Monty Hall Makes the Rounds
Herb Brown
>
>Monty Hall is making the rounds in Marily Mach vos Savant's
>column _Ask Marilyn_. She initially answered the question
>incompletely, and then three mathematicians wrote in correcting
>her incomplete answer with equally incomplete answers.
>
>Sigh.
>
>-Chris
What was the question? What was Marilyn's (incomplete) answer?
What were the mathematician's (incomplete) answers? And finally,
what is the CORRECT answer?
Herb
--
----------------------------------------------------------------------------
Herb Brown Math Dept The Univ at Albany Albany, NY 12222 (518) 442-4640
hib...@leah.albany.edu or hib...@cs.albany.edu or hb...@ALBNYVMS.BITNET
----------------------------------------------------------------------------
Bernie Cosell
}Monty Hall is making the rounds in Marily Mach vos Savant's
}column _Ask Marilyn_. She initially answered the question
}incompletely, and then three mathematicians wrote in correcting
}her incomplete answer with equally incomplete answers.
}Sigh.
That's OK --- better her column than HERE, no? [i had actually mused
the other day that we've in the midst of a remarkably long monty-less
stretch!]
/B\
Captain Trott
Monty Hall is making the rounds in Marily Mach vos Savant's
column _Ask Marilyn_. She initially answered the question
incompletely, and then three mathematicians wrote in correcting
her incomplete answer with equally incomplete answers.
Sigh.
-Chris
Kenneth Arromdee
>What was the question? What was Marilyn's (incomplete) answer?
>What were the mathematician's (incomplete) answers? And finally,
>what is the CORRECT answer?
I don't know the way the question was phrased in Marilyn's column, but here's
the general problem: (I _do_ have the Straight Dope column)
First a disclaimer. I know you're tired of seeing this. I also know that
_someone's_ going to post something, and I may as well get it right. (Up to
now, I have not yet seen any responses in this thread stating the problem and
its solution, so I _could_ be the first to post in this particular incarnation
of the Monty Hall thread.
I also think this should go in the rec.puzzles FAQ if it's not already. (What
ever happened to that list anyway?)
Monty Hall gives you a choice of 3 doors, one of which contains a prize. After
you make your choice, Monty picks an empty door other than yours, and shows
you it's empty. Now he asks if you want to switch to the remaining door.
Should you?
Solution: The probability that your original choice is right is 1/3, and the
probability that the remaining door is right is 2/3. Simplest way to figure
this out: The probability that your original choice is right _has_ to be
1/3. If your first choice has a prize, the last door does not, and if your
first door has no prize, the last door does. Thus, 1-1/3 = 2/3.
The catch comes in people's intuitive objections to this, which are wrong.
"The probability is 1/2. After Monty opens a blank door, you have 2 choices,
your original or the remaining door, and one of those has a prize. Thus, 1/2."
But, Monty did not pick the remaining empty door at _random_. If he had picked
a door to show you at random, there would have been some chance that he had
picked a door with a prize in it, denying you any prize at all. For the
subset of cases where the game does not stop (because Monty _happened_ to pick
an empty door), the probability would indeed have been 1/2 for switching or
not switching.
Since Monty does not pick the door at random, the "Monty picks a full door,
denying you the prize" case turns into "Monty is forced to pick an empty door,
leaving you a 100% chance of gaining a prize by switching, and a 0% chance of
doing it by not switching". When this case is combined with the other cases
(where switching has an overall 50% chance of success), it boosts up the
chance that switching works to 2/3, and reduces the chance of success by not
switching, to 1/3.
If you still think it's 1/2, consider this.
You say that if you never switch, the probability is 1/2, and decide ahead of
time never to switch.
Monty says "I'll change the rules: I won't tell you which door I picked and
won't show you that it's empty. Since you will never switch anyway, this
will not change your action in the least, and won't change the probability that
you get the prize".
Conclusion: If you just pick 1 out of 3 doors, and Monty just _imagines_ himself
opening an empty door, this changes the probability to 1/2.
Second conclusion: If you just pick 1 out of 3 doors, the probability is 1/2.
--
"Thinking small-minded is when you see your bus on the other side of the
street and wish you could teleport across to catch it."
Kenneth Arromdee (UUCP: ....!jhunix!arromdee; BITNET: arromdee@jhuvm;
INTERNET: arro...@cs.jhu.edu)
Chris Cole
>
>I also think this should go in the rec.puzzles FAQ if it's not already. (What
>ever happened to that list anyway?)
Since you ask, I have made such a list available via email. To get an index
and instructions for requesting articles (puzzles and solutions), send a letter
with the single line:
send index
to net...@peregrine.com.
The Monty Hall puzzle is indeed on the list, and both the problem and
solution from the list are given below (although it is the same as yours).
********
switch.p
********
Switch? (The Monty Hall Problem)
Two black marbles and a red marble are in a bag. You choose one marble from the
bag without looking at it. Another person chooses a marble from the bag and it
is black. You are given a chance to keep the marble you have or switch it with
the one in the bag. If you want to end up with the red marble, is there an
advantage to switching? What if the other person looked at the marbles remaining
in the bag and purposefully selected a black one?
********
switch.s
********
Generalize the problem from three marbles to n marbles.
If there are n marbles, your odds of having selected the red one are 1/n. After
the other person selected a black one at random, your odds go up to 1/(n-1).
There are n-2 marbles left in the bag, so your odds of selecting the red one
by switching are 1/(n-2) times the odds that you did not already select it
(n-2)/(n-1) or 1/(n-1), the same as the odds of already selecting it. Therefore
there is no advantage to switching.
If the person looked into the bag and selected a black one on purpose, then
your odds of having selected the red one are not improved, so the odds of
selecting the red one by switching are 1/(n-2) times (n-1)/n or (n-1)/n(n-2).
This is (n-1)/(n-2) times better than the odds without switching, so you
should switch.
The original Monty Hall problem (and solution) appears to be due to
Steve Selvin, and appears in American Statistician, Feb 1975, V. 29,
No. 1, p. 67 under the title ``A Problem in Probability.'' It should
be of no surprise to readers of this group that he received several
letters contesting the accuracy of his solution, so he responded two
issues later (American Statistician, Aug 1975, V. 29, No. 3, p. 134).
I extract a few words of interest, including a response from Monty
Hall himself:
... The basis to my solution is that Monty Hall knows which box
contains the prize and when he can open either of two boxes without
exposing the prize, he chooses between them at random ...
Benjamin King pointed out the critical assumptions about Monty
Hall's behavior that are necessary to solve the problem, and
emphasized that ``the prior distribution is not the only part of
the probabilistic side of a decision problem that is subjective.''
Monty Hall wrote and expressed that he was not ``a student of
statistics problems'' but ``the big hole in your argument is that
once the first box is seen to be empty, the contestant cannot
exchange his box.'' He continues to say, ``Oh, and incidentally,
after one [box] is seen to be empty, his chances are not 50/50 but
remain what they were in the first place, one out of three. It
just seems to the contestant that one box having been eliminated,
he stands a better chance. Not so.'' I could not have said it
better myself.
Actually, there are much earlier variations on this problem.
Professor J. Laurie Snell (who also pointed me to the American
Statistician letters noted above) told me of a conference held
sometime in the 50's in which one of the participants made up the
following problem:
There are three prisoners, one of which has been randomly selected
to be executed at dawn. One of the prisoners asks a guard, ``Since
only one of us is to be executed, at least one of the remaining two
prisoners will not be executed. Will you tell me the name of one of
the others that is not to be executed?'' The guard refuses saying,
``Right now your probability of being executed is one third, but if
I tell you the name, then the probability will increase to one half
and you will sleep no easier.'' Is the guard correct?
James R. Driscoll
Asst. Prof. of Mathematics
and Computer Science
Dartmouth College
James Smith
with this.
Kenneth Arromdee writes:
> Solution: The probability that your original choice is right is 1/3, and the
> probability that the remaining door is right is 2/3. Simplest way to figure
> this out: The probability that your original choice is right _has_ to be
> 1/3. If your first choice has a prize, the last door does not, and if your
> first door has no prize, the last door does. Thus, 1-1/3 = 2/3.
Monty asks you to choose a door. Your chance of choosing correctly
is 1/3. He then eliminates a door and asks you to choose a second.
At this stage the odds of you having the right door are 1/2, surely.
You know it is not behind the door eliminated, but you don't know
which of the remaining two doors it is behind.
You argument implies that the two choices are not independant of
each other, and I can't see that this is the case. Monty can always
eliminate one of the two remaining doors, no matter which door you
first choose, and it doesn't matter that he doesn't do so at random,
because it doesn't depend in any way on the correctness of your first
choice.
Therefore the odds that you have selected the right door must be 50-50.
Doing this by cases:
Call the door the prize is behind door A.
If you pick A and Monty eliminates B and you choose A you win.
If you pick A and Monty eliminates B and you choose C you lose.
If you pick A and Monty eliminates C and you choose A you win.
If you pick A and Monty eliminates C and you choose B you lose.
If you pick B and Monty eliminates C and you choose B you lose.
If you pick B and Monty eliminates C and you choose A you win.
If you pick C and Monty eliminates B and you choose C you lose.
If you pick C and Monty eliminates B and you choose A you win.
There are only eight possible cases, and so the odds are 50-50.
Jim
--
James Smith | Wherever the French girl with the green eyes
Computing Centre | passes, nothing remains save wreckage and dead
Newcastle University | bodies.
cc...@cc.nu.oz.au |
Alan Wm Paeth
>I also think this should go in the rec.puzzles FAQ if it's not already...
>Monty Hall gives you a choice of 3 doors, one of which contains a prize...
I've always liked "Three-Door Monty". My reducto ad absurdum for those
who refuse to believe they double their odds by switching goes like this:
---
There are 100 doors, one has a prize, Monty knows were it is. You decide on
"17" before a word is spoken. Monty says "OK, 17. Now let's look at what's
behind door #1, door #2..." etc. right through the high ninties, so that in
the end you still have your door. But he also skipped door #86 (for instance).
Specifically, he has opened all the doors save two: the door you preselected
plus "his" (this is the generalization of the n=3 case to n=100. Once again,
two doors remain with contents undisclosed).
Now he says "would you like to switch??!!!!".
---
Offered this clue, it is remarkable how many people with whom I've reasoned
would rather fight than switch.
/Alan Paeth
Computer Graphics Laboratory
University of Waterloo
Ken Kaufman
Also in the Washington Post's weekly Straight Dopeish column. The
column has been in existence about five weeks, and the well-traveled Mr.
Hall has shown up in three of them. Lots of readers are still
unconvinced, but the answers are getting more complete each week ...
>Sigh.
Ditto.
OB-Puzzle. There are twelve boxes, one of which contains fabulous
riches, and eleven of which contain goats. There is also a large
balance, on which you can weigh the boxes. The balance is surrounded by
53 bicycles. Three Monty Halls, one of whom always tells the truth, one
of whom always lies, and one of whom answers randomly, will answer a
single question. All three say, "I do not know the two numbers," and
then look at one another. What happened to the other dollar?
Answer (ROT-13): Vg jnf fcrag genafzvggvat gur nobir!
==Ken Kaufman (kau...@gmuvax2.gmu.edu)
Kenneth Arromdee
Not only that, Cecil Adams then answered the question in The Straight Dope.
Even more wrongly, of course.
JRAM...@enh.prime.com
To add to the list of questions in an earlier reply, who IS Monty Hall ?
J Ramsden
David Petry
Kenneth Arromdee writes:
>Monty Hall gives you a choice of 3 doors, one of which contains a prize. After
>you make your choice, Monty picks an empty door other than yours, and shows
>you it's empty. Now he asks if you want to switch to the remaining door.
>Should you?
>
>Solution: The probability that your original choice is right is 1/3, and the
>probability that the remaining door is right is 2/3.
The problem is more subtle than that.
Maybe, just maybe, Monty only gives you the choice of switching doors if you
have picked the right door the first time, in which case you definitely should
not switch doors (i.e. the probability that the remaining door is right is 0).
In light of such considerations, the "intuitive" answer of 1/2 is not so bad.
David Petry
pe...@math.washington.edu
On the other hand, I would probably switch doors (why not?).
Daniel Hay
|> You're talking to Monty. You've made your choice. He opens the
|> door that has nothing behind it, and offers you a chance to change you
|> mind. While you are thinking, you see someone being ushered into a
|> sound-proof booth at the back of the studio. Since you read about this
|> scenario on the net, you know that the odds favor changing your mind by
|> 2:1, so you change your mind, and you (either win or lose).
|>
|>
|> You're standing outside a door. Someone ushers you into a sound
|> proof booth, and you see a stage with 3 doors, one of them open. You
|> notice the contestant on stage glance your way. You are told that you
|> can play the same game as the contestant, but are not told which choice
|> the contestant made. Since there are two doors, you know you have a 50%
|> chance of guessing correctly.
|>
|>
|> Now. Would somebody like to explain how one event viewed two different
|> ways can have two different sets of odds? And for fun:
|>
They have different odds because they are not the same event. In the first case, someone has chose one of three doors, and been shown that one of the remaining two is blank..that leaves these possibilities...
Door 1 (chosen) Door 2 Door 3
Prize Blank Blank
Blank Prize Blank
Blank Blank Prize
Now this happens three times (just rename the chosen door).
Now..if Door 2 is shown to be blank, then you have two possible outcomes...
Door 1 is prize or door 3 is prize.
If door 3 is shown blank, then you have two possible outcomes...
Door 1 is prize or door 2 is prize.
Now, that is four cases, but two are the same case. Door 1 is prize.
So by not switching, you have a 2 out of 3 chance of winning.
Now, the second man walks up to two closed doors and an open one and is asked
to choose one of the doors (if you meant he has to choose to switch or not, it is
the same thing..he is saying whether he wants a pre-chosen door or the other one)
He has a 50/50 shot at guessing the right door because he does not know all of the
information the other person knows.
Your question was incorrectly founded :)
*****Daniel
James Mork
I posted this problem in rec.humor... I didn't know this newsgroup
existed. Many people wouldn't agree that a switch gives a 2/3
chance of winning. You might want to find the article there if
you are interested. Look for 'goats' or 'BRAIN TEASER' in the
subject line. For doubters, last night I wrote a simulation of
this in C. Please don't send me to a .sources newsgroup. But
here it is...
/* I KNOW THINGS COULD BE OPTIMIZED... I TRIED TO MAKE IT READABLE
DOORS ARE NUMBERED FROM 0-2... AFTER ALL... THIS IS C :-) */
#include <stdio.h>
/* THIS MUST BE DEFINED TO MAXIMUM RANDOM NUMBER IN YOU SYSTEM
SEE 'man random' for more information */
#define MAXRANDOM 0x7fffffff
/* FOUR STATES OF DOORS */
#define NOT_TOUCHED 0
#define CHOSENDOOR 1
#define GOATDOOR 2
#define SWITCHDOOR 3
/* TWO THINGS BEHIND DOORS */
#define GOAT 0
#define CAR 1
int randthird;
int outflag;
/* FOUR MESSAGES FOR STATES OF DOORS */
char *results[4] = {" ** ERROR ** ",
" Chosen 1st ",
"Revealed Goat ",
"Switched to "};
/* NAMES OF TWO THINGS BEHIND DOORS */
char *names[2]= {" GOAT ",
" CAR "};
/* this function should return a random number = 0,1,2 */
int random0_2() { return(random()/randthird);}
int simulate() {
int i, yourdoor, touched[3], behinddoor[3];
/* NO DOORS PICKED YET, GOATS BEHIND ALL */
for(i=0; i<3; i++) {
touched[i]=NOT_TOUCHED;
behinddoor[i]=GOAT;
}
/* ALWAYS CHOOSE THE FIRST DOOR (0)*/
touched[0] = CHOSENDOOR;
/* PUT THE CAR BEHIND A RANDOM DOOR */
behinddoor[random0_2()] = CAR;
/* OPEN A DOOR THAT HAS NOT BEEN CHOSEN AND HAS A GOAT BEHIND IT*/
for(i=0; i<3; i++)
if (behinddoor[i]==GOAT && touched[i]!=CHOSENDOOR)
touched[i]=GOATDOOR;
/* SWITCH TO THE DOOR THAT IS NOT THE CHOSEN DOOR AND NOT THE
DOOR WITH THE REVEALED GOAT... THE 'UN_TOUCHED' DOOR */
for(i=0; i<3; i++)
if (touched[i]==NOT_TOUCHED) {
touched[i]=SWITCHDOOR;
yourdoor = i;
}
/* IF RESULTS WERE REQUESTED, PRINT THEM */
if (outflag) {
for(i=0; i<3; i++) printf(names[behinddoor[i]]);
printf("\n");
for(i=0; i<3; i++) printf(results[touched[i]]);
printf("\n");
if (behinddoor[yourdoor]==GOAT) {
printf("---------------- LOST -------------------\n\n");
} else {
printf("-----------***** WIN ***** ---------------\n\n");
}
}
/* IF CAR BEHIND DOOR SWITCHED TO (YOURDOOR), RETURN A ONE */
return(behinddoor[yourdoor]==CAR ? 1 : 0);
}
main() {
int i,wins,runs;
srandom(1);
randthird = (int) (((unsigned) MAXRANDOM)/3);
printf("Watch output (y/n)");
outflag=((getchar()|0x20)=='y');
printf("How many runs ? ");
scanf("%d",&runs);
for(i=wins=0; i<runs; i++) wins+=simulate();
printf("\n\n\n\n wins: %d runs %d: perc %d:\n",wins,runs,(wins*100)/runs);
}
--
UUCP Bitnet Internet
uunet!ndsuvax!numork numork@ndsuvax num...@plains.nodak.edu
Eric Hjelmfelt
>To add to the list of questions in an earlier reply, who IS Monty Hall ?
Monty Hall was the (probably fictional) name of the announcer/host of
the old television game show "Let's make a deal." Sort of a cross
between Bob Barker and a used car salesman. You might remeber LMaD only
for the fact that the contestants wore ridiculously silly contumes.
----------------------------------------------------------------------------
Eric Hjelmfelt | disclaimer: What? They wouldn't let
hjelmflt@ | me speak for them even if I were
symcom.math.uiuc.edu | to pay them!
----------------------------------------------------------------------------
Peter van der Linden
up in Marilyn "world's greatest IQ & ego" vos Savant's column.
Quite right, and Marilyn can never admit she's wrong. She
got the "does a bullet come down with the same force it
went up" question horribly wrong. Then she compounded her
error by trying to prove that she was right after all (she
wasn't). Now she's trying the same thing again for the
Monty Hall puzzle, apparently.
Obligatory puzzle: Waldo asks "does a bullet come down with
the same force with which it ascended?"
----------------
Peter van der Linden lin...@eng.sun.com (415) 336-6206
Favorite forbidden fruits #237: the durian.
Bill Jefferys
#In article <1990Dec5.0...@watcgl.waterloo.edu> awp...@watcgl.waterloo.edu (Alan Wm Paeth) writes:
#>There are 100 doors, one has a prize, Monty knows were it is. You decide on
# ^^^-- make it a million
#>"17" before a word is spoken. Monty says "OK, 17. Now let's look at what's
#>behind door #1, door #2..." etc. right through the high ninties, so that in
#>the end you still have your door. But he also skipped door #86 (for instance).
#>Specifically, he has opened all the doors save two: the door you preselected
#>plus "his" (this is the generalization of the n=3 case to n=100. Once again,
#>two doors remain with contents undisclosed).
#>
#>Now he says "would you like to switch??!!!!".
#
# Are you now trying to say that the other door is 999,999 times more
#likely to be hiding the prize? How about the guy that walks up, sees
#999,998 open doors and two closed ones. How is it that his odds are
#50/50, yet yours are virtually certain?
#
#
#>Offered this clue, it is remarkable how many people with whom I've reasoned
#>would rather fight than switch.
#
# I don't see no steenkeen clue.
Instead of trying to convince you that Alan Paeth is correct,
I am going to ask you to convince yourself. I want you to
ACTUALLY DO the following experiment. If, AFTER YOU DO THE
EXPERIMENT, you still believe that switching makes no difference,
then we can talk further.
You will need a friend to help you. Take an ordinary pack
of cards. Discard the jokers. By definition, the PRIZE is
the ace of spades.
Have your friend shuffle the cards thoroughly. Cut the deck
as many times as you like. Have your friend deal the cards,
FACE DOWN, in an array on the table. After this is done,
you choose one of the cards (but do not look at it). Put
a quarter on top of the cards as a token.
Now your friend is to look through the remaining cards,
and place one of them face down on the table. If the ace
of spades is among the remaining cards, he MUST put the
ace on the table. Otherwise, he can put any card down on
the table. After doing this, he shows you the faces of
all the other cards, to prove to you that the ace is not
among them.
The ace is either the card you put the quarter on top of,
or the card that your friend has just put face down on the
table.
According to your reasoning, the chances are now 50-50 that
the card with the quarter on top of is the ace of spades. Turn
it face up. If it is the ace of spades, your friend must pay
you a quarter. If not, your friend keeps the quarter you
put on top of the card.
Now play the game, over and over. Your friend must shuffle
the cards thoroughly each time. Play until you have been
convinced that you are wrong. I guarantee that it will not
take too long.
---
Remember, I do not want to hear arguments from you as to
why Alan's analysis is wrong, until AFTER you have played
a few hands of this game. Ten should suffice.
Bill Jefferys
--
If you meet the Buddha on the net, put him in your kill file
--Robert Firth
Bill Jefferys
#
#Have your friend shuffle the cards thoroughly. Cut the deck
#as many times as you like. Have your friend deal the cards,
#FACE DOWN, in an array on the table. After this is done,
#you choose one of the cards (but do not look at it). Put
#a quarter on top of the cards as a token.
The last sentence should read "Put a quarter on top of the
card you have selected to mark it."
Sorry for the confusion,
Eric Postpischil (Always mount a scratch monkey.)
> I utterly reject the idea that someone's attitude can change the
>probalility of a situation as described.
Like the rest of the problem, this has also been raised before. The fact is, it
is not attitude that is relevant, but knowledge. Two people with different
knowledge about a situation will assign different probabilities to the same
physical events.
We have three cards, A, B, and C. I shuffle them. (Assume a uniform
distribution.) I deal you one card. What do you think is the chance it is an
A? (1/3 I hope.) I look at the card. I know whether it is an A or not. I
don't tell you. What do you think is the chance it is an A?
As far as you are concerned, you don't have any more information; your
assessment can't change because you don't have any other choice. But I know
with certainty; my assessment is either 1 or 0, depending on what I saw.
-- edp
ee...@cc.nu.oz.au
remain unconvinced by the explanations (some of them very good) that have
been posted so far. For example, in article <1990Dec5.1...@demott.com>,
> ways can have two different sets of odds?
>
problems:
PROBLEM #1: I have a coin which is weighted so that heads will come up
twice as often as tails. You know that the coin is biased. When I toss the
coin, what is the probability of getting heads?
SOLUTION: If you accept that the coin is unlikely to stand on its edge,
there are two possible outcomes:
(a) The coin comes up heads
(b) The coin comes up tails.
Therefore the probability is 50%.
MORAL #1: Working out probabilities by listing all the possible outcomes is
fine (provided that there are not too many possible outcomes), but you have
to watch out for a hidden assumption: are you sure that all the outcomes that
you listed are equally likely?
PROBLEM #2: Monty Hall (whoever he was ... the man ain't got no culture)
tells you that a prize is hidden behind one of three doors, and invites you
to choose a door. You open a door and see the prize. Then Monty opens
another door and shows you that the prize is not behind the second door, and
gives you the option of changing your mind. Should you switch?
SOLUTION: The probability of the prize being behind the first door you
chose is 1/3. (The easiest way to see this is that the probability was 1/3
before you opened the door.) Therefore the probability that it's behind
one of the other doors is 2/3, and you can double your chances by switching.
MORAL #2: A probability is a measure of your uncertainty about something.
(I like to express this, for those who know the jargon, as "every
probability is a conditional probability".) A probability is *not* an
objective function of the thing being observed. In fact, it depends much
more on the observer than on the thing being observed. When the information
available to the observer changes, then the probabilities change too.
Furthermore, when two observers observe the same thing, and they have
different amounts of information available, then the probabilities are
different for the two observers. There is no paradox about this once you
accept that probabilities are in the mind of the observer, and not in the
phenomenon being observed.
Of course, this depends on an axiom set that not everyone will accept, so
let's consider an ...
ALTERNATIVE EXPLANATION FOR PHYSICISTS: Before any of the doors was opened,
there was a fuzzy prize behind every door. However the act of making any
observation causes, by an action-at-a-distance mechanism which is not yet
fully understood, a partial collapse of the wave functions. (Of course,
Monty has already seen the prize, so in his frame of reference the wave
functions have already collapsed. But in your frame of reference they
haven't yet collapsed.) By the time all three doors are open, the wave
function has collapsed completely, and a dead cat appears in the tree in
the quad.
Peter Moylan ee...@cc.nu.oz.au
Michael A. Greene
> LOTS of text deleted
>Actually, there are much earlier variations on this problem.
>Professor J. Laurie Snell (who also pointed me to the American
>Statistician letters noted above) told me of a conference held
>sometime in the 50's in which one of the participants made up the
>following problem:
>
> There are three prisoners, one of which has been randomly selected
> to be executed at dawn. One of the prisoners asks a guard, ``Since
> only one of us is to be executed, at least one of the remaining two
> prisoners will not be executed. Will you tell me the name of one of
> the others that is not to be executed?'' The guard refuses saying,
> ``Right now your probability of being executed is one third, but if
> I tell you the name, then the probability will increase to one half
> and you will sleep no easier.'' Is the guard correct?
>
>James R. Driscoll
>Asst. Prof. of Mathematics
>and Computer Science
>Dartmouth College
This problem (not coincidentally) is in Kemeny, Snell and Thompson's
Finite Mathematics book where it delighted (tortured) students.
The relationship to Monte is obvious. If the correct strategy
in Monte is to switch, then the prisoner doesn't get insomnia, i.e.
his probability of being executed remains at 1/3. The person
mentioned by the guard as not being executed however, has his
chance elevated to 2/3. He shouldn't sleep.
After following the Monte and it's variants on the net, I am
surprised to see that nobody has mentioned that this whole issue
is a simple application of the Bayes formula.
P (A hangs | guard says B) =
P (guard says B | A hangs) P (A hangs)
_________________________________________________
P (g says B|A hangs)P(A hangs) + P(g says B|C hangs)P(C hangs)
.5 * (1/3)
_____________________ = 1/3 = P (A hangs)
.5 * (1/3) + 1 * (1/3)
The calculations use the original formulation of the problem where
the guard is instructed to say B goes free should C hang and
the guard flips a coin (.5) and says B or C should A hang. This
is the same as Monte revealing doors which do not have prizes.
Department of Mathematics and Statistics, The American University
Wayne Folta
seems to take a turn at some point that loses us non-probablilty types.
I think I've simplified it enough so that even I would be convinced...
The way the game is actually played is this:
1. You choose a door at random.
2. Monty chooses a door, according his perfect knowledge, and the rule:
a. If you chose the prize door, he randomly choses an empty door;
b. Otherwise, he must chose the prize door.
3. All doors that were not chosen by Monty or you are opened. In the
standard problem, this results in one door being opened.
This description of the game is totally equivalent to the standard
description. Monty's rules prevent him from choosing your door (and thus
causing the other two doors to be opened), and from allowing the prize to
be revealed (he must "protect" the prize from being revealed).
Another way of putting this is that if you do not choose the prize, you
force him to choose it--and since he has perfect knowledge, he will. But if
you choose it, he just makes his pick at random, to "spoof" you.
With three doors, the odds that you choose correctly are 1/3, so 1/3 of the
time Monty will spoof you, and switching will lose. However, 2/3 of the time,
you will be wrong, forcing Monty to show you the correct door, and switching
will win. QED
How can two different people have different odds for the same event?
Imagine that Monty starts out with only two doors. He then gives you a hint
as to the correct door to choose: 1/3 of the time he lies, and 2/3 of the
time he tells the truth. Obviously, you would take his advice, even though
it is not totally reliable, and you would win 2/3 of the time. On the other
hand, if he gave no suggestion, you will win only 1/2 the time. This is
completely equivalent to the original problem and the scenario where another
contestant is placed in a sound-proof booth.
Wayne
--
Wayne Folta (fo...@cs.umd.edu 128.8.128.8)
Keith Goldfarb
(analysis deleted)
>There are only eight possible cases, and so the odds are 50-50.
>
>Jim
Arrgh!!!!!
Why do people so often assume that all possibilities are equally likely?
Example: There are only TWO POSSIBILITIES: Either the world will
end tomorrow, or it won't. Would you say that there is therefore
a 50% chance that it will?
I wonder how many of the world's problems are caused by this fundamental
misunderstanding of probabilities. Perhaps we could raise a better
generation of humans if we replaced the "Pledge of Allegiance" with
the statement "All possible outcomes are not necessarily equally likely."
and had all the school children recite it at the beginning of each day.
K.
PS: Anybody else ever see a high-school film where a young man
is avoiding sex with his girlfriend because he isn't convinced
that he doesn't have VD? He (incorrectly) reasons that even though
the test came back negative, it's only 90% accurate, so he must
therefore have a 10% chance of having the disease! She (the girlfriend)
then starts _singing_ about Bayes' theorem and how foolish he is being.
Quite an entertaining film.
PPS: I am trying to accumulate a list of the world's most misunderstood
mathematical problems. (Monty is at the top of the list, of course.)
Any suggestions will be greatly appreciated. (Yes, I've read the book
"Innumeracy" -- it has several excellent ones.)
--
Keith Goldfarb Rhythm & Hues Claudia is Everywhere.
celia!ke...@usc.edu celia!ke...@tis.llnl.gov ...mlogic!celia!keith
I got nothing. Too bad.
But I'm happy 'cause that's all I have.
ger...@merrimack.edu
>
> Example: There are only TWO POSSIBILITIES: Either the world will
> end tomorrow, or it won't. Would you say that there is therefore
> a 50% chance that it will?
No, no, no. It means that exactly 50% of the world will end tomorrow :-)
-[mpg]
...!samsung!hubdub!gerlekm
ger...@merrimack.edu
"Log on, page in, swap out!"
Russell Turpin
In article <28...@mimsy.umd.edu> fo...@tove.cs.umd.edu (Wayne Folta) writes:
> How can two different people have different odds for the same event?
Mr Folta provides a correct analysis, but I would like to state
this one important fact more directly.
The probability assigned an event depends on what one
knows. Two people with different knowledge can correctly
assign different probabilities to the same event.
A simple example: Joe flips a coin and observes it lands heads. The
result is hidden from Sam. For Joe, the probability of heads is 1.
For Sam, 1/2. Same event. Different knowledge. Different
probabilities.
Simple, right? Nevertheless, the subtlety of dealing with this
central fact about probability causes people to misstate and
wrongly analyze the Monty Hall puzzle.
Russell
Kevin D. Quitt
Gods, but I hate it when I'm wrong. Maybe it's because I've put in
160 hours of work the last 10 days. To do pennance, if *anyone* out
there *ever* disagrees that the best strategy is to switch, send me
email, and I'll patiently explain how it all works.
(I don't have to always be right, but I *HATE* it when I'm wrong.
Just last december I was wrong - I thought I'd made a mistake and I
hadn't). 8-{)}
--
_
Kevin D. Quitt demott!kdq k...@demott.com
DeMott Electronics Co. 14707 Keswick St. Van Nuys, CA 91405-1266
VOICE (818) 988-4975 FAX (818) 997-1190 MODEM (818) 997-4496 PEP last
Tom Gray
>
>\
>To anyone who still believes that there is nothing to be gained by switching
>I offer the following challenge. I will deal 3 cards, one of which is the
>ace of spades. I will look at the cards, but not show them to you. You
>will then pick a card. I will then turn over a card which is not the ace.
>You will then turn over your card. (Since you believe there is nothing
>to be gained by switching at this point you shouldn't mind not being able to.)
> If it's the ace I will give you $3!!!
>If it is not the ace you must give me $2. That's 3:2 odds on an even bet,
>folks! Come make some money!
>
There is something called conditional probability. With three cards
P( ace of spades) = 1/3
since possibilites are:
My hand Table
Ace non-ace non-ace
non-ace Ace non- ace
non-ace non-ace Ace
One non-ace of spades is removed
Possibilites now are:
My Hand Table
Ace non-Ace
non-Ace Ace
Probability that my hand now contains an ACE is 1/2
P(ace of spades/one non-ace removed)= 1/2
Note that all of the possibilities have been enumerated. The frequency of each
outcome has been accounted for. Rememeber that probability is just counting.
If some of the outcomes are removed, the probailities change but this is
accounted for by not counting the removed outcomes. They don't return like
Banquo's ghost.
V. J. Burkley
>
> Gods, but I hate it when I'm wrong. Maybe it's because I've put in
>160 hours of work the last 10 days. To do pennance, if *anyone* out
>there *ever* disagrees that the best strategy is to switch, send me
>email, and I'll patiently explain how it all works.
>Kevin D. Quitt demott!kdq k...@demott.com
>DeMott Electronics Co. 14707 Keswick St. Van Nuys, CA 91405-1266
>VOICE (818) 988-4975 FAX (818) 997-1190 MODEM (818) 997-4496 PEP last
Okay all you Smarty pants, Kevin is right. The problem is that it is all
a statistical abberation that you don't get 50/50 odds. If there are two
doors, one has the car the other don't. 1/2 times you win so 50/50. As
a practical experiment, I have been opening garage doors at random. More
than half the time I find a car. I can't beleive how stupid all you
"educated" people are. If you would get off your butts and try and
put your theory in practice, you would find your errors. Jeez!!
Anafin Hargola, Esquire
Heine Rasmussen
>Call the door the prize is behind door A.
>If you pick A and Monty eliminates B and you choose A you win.
>If you pick A and Monty eliminates B and you choose C you lose.
>If you pick A and Monty eliminates C and you choose A you win.
>If you pick A and Monty eliminates C and you choose B you lose.
>If you pick B and Monty eliminates C and you choose B you lose.
>If you pick B and Monty eliminates C and you choose A you win.
>If you pick C and Monty eliminates B and you choose C you lose.
>If you pick C and Monty eliminates B and you choose A you win.
>There are only eight possible cases, and so the odds are 50-50.
It should bother you that you implicitly state that all cases have
equal probabilities, but in four of them, you have started with
picking A!
--
Heine Rasmussen --- <sam...@debet.nhh.no> or <he...@makro.nhh.no>
___________________________________________________________________________
Center for Applied Research, Norwegian School of Economics
Mike Perry
In article <4958.2...@cc.nu.oz.au>, cc...@cc.nu.oz.au (James Smith) writes:
> Doing this by cases:
>
> Call the door the prize is behind door A.
>
> If you pick A and Monty eliminates B and you choose A you win.
> If you pick A and Monty eliminates B and you choose C you lose.
> If you pick A and Monty eliminates C and you choose A you win.
> If you pick A and Monty eliminates C and you choose B you lose.
> If you pick B and Monty eliminates C and you choose B you lose.
> If you pick B and Monty eliminates C and you choose A you win.
> If you pick C and Monty eliminates B and you choose C you lose.
> If you pick C and Monty eliminates B and you choose A you win.
>
> There are only eight possible cases, and so the odds are 50-50.
Ignoring the fact that you pick the right answer half the time,
consider this case:
I set out a deck of cards (52) in front of you, your job is to pick out
the ace of spades. You pick out a card and set it in front of you
without looking at it. I then turn over 50 of the remaining 51 cards
which are NOT the ace of spades. I ask you if you want to switch.
Common sense dictates that you should. This is merely the goat/car
problem on a larger, easier to see scale. The chances of you picking
a useless card on first try is 51/52. Therefore, chances are that
50 of them have been turned over, you have the other one, and the
last one is the Ace, you should swithc. On the odd chance that you did
pick the ace (1/52) you will lose. QED
ciao,
Mike.
/**********************************************************************
* This is a test, this reality is only a test. If this had been a *
* "real" reality, you would have been given instructions about *
* to do. *
**********************************************************************/
--
ciao,
Mike (the tyke)
Ever notice how fortune cookies are much more humorous if you add
'...in bed.' to the end of the messages found inside. For example 'You
will achieve great conquests.' or 'Close friends will soon wish their
debts repaid.'
Oleary
In article <4958.2...@cc.nu.oz.au> you write:
>Forgive me if I'm flogging a dead horse, but I'm having some trouble
>with this.
>
[filler deleted . . .]
>
>Doing this by cases:
>
>Call the door the prize is behind door A.
>
>If you pick A and Monty eliminates B and you choose A you win.
>If you pick A and Monty eliminates B and you choose C you lose.
>If you pick A and Monty eliminates C and you choose A you win.
>If you pick A and Monty eliminates C and you choose B you lose.
>If you pick B and Monty eliminates C and you choose B you lose.
>If you pick B and Monty eliminates C and you choose A you win.
>If you pick C and Monty eliminates B and you choose C you lose.
>If you pick C and Monty eliminates B and you choose A you win.
>
>There are only eight possible cases, and so the odds are 50-50.
>
>Jim
>--
>James Smith | Wherever the French girl with the green eyes
>Computing Centre | passes, nothing remains save wreckage and dead
>Newcastle University | bodies.
>cc...@cc.nu.oz.au |
What you are failing to consider in those cases is the initially probability
that you have chosen the correct door, A, which is only 1/3.
OK, enough of this crap, let's use math, let's use proofs, let's use . . .
BAYES THEOREM!!!
For those that don't know it, and that seems to be the whole net (please don't
flame me if you do and were to lazy to "enlighten" the rest of us), Bayes
Theorem takes into account those nasty initial probabilities. In general,for
a two result problem (win/lose) such as this, Bayes theorem states:
p(x|w)p(w)
p(w|x) = ------------------------
p(x|w)p(w) + p(x|~w)p(~w)
where ~w is "not w" or "1 - w" and p(A1|A2) is the probability of A1 given
the knowledge of A2 (yes, knowledge *is* important in probabilities).
(The above is correct, barring typos. I believe the application to be correct,
though its been awhile since I've used Bayes Theorem . . .)
If we apply the general case to the Monty case, we will assign x to be the fact
that one door has been eliminated (leaving two doors) and w to be the "win"
condition). This gives:
p(2 doors|win)p(win)
p(win|2 doors) = ---------------------------------------------
p(2 doors|win)p(win) + p(2 doors|lose)p(lose)
which basicly says "We can determine the probability that we have won (picked
the correct door initially) if we know the the probability of there being two
doors remaining given that we have won (don't be confused, this will be
resolved) and the probability of picking a winning door (the other
probabilities follow from those two).
-The probability of picking a winning door (initially) is 1/3, so p(win) = 1/3.
-The probability of picking a losing door (initially) is 1 - 1/3 = 2/3, so
p(lose) = 2/3.
-The probability of there being 2 doors remaining given that we have won/lost
is 1, since Monty always eliminates one of the three doors, so
p(2 doors|win) = p(2 doors|lose) = 1.
Substitution into the above equation yields:
1 * 1/3 1/3 1/3
p(win|2 doors) = ----------------- = --------- = --- = 1/3
1 * 1/3 + 1 * 2/3 1/3 + 2/3 1
Translated, this means that the probability of us having chosen the correct
door initially is 1/3!!! *That* means that the probability of the last
remaining door being correct is 1 - 1/3 = 2/3!!!!!!
If in doubt, the above theorem can be proven using p(lose|2 doors). This is
left as an exercise for the reader :-).
So, please, please, please drop this subject!!! I have finals to fail and
making extraordinarily long posts to the net only helps.
Those wishing to continue the "discussion" may do so in alt.* (they're
planning on removing our feed to the alt's soon, so I don't care which one).
Thank you . . . and goodnight.
***************** Signature Block : Vaporware Version *******************
* | *
* | *
* | *
* | *
****************** Copyright (c) 1990 by Doc O'Leary ********************
Erann Gat
>
> Call the door the prize is behind door A.
>
> If you pick A and Monty eliminates B and you choose A you win.
> If you pick A and Monty eliminates B and you choose C you lose.
> If you pick A and Monty eliminates C and you choose A you win.
> If you pick A and Monty eliminates C and you choose B you lose.
> If you pick B and Monty eliminates C and you choose B you lose.
> If you pick B and Monty eliminates C and you choose A you win.
> If you pick C and Monty eliminates B and you choose C you lose.
> If you pick C and Monty eliminates B and you choose A you win.
>
> There are only eight possible cases, and so the odds are 50-50.
By this reasoning there is a 50-50 chance that you picked door A
to begin with! So much for free will...
You can't figure the probabilities by simply listing all possible
cases because not all the cases are equally likely. Specifically,
the first four cases are only half as likely as the second four.
This is because if you pick the car the first time, then Monty has
a CHOICE which door to pick. If you pick wrong, his choice is
forced. Therefore, 1/3 of the time he will be forced to pick B (because
you chose C), 1/3 of the time he will be forced to choose C (because you
chose B) and 1/3 of the time he'll have a choice. Half the time he has
a choice he will choose B, or 1/6 of the total times. Likewise for C.
\
To anyone who still believes that there is nothing to be gained by switching
I offer the following challenge. I will deal 3 cards, one of which is the
ace of spades. I will look at the cards, but not show them to you. You
will then pick a card. I will then turn over a card which is not the ace.
You will then turn over your card. (Since you believe there is nothing
to be gained by switching at this point you shouldn't mind not being able to.)
If it's the ace I will give you $3!!!
If it is not the ace you must give me $2. That's 3:2 odds on an even bet,
folks! Come make some money!
Kevin D. Quitt
>
>I've just gotta *GLOAT* 'cause I'm the one that converted him. And
>160 hours in 10 days is no excuse. (I work for him & I've put in just as
>many)
I'd like to point out that he knew the answer berfore those 160 hours.
So I will.
--
Brian R. Hunt
Let me offer an attempt at a reasonably precise statement of a Monty
Hall scenario in which the "switching wins 2/3 of the time" answer is
correct.
"You will play a game with a fellow named Monty Hall, and he has to
play by the rules I'm about to describe. A prize will be randomly
placed behind one of three doors, and Monty will be told which door
it's behind. You will then get to choose one of the doors. No matter
which door you choose, Monty must then open a door other than yours
which does not have the prize behind it (if he has a choice, he'll
pick randomly--while maintaining his poker face, of course.) Monty
will then offer you the choice between the two remaining doors, and
you will win whatever is behind the door you pick this time. Given
this information only, what is your best strategy, and what odds does
it give you of winning the prize?"
The main source of disagreement on this problem seems to be the lack
of precision with which it is initially posed, which leads to
differing assumptions about what Monty's role is, which in turn
leads to different answers. My next message contains a more
long-winded form of my argument, and some advice on how better to
explain the answer to the above question, for anyone who cares.
Brian Hunt
bh...@nswc-wo.navy.mil
Brian R. Hunt
Recently, I have explained the Monty Hall problem to a few people with
relatively (to people who post here) little mathematical background.
I did not have to (nor could I really) beat them over the head with
Bayes' formula or what-have-you to get them to agree with me about the
"right" answer. I am convinced that the primary cause of disagreement
about this problem is not people's lack of understanding of
probability, but rather people's lack of precision in posing the
problem.
I usually hear the Monty Hall problem stated more or less as follows:
"You're on Let's Make a Deal, and Monty Hall shows you three doors,
behind one of which is a prize. You pick door #1, and Monty proceeds
to show you that the prize is not behind door #2. Monty then asks
you whether you would like to stick with door #1 or switch to door #3.
What should you do?"
Often Person A asks such a question, and Person B thinks briefly
and responds, "It doesn't matter," whereupon Person A says, "You're
wrong." Person A then trys to explain why switching is the best
strategy, probably revealing in the process the assumptions that are
necessary to correctly draw this conclusion. However, Person B may
not have made the same assumptions, and, being on the defensive, may
not catch the distinction. Person B becomes more attached to his/her
original interpretation, and becomes harder to convince that there is
a scenario under which Person A's answer is correct.
If Monty were standing on a street corner and playing with
three cards (and I had not observed anyone else play against Monty and
so forth), I might be inclined to think I had guessed right originally
and that Monty is trying to cheat me. When the problem is presented
in the context of a game show, people tend to assume the host is not
deliberately trying to cheat the contestant, but I think most of us
haven't seen Let's Make a Deal recently enough to know what Monty
does as a general rule. So let me offer a more precise statement of the
problem, in hopes that it will inspire people to be more careful in the
future when bringing up the problem. I do not claim this statement will
instantly lead everyone to get the right answer, just that it is well-
posed enough to have a right answer, and that by posing the problem
this way in the first place one should have an easier time convincing
people the answer is correct.
"You will play a game with a fellow named Monty Hall, and he has to
play by the rules I'm about to describe. A prize will be randomly
placed behind one of three doors, and Monty will be told which door
it's behind. You will then get to choose one of the doors. No matter
which door you choose, Monty must then open a door other than yours
which does not have the prize behind it (if he has a choice, he'll
pick randomly--while maintaining his poker face, of course.) Monty
will then offer you the choice between the two remaining doors, and
you will win whatever is behind the door you pick this time. Given
this information only, what is your best strategy, and what odds does
it give you of winning the prize?"
I suppose if one is interested in a more terse statement of the
problem, one could cut out the parts about randomly choosing things
(the "given this information only" phrase comes in handy), but if for
instance Monty always opens the highest numbered empty door, one's
perspective on the final choice is a bit different (though switching
is never a bad choice, and still gives you a 2/3 chance a priori.) In
any case, the points to emphasize are that Monty knows which door the
prize is behind and that he is required to show you an empty door
whether or not you initially choose the door with the prize behind it.
Finally, I have a suggestion to people who must resort to simulations
to convince somebody that switching can be advantageous. Don't offer
to play the Monty Hall role, make the person you're trying to convince
play Monty. For one thing, if you play Monty and the other person
decides to be random about sticking vs. switching, he/she will
probably win about half the time and likely not be convinced of
anything. Also, by playing Monty he/she should more quickly realize
the constraints Monty is under and why Monty is forced in some cases
to lead you to the right answer. If the other person does not play
the Monty role properly, then you haven't explained the problem
properly.
Brian Hunt
bh...@nswc-wo.navy.mil
Robert E. Stampfli
> to convince somebody that switching can be advantageous. Don't offer
> to play the Monty Hall role, make the person you're trying to convince
> play Monty.
The best way I have found to convince people is to skew the odds to the
point of absurdity: Take a deck of cards. Find the ace of spades.
Place all the cards face down on the table. Ask the "believer" to pick a
card. Ask what the probability of it being the ace of spades is. Start
turning over cards that are not the ace of spades until only two are
left -- the one originally selected and the true ace. Very few people at
this point will contend that there is a 50-50 chance they selected
the ace of spades.
--
Rob Stampfli / att.com!stampfli (uucp@work) / kd8wk@w8cqk (packet radio)
614-864-9377 / osu-cis.cis.ohio-state.edu!kd8wk!res (uucp@home)
Gregory J. E. Rawlins
I'm not surprised to see Monty on the net yet again. I don't think humans are
evolutionarily tailored to make good probabilistic inferences, that's why god
made mathematics. Every year i have to quash some student's belief that if
something either happens or doesn't happen then the probability of it happening
is one-half. The following is usually enough to stifle dissent:
You will either sneeze or not sneeze as you read this sentence; does that
mean that the probability is one-half that you will sneeze before you get
to the end of this sentence?
Most students pipe down after that.
gregory.
--
ps. the following is adapted from von Mises' wonderful
_Probability, Statistics, and Truth_, it deserves to be better known:
We are given a glass containing a mixture of water and wine. All we know is
that it contains at least as much water as wine and at most twice as much
water as wine. So if r is the (unknown) ratio of water to wine then
1 <= r <= 2
If we assume, based on our ignorance of the real value of r, that r is equally
likely to be any number in the range between 1 and 2 then
P(1 <= r <= 3/2) = 1/2
Which implies that
P(3/2 < r <= 2) = 1/2
Also, since 1 <= r <= 2 then 1/2 <= 1/r <= 1. And, arguing as before we infer
that
P(1/2 <= 1/r <= 3/4) = 1/2
Which implies that
P(4/3 < r <= 2) = 1/2.
Therefore
P(4/3 < r <= 3/2) = 0
So without knowing the real ratio we have concluded that it cannot be between
4/3 and 3/2!
Michael Foster
In analyzing this problem we must consider Monty Hall's strategy. It might be
anything, since the problem statement is silent on the subject. Previous
solutions have assumed that Monty will always show you a goat, but suppose that
Monty only shows you a goat when you have picked the car, and otherwise doesn't
show you anything? Then switching is exactly the wrong thing to do, since
every time the player has an opportunity to switch, he should refuse.
This problem should be analyzed a la von Neumann and Morganstern. Monty has
three strategies that may result in showing a goat: always show a goat, show a
goat only when the player has picked a car, or show a goat only when the player
has picked a goat.
The player's strategies are: switch or don't switch.
Here is the player's payoff matrix, in units of cars:
Player switch don't switch
Monty
always show 2/3 1/3
show only when player has car 0 1
show only when player has goat 1 0
The player's optimal mixed strategy is to switch with probability 1/2, giving
him an expected payoff of 1/2, no matter what Monty is trying to do. This is
also the optimal strategy for n-door Monty.
Note also that if you erroneously believe that the payoffs in the "always
show" case are 1/2-1/2, the optimal mixed strategy is still to flip a coin.
Lex Mierop
>
> Gods, but I hate it when I'm wrong. Maybe it's because I've put in
>160 hours of work the last 10 days. To do pennance, if *anyone* out
>there *ever* disagrees that the best strategy is to switch, send me
>email, and I'll patiently explain how it all works.
AH! Sinner be gone!!!
*GLOAT*GLOAT*GLOAT*
I've just gotta *GLOAT* 'cause I'm the one that converted him. And
160 hours in 10 days is no excuse. (I work for him & I've put in just as
many)
--
Lex Mierop - DeMott Electronics | "Break the code, solve the crime!"
Internet/uucp: l...@demott.com | - Agent Dale B. Cooper, Twin Peaks
US Snail: 14707 Keswick St.; Van Nuys, CA 91405 Tel. 818-988-4975
Kevin D. Quitt
Monty is a good demonstration of the gambler's paradox. Imagine the
following situation: a pair of lights, red and blue, are flashing
rapidly enough so as to be indistinguishable from steady. There is a
blue button and a red button. When you press either button, one of the
lights stays on while the other turns off. What you don't tell the person
is that the red light will come on 75% of the time.
A human playing this game will pretty rapidly settle into pressing
the red button 75% of the time, thereby getting it right 62.5% of the
time. On the other hand, a monkey will very rapidly start pressing the
red light 100% of the time, maximizing his correct guesses at 75%. I'm
told that the next highest animal that shows the gambler's paradox
behavious is the goldfish, and the next below that the cockroach.
(Personally I'd give them more credit than that).
If there are two choices, a naive human will choose a less-than-
optimal strategy.
Engbert Gerrit IJff
> Doing this by cases:
>
> Call the door the prize is behind door A.
>
> If you pick A and Monty eliminates B and you choose A you win.
> If you pick A and Monty eliminates B and you choose C you lose.
> If you pick A and Monty eliminates C and you choose A you win.
> If you pick A and Monty eliminates C and you choose B you lose.
> If you pick B and Monty eliminates C and you choose B you lose.
> If you pick C and Monty eliminates B and you choose C you lose.
> There are only eight possible cases, and so the odds are 50-50.
Actually there are 4 more possible cases,
which are improbable to happen.
^^^^^^^^^^
Moreover, the cases mentioned above are not equally probable.
* If you pick B and Monty eliminates A and you choose B you loose.
* If you pick B and Monty eliminates A and you choose C you loose.
* If you pick C and Monty eliminates A and you choose B you loose.
* If you pick C and Monty eliminates A and you choose C you loose.
This makes a total of 12 cases,
the probability for each of which is,
on first sight:
P(case) = P(1st) * P(eli) * P(2nd) =
1/3 * 1/2 * 1/2 = 1/12
However, in the last four possibilities, you wouldn't get
a second choice: you always lose these.
The odds are 1/3 to win.
Monty lets P(elimination) depend on the first choice.
The probabilities of the six possible cases after
your first choice are:
P(1st) * P(eli) Monty
P(A B) = 1/3 * 1/2 = 1/6 has free choice
P(A C) = 1/3 * 1/2 = 1/6 has free choice
P(B A) = 1/3 * 0 = 0 MUST take C
P(B C) = 1/3 * 1 = 1/3 MUST take C
P(C A) = 1/3 * 0 = 0 MUST take B
P(C B) = 1/3 * 1 = 1/3 MUST take B
--- +
1
For each of the four remaining cases, you can
either stick to your choice or change,
which makes you win or loose with the probability
of that case.
stick | change
win loose | win loose
P(A B) = 1/6 1/6 | 1/6
P(A C) = 1/6 1/6 | 1/6
P(B C) = 1/3 1/3 | 1/3
P(C B) = 1/3 1/3 | 1/3
___ ___ | ___ ___
1/3 2/3 | 2/3 1/3
So, if you decide (in advance) to always change
your choice, the odds to win are 2 to 1.
However, if you let the audience decide for you
what to do, the odds are 50/50!! :-)
I hope this cleared the sky
Bert
Milton Tinkoff
I just wanted to post a more obvious solution to this puzzle.
I don't think anyone has posted it yet.
There are three doors: prize, empty1, empty2.
Here are the possible situations:
Door you pick Door Monty opens Door if you switch
------------- ---------------- ------------------
prize empty1 empty2) These situations
prize empty2 empty1) are equivalent.
empty1 empty2 prize
empty2 empty1 prize
There. If you always switch, 2 times in 3 you will get the prize.
--
-------------------------------------------------------------------------------
Milt Tinkoff | "The average man is a
Silicon Graphics Inc. | stupid man."
mi...@waynes-world.esd.sgi.com | -Ed Mao
Bernie Cosell
}Flip a coin.
}In analyzing this problem we must consider Monty Hall's strategy. It might be
}anything, since the problem statement is silent on the subject. Previous
}solutions have assumed that Monty will always show you a goat, but suppose that
}Monty only shows you a goat when you have picked the car, and otherwise doesn't
}show you anything? Then switching is exactly the wrong thing to do, since
}every time the player has an opportunity to switch, he should refuse.
The problem is that this line of reasoning depends on the fact that the problem
incompletely describes the situation --- that is, instead of presuming that
what you've been given [in the informal, and admittedly usually not rigorous
and precise statement of the problem] is *complete* as given, and so you should
assume that you don't need to know a whole lot else to solve the problem, you
instead assume that each ambiguity in the statement of the problem can take an
arbitrary turn, and so your analysis *requires* that more information be given.
Isn't it easier just to assume that you have enough information at hand? In
particular, if one takes the phrase in the informal statement of the problem
"and then Monty shows you a zonk" [I think 'zonk' is that actual term they used
to use on LMaD for the non-winners] that Monty *always* shows you a zonk?
...and all this ...
}This problem should be analyzed a la von Neumann and Morganstern. Monty has
}three strategies that may result in showing a goat: always show a goat, show a
}goat only when the player has picked a car, or show a goat only when the player
}has picked a goat. ...
because it is bogus in exactly the same way: since you're making your
own assumptions about how the problem reads. Who ever said that Monty
has to adhere to a *single*, *simple* strategy in what he shows you?
[once, of course, you make the rather arbitrary, and unwarranted,
extension from the problem, that he EVER woudl should you the actual
prize]. And further, on what basis do you conclude that you would LOSE
if he revelaed the prize? maybe [since you're freelancing the rules to
some OTHER, unstated, problem], you would *win* the prize if Monty
reveals it?
Altogether, it is a pandora's box that belies these fancy-seeming
analyses: EITHER you solve it in the normal way, as generally
intended, *OR* you refuse to solve it because it is so underspecified
[to your reading] so as to be unsolvable. Trying to make up your own
unstated 'rules' for unmentioned events, and then base an elaborate
analysis on those assumptions is bogus.
/Bernie\
James Smith
following argument might help make things a little clearer. It's what
convinced me.
Monty has three cards. You choose one, and he keeps the other two. He
then asks you whether you want to keep the one you have chosen, or swap
it for the two he has. Therefore, the odds are 2:1. The catch is in
the phraseology. Although you are picking both of Monty's cards, he
displays one to you and you select the other one, which makes it seem
as though you are only picking one.
Robert Ebert
Not that I want to contribute more to the Monty-mania that's sweeping the
net yet another time, but I need a snappy comeback for the dweebs in my
office who ask me to solve the Monty Haul problem. Here's an (easy)
example:
Okay, so you figured out that switching is better. Now, what if the management
at the network decides that Monty's giving out too much good stuff, so they
tell him to sometimes not give the contestants a chance to switch, based
on a separate probability problem. (Flipping a coin will do, but in the
real game this could be related to an earlier deal...)
In any case, what allow:disallow ratio does the coin need to produce in order
to bring the odds of winning back to 50:50 assuming the contestant will always
switch, if given the chance.
I'll illustrate with a "fair" coin:
I (the contestant) pick door A. Monty exposes door C. He then flips a coin,
and if it comes up heads, he allows me to switch. If it comes up tails he
doesn't allow me to switch, and exposes door B, then my door.
Q1: What are my odds of winning the prize?
Q2: How "unfair" does the coin have to be to make the whole game even.
Q3: Do you have any other variations that might make for interesting puzzles.
Q4: Would it be better to just punch my witty office-mates in the mouth?
--Bob
James Smith
>>If you pick A and Monty eliminates B and you choose C you lose.
>>If you pick A and Monty eliminates C and you choose A you win.
>>If you pick A and Monty eliminates C and you choose B you lose.
>>If you pick B and Monty eliminates C and you choose B you lose.
>>If you pick B and Monty eliminates C and you choose A you win.
>>If you pick C and Monty eliminates B and you choose C you lose.
>>If you pick C and Monty eliminates B and you choose A you win.
>>There are only eight possible cases, and so the odds are 50-50.
Doc O'Leary writes:
> What you are failing to consider in those cases is the initially probability
> that you have chosen the correct door, A, which is only 1/3.
It amazes me that so many people have pointed this out. One could add the
four extra cases
>>If you pick B and Monty eliminates C and you choose B you lose.
>>If you pick B and Monty eliminates C and you choose A you win.
>>If you pick C and Monty eliminates B and you choose C you lose.
>>If you pick C and Monty eliminates B and you choose A you win.
to give equal chances of starting with A, B, or C and the odds would
still be 50-50. My point was that for each "you win" there is an equally
likely "you lose".
I had considered this. My error in reasoning lay elsewhere.
Michael Foster
In analyzing this problem we must consider Monty Hall's strategy. It might be
anything, since the problem statement is silent on the subject. Previous
solutions have assumed that Monty will always show you a goat, but suppose that
Monty only shows you a goat when you have picked the car, and otherwise doesn't
show you anything? Then switching is exactly the wrong thing to do, since
every time the player has an opportunity to switch, he should refuse.
This problem should be analyzed a la von Neumann and Morganstern. Monty has
three strategies that may result in showing a goat: always show a goat, show a
goat only when the player has picked a car, or show a goat only when the player
has picked a goat.
The player's strategies are: switch or don't switch.
David Chesler
Since I've spent some time on this (including an hour have the correct
answer beaten into my skull by my ex-roommate who never took prob and stat
[I've taken it about three times]) here are some thoughts, and a script
of a simulation, and another simlulation written in C. I've just gone
through what was written, my simulation is better, so there, and thanks
to whomever on the coin example.
It's eay to show that if you act randomly you get the car 50% of the time.
So what?
Counting all events (which is what Marilyn Mach vos Savant Jarvik did in
the Parade, I think) is not valid. What if you flip a coin. If it comes
up heads Monty Hall shows you the Ace of Spades. If it comes up coins
Monty Hall shows you a random card. Since he can show you 52 cards if
you get tails, but only one card if you get heads, the chances must be
only 1 out of 53 that you get heads, right?
Conversely, someone on the net has a coin that comes up heads 2/3 of
the time. Like he said, there are only 2 sides, so it's 50-50, right?
Marilyn is a good example of what IQ tests don't measure, but she's
got the credited answer here.
----------------------------------------------------------------------
(On Friday I mailed to my co-workers that the chances are 50-50, but
that was based on random counting. This morning I mailed them:)
Cancel last message. My friends set me straight over the weekend.
The first guess is a partitioning of the set of doors into the set
of one you picked and all the rest. When Monty shows you the goat
he is collapsing all the value behind all the doors in the remainder
set into the remaining door. There is a 1/3 chance the car was in
the small set, a 2/3 chance it was in the big set.
If the answer were (as we thought) 50-50, then if there were a million
doors then the car would be no more likely to be in the first door you
pick than in all the remaining doors. If Monty shows you 999,998 goats
then the car is behind either your door or the remaining door. If you
guessed right the first time, you shouldn't swap. But this will only
happen one time in a million. (In that case the remaining door also
has a goat.) In all the other cases the remaining door, which represents
the top door out of 999,999, is more likely to have the car than "your"
door, which is the top door out of 1.
(The statement of the question forces Monty to show you all the goats,
which he might not do in real life.)
Do we need to write a simulation?
Consider a boxing match between the champion of Lower Slobovia and the
champion of Upper Slobovia. On whom should you bet? A priori, with no
further knowledge, it's even money. But what if you know that Lower Slobovia
has a population of 10, and Upper Slobovia has a population of 10,000,000?
Add further that the former colonial power partitioned the population by
lot. Even though there are only two boxers in the ring, the boxer who
is no worse than all the others in his country will be more likely to be
the winner if he is from the larger country. (There is a chance that
out of the 10 people who were partitioned into Lower Slobovia, the
champion boxer of all of Slobovia is among them, but this is a low
probability event.)
Yes, if you swap or don't swap at random you'll get the car half the
time, but that's not the question.
And I still don't think Marilyn is wise.
----------------------------------------------------------------------
(One of my co-workers still wasn't convinced, so I sent out:)
There is a simulation [reproduced below] which runs the
"Let's Make a Deal" game, with the rules that for any number of doors
you pick a door, then Carol Merrill opens all but one of the remaining
doors (if you picked the car, she leaves one at random; if you did not
pick the car she leaves the car) and Monty Hall lets you swap from your
current choice to the remaining door.
The code and its partial output should make it clear that this is
what is going on. Running one million trials for each of three, four,
five, ten, one hundred, and one million doors, it's pretty clear that
the chance that the car is behind the remaining door (that swapping wins)
is (n-1)/n, where n is the number of doors.
My mathematically inclined ex-roommate says yes, this is a consequence
of Bayes Theorem. My practically inclined ex-roommate says it's thus
because of the collapsing of all the cars behind the larger of the two
sets into which the doors were partitioned into the "most car-full example"
door.
If you are Mr. Short-term Memory, and forget which door is which, then
the chances are 50%, that is there are two doors and the car is behind
exactly one of them, so with no a priori knowledge the best you can do
is guess.
My more pragmatic ex-roommate says of course this isn't how the real-life
case works anyway -- Monty Hall is more likely to offer you the swap only if
he wants you be dissuaded from your original choice, that is only if you
guessed the car on the first choice. Field work should determine how
often Mr. Hall offers beneficial and detrimental swaps.
Here are the results of the simulation:
Summary: (in 1000000 trials)
Number of Doors Swapping wins
--------------- -------------
3 67%
4 75%
5 80%
10 90%
100 99%
1000000 100%
----------------------------------------------------------------------
Here is more complete output of the simulation: (Yes, it only shows
the trials for the first few trials of each number of doors, and
doors are numbered from zero.)
The output of the simulator is the next 180 lines. The C source
follows.
Test the random number generator.
800,000 trials, find value between 0 and 7 inclusive.
-1 0 1 2 3 4 5 6 7 8
0 100419 99749 100638 99827 100059 99519 100118 99671 0
1000 trials, find value between 0 and 2 inclusive, two times.
Distribution:
0 1 2
Car was in: 308 368 324
First guess: 349 307 344
First guess won on 379 times.
Run a trial with 3 doors:
(The car is behind door number 1)
You guess door number 2
Monty says, "You can keep door number 2, or swap for door number 1."
*** Swapping would have yielded a car! ******
(The car is behind door number 0)
You guess door number 0
Monty says, "You can keep door number 0, or swap for door number 1."
--- Swapping would have yielded a goat. -----
(The car is behind door number 2)
You guess door number 2
Monty says, "You can keep door number 2, or swap for door number 1."
--- Swapping would have yielded a goat. -----
(The car is behind door number 1)
You guess door number 2
Monty says, "You can keep door number 2, or swap for door number 1."
*** Swapping would have yielded a car! ******
(The car is behind door number 1)
You guess door number 0
Monty says, "You can keep door number 0, or swap for door number 1."
*** Swapping would have yielded a car! ******
(The car is behind door number 2)
You guess door number 0
Monty says, "You can keep door number 0, or swap for door number 2."
*** Swapping would have yielded a car! ******
(The car is behind door number 1)
You guess door number 2
Monty says, "You can keep door number 2, or swap for door number 1."
*** Swapping would have yielded a car! ******
(The car is behind door number 0)
You guess door number 2
Monty says, "You can keep door number 2, or swap for door number 0."
*** Swapping would have yielded a car! ******
(The car is behind door number 2)
You guess door number 1
Monty says, "You can keep door number 1, or swap for door number 2."
*** Swapping would have yielded a car! ******
(The car is behind door number 1)
You guess door number 0
Monty says, "You can keep door number 0, or swap for door number 1."
*** Swapping would have yielded a car! ******
Swapping won in 666264 times out of 1000000 trials, or 67%
Distribution:
0 1 2
Car was in: 332814 333718 333468
First guess: 332846 333674 333480
Run a trial with 4 doors:
(The car is behind door number 2)
You guess door number 1
Monty says, "You can keep door number 1, or swap for door number 2."
*** Swapping would have yielded a car! ******
(The car is behind door number 0)
You guess door number 3
Monty says, "You can keep door number 3, or swap for door number 0."
*** Swapping would have yielded a car! ******
(The car is behind door number 2)
You guess door number 1
Monty says, "You can keep door number 1, or swap for door number 2."
*** Swapping would have yielded a car! ******
(The car is behind door number 3)
You guess door number 2
Monty says, "You can keep door number 2, or swap for door number 3."
*** Swapping would have yielded a car! ******
Swapping won in 750272 times out of 1000000 trials, or 75%
Distribution:
0 1 2 3
Car was in: 250416 249943 249686 249955
First guess: 249364 249935 251144 249557
Run a trial with 5 doors:
(The car is behind door number 3)
You guess door number 0
Monty says, "You can keep door number 0, or swap for door number 3."
*** Swapping would have yielded a car! ******
(The car is behind door number 0)
You guess door number 0
Monty says, "You can keep door number 0, or swap for door number 1."
--- Swapping would have yielded a goat. -----
(The car is behind door number 1)
You guess door number 0
Monty says, "You can keep door number 0, or swap for door number 1."
*** Swapping would have yielded a car! ******
(The car is behind door number 0)
You guess door number 1
Monty says, "You can keep door number 1, or swap for door number 0."
*** Swapping would have yielded a car! ******
Swapping won in 799690 times out of 1000000 trials, or 80%
Distribution:
0 1 2 3 4
Car was in: 199647 200138 200390 199996 199829
First guess: 200211 199343 199944 200298 200204
Run a trial with 10 doors:
(The car is behind door number 4)
You guess door number 1
Monty says, "You can keep door number 1, or swap for door number 4."
*** Swapping would have yielded a car! ******
(The car is behind door number 4)
You guess door number 7
Monty says, "You can keep door number 7, or swap for door number 4."
*** Swapping would have yielded a car! ******
(The car is behind door number 4)
You guess door number 3
Monty says, "You can keep door number 3, or swap for door number 4."
*** Swapping would have yielded a car! ******
Swapping won in 900195 times out of 1000000 trials, or 90%
Distribution:
0 1 2 3 4 5 6 7 8 9
Car was in: 100058 100141 99642 100036 100312 99718 100230 99821 99973 100069
First guess: 99556 100043 100107 99961 100268 99735 100337 100193 100257 99543
Run a trial with 100 doors:
(The car is behind door number 16)
You guess door number 41
Monty says, "You can keep door number 41, or swap for door number 16."
*** Swapping would have yielded a car! ******
(The car is behind door number 71)
You guess door number 93
Monty says, "You can keep door number 93, or swap for door number 71."
*** Swapping would have yielded a car! ******
Swapping won in 990135 times out of 1000000 trials, or 99%
Run a trial with 1000000 doors:
(The car is behind door number 17386)
You guess door number 190634
Monty says, "You can keep door number 190634, or swap for door number 17386."
*** Swapping would have yielded a car! ******
(The car is behind door number 320150)
You guess door number 883438
Monty says, "You can keep door number 883438, or swap for door number 320150."
*** Swapping would have yielded a car! ******
Swapping won in 999999 times out of 1000000 trials, or 100%
Summary: (in 1000000 trials)
Number of Doors Swapping wins
--------------- -------------
3 67%
4 75%
5 80%
10 90%
100 99%
1000000 100%
----------------------------------------------------------------------
Here is the C source to the Monty Hall simulation. It is portable,
but uses the newer BSD random() function. I have built in code to
check the random number generator, so swap in whatever random number
generator you want.
/* -------------------- cut here ----------------------------*/
/* This routine plays Let's Make a Deal - should you swap or */
/* keep your original choice if Monty Hall shows you a goat */
/* behind one of the remaining doors? */
/* To keep this simple and very easy to verify, the simulation is */
/* incredibly naive, and coded so. */
/* David Chesler 12/10/90 */
#define NUM_OF_TRIALS 1000000
#define NUM_TYPES 6
#define NUM_OF_DOORS0 3
#define NUM_OF_DOORS1 4
#define NUM_OF_DOORS2 5
#define NUM_OF_DOORS3 10
#define NUM_OF_DOORS4 100
#define NUM_OF_DOORS5 1000000
#include <stdio.h>
long random();
long getpid();
int rand_int(range)
/* Returns random number from 0 to range-1 */
int range;
{
return (int)(random() % range);
}
int merrill(where_car, where_guess, num_doors)
int where_car;
int where_guess;
int num_doors;
{
int door_to_leave;
if (where_car == where_guess)
/* In this case the player guessed the car in the first place, */
/* so we can leave any door except the guessed door. */
do
door_to_leave = rand_int(num_doors);
while (door_to_leave == where_guess);
else
/* In this case the player guessed a loser, so we must leave */
/* the door with the car. */
door_to_leave = where_car;
return door_to_leave;
}
main()
{
int i;
register int j, wins;
register int num_doors;
register int print_now;
register int dist_now;
register int guess, car_is_here, swap_option;
int num_door_array[NUM_TYPES];
int percentage_win[NUM_TYPES];
int rand_array[10];
int car_is_array[10];
int first_guess_array[10];
srandom((int)getpid()); /* <- I got this from horny.c, off the net */
num_door_array[0] = NUM_OF_DOORS0;
num_door_array[1] = NUM_OF_DOORS1;
num_door_array[2] = NUM_OF_DOORS2;
num_door_array[3] = NUM_OF_DOORS3;
num_door_array[4] = NUM_OF_DOORS4;
num_door_array[5] = NUM_OF_DOORS5;
/* First some sanity checks on the random number generator */
printf("Test the random number generator.\n");
printf(" 800,000 trials, find value between 0 and 7 inclusive.\n");
for (i = 0; i < 10; i++)
rand_array[i] = 0;
for (i = 0; i < 800000; i++)
rand_array[rand_int(8)+1]++;
for (i = 0; i < 10; i++)
printf("%4d ",i-1);
printf("\n");
for (i = 0; i < 10; i++)
printf("%6d ",rand_array[i]);
printf("\n\n");
printf(" 1000 trials, find value between 0 and 2 inclusive, two times.\n");
for (i = 0; i <= 2; i++)
{
rand_array[i] = 0;
car_is_array[i] = 0;
first_guess_array[i] = 0;
}
wins = 0;
for (i = 0; i < 1000; i++)
{
car_is_here = rand_int(3);
rand_array[car_is_here]++;
car_is_array[car_is_here]++;
guess = rand_int(3);
rand_array[guess]++;
first_guess_array[guess]++;
if (car_is_here == guess)
wins++;
}
printf(" Distribution:");
printf("\n ");
for (i = 0; i < 3; i++)
printf("%3d ",i);
printf("\n Car was in:");
for (i = 0; i < 3; i++)
printf(" %3d",car_is_array[i]);
printf("\nFirst guess:");
for (i = 0; i < 3; i++)
printf(" %3d",first_guess_array[i]);
printf("\n\nFirst guess won on %d times.\n\n",wins);
/* Simulation part begins here */
for (i = 0; i < NUM_TYPES; i++)
{
num_doors = num_door_array[i];
wins = 0;
printf("\nRun a trial with %d doors:\n",num_doors);
dist_now = num_doors <= 10;
if (dist_now)
for (j = 0; j < num_doors; j++)
{
car_is_array[j] = 0;
first_guess_array[j] = 0;
}
for (j = 0; j < NUM_OF_TRIALS; j++)
{
/* Print a decreasing number of examples */
print_now = (num_doors == 3)?(j < 10):(j < (10 / num_doors + 2));
/* In this phase, the car is randomly placed behind a door */
car_is_here = rand_int(num_doors);
if (dist_now)
car_is_array[car_is_here]++;
if (print_now)
printf("\n (The car is behind door number %d)\n",
car_is_here);
/* In this phase, contestant picks a door */
guess = rand_int(num_doors);
if (dist_now)
first_guess_array[guess]++;
if (print_now)
printf(" You guess door number %d\n",guess);
/* In this phase Carol Merrill reveals goats behind all but */
/* one of the doors contestant didn't pick. */
swap_option = merrill(car_is_here, guess, num_doors);
if (print_now)
{
printf(" Monty says, \"You can keep door number %d,",
guess);
printf(" or swap for door number %d.\"\n",swap_option);
}
if (swap_option == car_is_here)
wins++;
if (print_now)
/* Now contestant could swap or remain. */
if (swap_option == car_is_here)
printf(" *** Swapping would have yielded a car! ******\n");
else
printf(" --- Swapping would have yielded a goat. -----\n");
} /* For the trials */
percentage_win[i] = (int)((double)(wins)/NUM_OF_TRIALS*100+0.5);
printf("\nSwapping won in %d times out of %d trials, or %d%%\n",
wins,j,percentage_win[i]);
/* Sanity check: show distributions of cars and guesses. */
if (dist_now)
{
printf(" Distribution:");
printf("\n ");
for (j = 0; j < num_doors; j++)
printf("%3d ",j);
printf("\n Car was in:");
for (j = 0; j < num_doors; j++)
printf(" %3d",car_is_array[j]);
printf("\nFirst guess:");
for (j = 0; j < num_doors; j++)
printf(" %3d",first_guess_array[j]);
printf("\n");
}
} /* Next type */
printf("\n\n Summary: (in %d trials)\n\n",NUM_OF_TRIALS);
printf("Number of Doors Swapping wins\n");
printf("--------------- -------------\n");
for (i = 0; i < NUM_TYPES; i++)
printf(" %7d %3d%%\n",
num_door_array[i],percentage_win[i]);
} /* main */
/* -------------------------------- cut here -------------------------- */
/* David Chesler <che...@netrix.enet.dec.com> formerly da...@prism.tmc.com */
/* Working at DEC Littleton Mass, speaking for myself */
Dan Hoey
>Let me offer an attempt at a reasonably precise statement of a Monty
>Hall scenario in which the "switching wins 2/3 of the time" answer is
>correct.
>"You will play a game with a fellow named Monty Hall, and he has to
>play by the rules I'm about to describe....
and goes on to describe such a scenario. The problem is that in the real world
we are never given a set of rules that Monty must follow. Well, actually he
says there's a prize behind one of those doors, so I guess that's a rule. And
he says he'll give it to us if we pick it. But this bit about being given a
second choice has never been promised. I suppose a contestant who has just
been selected to play the game might ask whether Monty is going to open an
empty, unchosen door and offer a chance to switch. Monty might just say,
``you're not the contestant we're looking for'' and take the next contestant.
The audience is full of people who would like a chance to win a prize.
I've heard that Monty has an evil twin who sometimes stands in on the show.
The evil twin offers you a choice of three doors, behind one of which is a
prize. If you pick an empty door, he opens up the door with the prize (or the
door you chose, or both) and says, ``Hard cheese, you lose.'' If you pick the
door with the prize, he opens one of the empty doors and offers you a chance to
switch. This is a scenario in which switching never wins, and not switching
wins one third of the time.
Then there's Saint Monty, who offers you a choice of doors and only offers you
a chance to switch when you didn't pick the prize. When Saint Monty runs the
show, switching wins every time, and not switching wins one third of the time.
The problem is that you can't tell which Monty you've got. Has anyone got firm
statistics on how often the contestants on Let's Make a Deal got a second
chance, correlated to the accuracy of their first guess? Was there any
correlation with the value of the prize? Did Monty's behavior change when the
contestant had been through a few rounds, perhaps in response to the
contestant's strategy?
I solicit reliable answers to these questions via email.
Dan Hoey
Ho...@AIC.NRL.Navy.Mil
Sam Needham
>Puzzlers:
>
>Not that I want to contribute more to the Monty-mania that's sweeping the
>net yet another time,
>Q4: Would it be better to just punch my witty office-mates in the mouth?
>
> --Bob
Okay, Okay, Okay, I'm a reasonable person, I'll try and keep my voice down.
I think it's just fine that people with short memories,or new to the net
want to discuss Monty.
I think rec.puzzles - and not, say, sci.math - is the right place.
I applaud the well-meaning and well-informed people trying to convince
the less well-informed of the correct answer.
I sympathise with your office mental hygiene problem.
But I don't want to read any of it. I CAN'T TAKE ANY MORE.
This is a plea for understanding.
PLEASE DON'T CHANGE THE SUBJECT LINE OF THIS THREAD ANY MORE. THERE ARE
NOW AT LEAST
****>>>>> SIX <<<<<****
DIFFERENT SUBJECT LINES ABOUT THE SAME THREAD. I KILL IT BUT IT
***** JUST
****** WON'T
****** DIE!
Please.
Just follow-up to one of the many fine subject lines about Monty.
Don't amaze me with your creativity.
Thank you.
And note the follow-up: line, please.
Sam
Smug de-facto critics in their movie backdrop cities snarling
"Shut up and listen! Life's a lonely escalator, it's a fool who doesn't know
you have to leap off at the end". Cold Chisel
Sam Needham UCLA Math Grad Student
Phil Servita
>>"You will play a game with a fellow named Monty Hall, and he has to
>>play by the rules I'm about to describe....
>
>and goes on to describe such a scenario. The problem is that in the real world
>we are never given a set of rules that Monty must follow. Well, actually he
>says there's a prize behind one of those doors, so I guess that's a rule. And
>he says he'll give it to us if we pick it. But this bit about being given a
>second choice has never been promised. I suppose a contestant who has just
>been selected to play the game might ask whether Monty is going to open an
>empty, unchosen door and offer a chance to switch. Monty might just say,
>``you're not the contestant we're looking for'' and take the next contestant.
>The audience is full of people who would like a chance to win a prize.
>
>I've heard that Monty has an evil twin who sometimes stands in on the show.
>The evil twin offers you a choice of three doors, behind one of which is a
>prize. If you pick an empty door, he opens up the door with the prize (or the
>door you chose, or both) and says, ``Hard cheese, you lose.'' If you pick the
>door with the prize, he opens one of the empty doors and offers you a chance to
>switch. This is a scenario in which switching never wins, and not switching
>wins one third of the time.
>
>Then there's Saint Monty, who offers you a choice of doors and only offers you
>a chance to switch when you didn't pick the prize. When Saint Monty runs the
>show, switching wins every time, and not switching wins one third of the time.
>
>The problem is that you can't tell which Monty you've got.
This isnt a problem. Since we have no information on which monty we have,
but we know that he asked "Wanna Switch", we then can compute our
probability of winning if we switch:
P(win if switch) = P(Hell Monty) * P(win if switch | Hell Monty)
+ P(Normal Monty) * P(win if switch | Normal Monty)
+ P(Saint Monty) * P(win if switch | Saint Monty)
But: the probabilities of the given monties are NOT EQUAL! Since we have no
information on which monty showed up for our guessing game, we have no choice
but to assign 1/3 to each for the probability of doing our game. BUT, once
he asks us to switch, we have more information! Normal monty will offer a
switch every time, Hell Monty will offer a switch 1/3 of the time (only if
we picked the right door to begin with), and Saint Monty will offer a switch
2/3 of the time (only if we guessed wrong). Simple applications of Bayes
Theorem gives us probabilities of 1/6, 1/2, and 1/3 for Hell Monty, Normal
Monty, and Saint Monty, respectively.
Example:
P (Hell Monty | AskToSwitch) = P (HM | ATS) =
P(ATS | HM) * P(HM)
------------------------------------------------------------------ ==
P(ATS | HM) * P(HM) + P(ATS | NM) * P(NM) + P(ATS | SM) * P(SM)
1/3 * 1/3
--------------------------------------------------- == 1/6
1/3 * 1/3 + 1 * 1/3 + 2/3 * 1/3
Now we can do the final computation:
P(WinIfSwitch) = P(HM) * P(WIS | HM) + P(NM) * P(WIS | NM) + P(SM) * P(WIS |SM)
= 1/6 * 0 + 1/2 * 2/3 + 1/3 * 1
= 2/3 , exactly the same as if people assume only
Normal Monties.
QED.
-phil, with apologies to any
non-bayesians out there.
(there must be some of you...)
PS. This *should* end up being the final difinitive treatment of this
problem. Of course, Hell Monty and Saint Monty are not the only
theoretical possibilities for alternate Monty strategies. Eg:
Pseudo-Hell Monty, if you guess the right door, he offers you
a switch 50% of the time. There are at least countably infinite
possible Monty strategies (to avoid another flame fest...), each
one having a "dual", which ends up essentially cancelling it out
with respect to the Normal Strategy. Have fun with the integral...
rpj...@ccu.umanitoba.ca
>In article <10...@emanon.cs.jhu.edu> arro...@cs.jhu.edu (Kenneth Arromdee) writes:
>>
>>I also think this should go in the rec.puzzles FAQ if it's not already. (What
>>ever happened to that list anyway?)
>
>Since you ask, I have made such a list available via email. To get an index
>and instructions for requesting articles (puzzles and solutions), send a letter
>with the single line:
>send index
>to net...@peregrine.com.
>
>The Monty Hall puzzle is indeed on the list, and both the problem and
>solution from the list are given below (although it is the same as yours).
>
At the risk of making a fool of myself, let me describe another way of
looking at this (the risk of embarrassing myself is that I am a sporadic
reader of this group and I imagine this view has already been posted a
number of times).
Take 100 doors. You pick one at random, so your chances of getting the
prize are 1 in 100. Now Monty says, "You can keep that door, or I'll allow
you to switch to THE ENTIRE SET OF ALL OF THE OTHER DOORS (that is, all
99 of the other doors)." Of course, you'd have to be a total loon
(or a Larouchite) to not switch, since this changes your odds to 99 in 100.
Note, however, that this is exactly what is happening. The fact that Monty
exposes 98 of those 99 doors is irrelevant. In effect, he is allowing you
to take the set of 99 other doors; he just happens to be showing you that
at least 98 of those 99 doors have nothing behind them, but this is something
you already know.
Comments?
R. Day
... too lazy for a sig file ...