon become boring to the viewers of the show, and
    would fail to fool contestants.  If the host occasionally offers a
    switch away from a goat, it spices up the game and has a chance
    to fool more contestants into switching away from cars.

In other words, historical data and psychological effects are relevant
to the real-world case.  But I don't know how to analyze those without
a lot of hard data from the actual game show.  (Such an analysis might
be interesting if the data were collected.  I have read an interesting
article that analyzed a series of actual Jeopardy! games to decide on
optimal betting strategies.)

-- David A. Karr (k...@cs.cornell.edu)
-- <URL:http://www.cs.cornell.edu/Info/People/karr/home.html>

This is the incorrect sentence:

That depends on Monty's criterion of choice. For instance, if his strategy
is never to offer you a choice when you guessed wrong, but only when you
guessed right, then given that he shows you a goat the door you chose has
no chance of having the car!

How the probability 'lost' by the open door gets slooshed over onto the
other two doors depends on (your views about) how Monty decides what to do
next.

-Jamie

On 4 Dec 95 09:24pm, RICK DICKINSON wrote to ALL:

And her logic in this made no sense at all.

 RD> For those unfamiliar with this, it is the (mistaken) belief that glass
 RD> is a liquid, and actually flows over time.  Unevenness and varying
 RD> thickness of extremely old glass in windowpanes is often cited as
 RD> "proof".  However, such irregularities in thickness were due to the
 RD> manufacturing processes in use at the times.

Are all of the panes of glass thicker at the bottom?  Surely
they must have checked this.

Jud McCranie  jud.mccra...@swsbbs.com

 * Silver Xpress V4.02B03P SW20178

Yes, your analysis looks right. I agree that under the given hypotheses,
you maximize your minimum expected payoff by sticking with the door you
chose.

The classic sort of game theoretic analysis doesn't generally go by
maximizing minimum *expected* payoffs, but we can certainly extend the
idea here.

But now the question is, what is the relevance of this maximin? Is there
any reason to suppose it is the 'best' move in any sense? I can't see any
reason whatsoever.

Right--you've left out strategies that mention the numbering of the doors,
as you yourself note below. (I'm not sure you have this in mind: consider
the strategy for Monty, "Don't reveal anything if the contestant picks the
lower-numbered incorrect door, but do reveal the remaining goat if the
contestant picks the higher-numbered incorrect door.)
There's nothing wrong with dividing things this way, though, and treating
the other strategies as equivalent. Depends on your priors, I guess.

Maximin strategies are important in very, very limited contexts, I think.

-Jamie

In article <4a50ik$...@engnews2.eng.sun.com>,
Peter van der Linden <lin...@positive.eng.sun.com> wrote:

It sounds to me like she's just a plain old run-of-the-mill quack.
And let's not forget the Fundamental Law of Human Behaviour Analysis:

"Never attribute to malicious conspiracy that which can be adequately
 explained by simple stupidity."

-TomM

The world may be wacky enough these days for such a thing to exist, but I don't
see that it has any bearing on someone's intelligence.

G*D what a stupid idea.  That's an even more stupid idea than IQ tests.

-ToMm