mathematics contest
Ruth Bevan
contest for high school students went something like this:
If there are six boxes under one of which a key is hidden, and four of
those boxes have been overturned, which is the probability that I will
discover the key when the fifth box is overturned?
the answer apparently was 1/6
this caused a raging argument over dinner saturday, and I was wondering if
anyone in net-land knew if this actually was the question and answer that
was given at the competition? (Everyone I ate with thought the answer should
be 1/2)
thanks!
Ken Arromdee
>those boxes have been overturned, which is the probability that I will
>discover the key when the fifth box is overturned?
>the answer apparently was 1/6
How can it possibly be 1/2? Nobody said the key couldn't be under one of
the 6 boxes which were overturned.
--
"I wonder how the children of a Carggian/Winathian marriage would look at
reality?"
Kenneth Arromdee (UUCP: ....!jhunix!ins_akaa; BITNET: g49i0188@jhuvm;
INTERNET: arro...@crabcake.cs.jhu.edu) (please, no mail to arrom@aplcen)
Charles Cleveland
)I heard last week that the winning question in the national mathematics
)contest for high school students went something like this:
)
)If there are six boxes under one of which a key is hidden, and four of
)those boxes have been overturned, which is the probability that I will
)discover the key when the fifth box is overturned?
)
)the answer apparently was 1/6
)
)this caused a raging argument over dinner saturday, and I was wondering if
)anyone in net-land knew if this actually was the question and answer that
)was given at the competition? (Everyone I ate with thought the answer should
)be 1/2)
Of course the answer is 1/6. All we know is that four boxes have been
overturned. We do not know whether the key has already been revealed by
any of these four overturnings. Perhaps the box turner doesn't care about
keys and is, in fact, looking for a pair of sox or some food. (I look for
sox and food more often than I look for keys. That's why the answer was
obvious to me.)
Of course, if I were taking the test I would say 1/2. I may be clever,
but that doesn't mean I'm an idiot. And if 1/6 was the 'correct' answer
to THIS question, its author should have his or her gonads ripped off or
out. While being held down by the students who had to take the test and
who probably took it a lot more seriously than the aforesaid author.
Bernie Cosell
}those boxes have been overturned, which is the probability that I will
}discover the key when the fifth box is overturned?
}
}the answer apparently was 1/6
}
}be 1/2)
I'm afraid I concur that it shoudl be 1/2. Consider if you change the puzzle
so that *five* of the six boxes are overturned. *now* what is the prob that
it is under the last box? Still 1/6? What happens the other 5/6 of the
time? If not 1/6, what happened between overturning the fourth box and
overturning the fifth box to get the probabilities to "collapse".
I think that there must be a PILE of "how you reached this point in the
game" info that is missing. [just as with Monty Hall: unless you have the
_complete_ story of who picks what, when you can't figure out the final
answer]
__
/ ) Bernie Cosell
/--< _ __ __ o _ BBN Sys & Tech, Cambridge, MA 02238
/___/_(<_/ (_/) )_(_(<_ cos...@bbn.com
Alan Craig
>those boxes have been overturned, which is the probability that I will
>discover the key when the fifth box is overturned?
>
>the answer apparently was 1/6
Note that the question doesn't say that the key wasn't under
one of the already overturned boxes
Alan
Steve Masticola
them well. Blindfold yourself and turn over four boxes. Now turn over
a fifth box and feel around for the key. What is the probability that
you'll find it under the fifth box?
This is really an exercise in thinking about what is _not_ given (or
assumable) from the problem.
- Steve (mast...@paul.rutgers.edu)
Bryan Putnam
>those boxes have been overturned, which is the probability that I will
>discover the key when the fifth box is overturned?
>the answer apparently was 1/6
>
outcome for the first 4 turn-overs, then the anwswer is either
0 or 1/2, depending on whether the key has or has not already been
found. If you do not know the outcome of the first 4 turn-overs,
then the probability of seeing the key on the 5th turn-over is 1/6.
Bryan F. Putnam User Services Programmer
Internet: a...@s.cc.purdue.edu Purdue University Computing Center
Paul Campbell
>those boxes have been overturned, which is the probability that I will
>discover the key when the fifth box is overturned?
>
>the answer apparently was 1/6
The problem description doesn't say whether or not the key was found
under any of the first 4 boxes (maybe it was discovered already).
It seems like the wording is very important, such a 'trick' question
(with ambiguous wording or one that requires one to make assumptions
without evidence) probably doesn't have a place in a contest.
Paul
--
Paul Campbell
Taniwha Systems Design UUCP: ..!mtxinu!taniwha!paul
Oakland CA AppleLink: D3213
Achtung! Ve are from ze Interface Police! Ve vant to look und feel!
Ramsey W. Haddad
>contest for high school students went *SOMETHING* like this:
^^^^^^^^^ [emphasis added]
>
>If there are six boxes under one of which a key is hidden, and four of
>those boxes have been overturned, which is the probability that I will
>discover the key when the fifth box is overturned?
>
>the answer apparently was 1/6
People are claiming that either
(1) You can't assume that the overturned boxes haven't revealed the key.
Hence prob = 1/6.
(2) You can assume that the overturned boxes haven't revealed the key.
Hence prob = 1/2.
Given that this is not the exact wording of the original problem, I
think that claim (1) is tenuous at best, and only plausible at all
because it yields the claimed correct answer.
One person mentioned that this problem is not like the Monty Hall
problem --- where the answer *would be* 1/6. He is right that *as
stated in the message* this is not like the Monty Hall Problem.
My *conjecture* is that the original problem was indeed a variant of
the Monty Hall problem. In telling and re-telling, crucial wording in
the original was dropped by people who didn't realize that it was
crucial.
--
Ramsey W Haddad
Donald Benson
I watched you turn the boxes over, I could answer with certainty, "0" or
"1/2" depending on whether I had seen the key yet. I predict the students who
answered "wrong" are better at applying their skills to the real world, and
those who answered "right" will be condemned to a life of usless, theoretical
mathematics.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
There. Did I fail to offend anyone?
DonB
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Rick Wilson
>those boxes have been overturned, which is the probability that I will
>discover the key when the fifth box is overturned?
>
>the answer apparently was 1/6
>
There is nothing in the question as it is stated above that says the
key wasn't under any of the overturned boxes. So, overturning a box
doesn't remove it from the problem.
The question boils down to: 6 boxes, 1 key -> 1/6.
What bothers me is that unless you know the answer they want you can't
tell if the deceit was intended, or if it was a badly written question.
--
Rick Wilson
ri...@tekfdi.TEK.COM
Michael D. Kersenbrock
<
<If there are six boxes under one of which a key is hidden, and four of
<those boxes have been overturned, which is the probability that I will
<discover the key when the fifth box is overturned?
<
<the answer apparently was 1/6
<
<this caused a raging argument over dinner saturday, and I was wondering if
<anyone in net-land knew if this actually was the question and answer that
<was given at the competition? (Everyone I ate with thought the answer should
<be 1/2)
<
<thanks!
It's 50-50 to be under the fifth box ONLY if it isn't showing already
(and you know it). The status of already showing (or not) ISN'T a given in
the problem. The question (above) specifically said "discover the key
when the fifth box is overturned". There is a 2/3 chance that it is
already discovered, a 1/6 chance that it is under #5, and a 1/6 chance
being under #6.
--
Mike Kersenbrock
Tektronix Microprocessor Development Products
mich...@copper.MDP.TEK.COM
Aloha, Oregon
Jeffrey Naiman
Not too long ago, I was taking these tests, and they were much too careful
to put in a question like this. Now the original question did not say that
the wearer was blindfolded, nor did it say that the opener had eyes. It
did not state whether or not the key was invisible.
I believe the correct answer should be that the probability is 0 2/3 of the
time and 1/2 1/3 of the time. It is not proper to say that the probability
of finding it under the fifth box is 1/6, because this does not take into
account the information available to us: namely that we could have looked
to see whether the key has appeared or not. Now, is it an assumption that
the contents of the first four boxes are visible? This whole thing sounds like
a quantum mechanics problem....
- Jeff Naiman (nai...@yale.edu)
Steven Den Beste
> I think a question like this is important for teaching programmers to deal
> with unexpected situations (Aircraft have crashed due to false assumptions;
> see COMP.RISKS for more info.)
>
> If asked, "Which would fall faster, a 5lb brick or a 10lb brick, assuming
> no atmosphere?", I would say, "The 10lb brick".
>
Actually, the way to ask the question is "Which would touch the ground first, a
5lb brick or a 10lb brick, assuming no atmosphere?"
Given they are dropped at different times, they'll fall at the same speed, but
the 10lb brick will hit the ground sooner. (!!)
(Did someone mention something about "immovable objects"?)
Steven C. Den Beste, BBN Communications Corp., Cambridge MA
denb...@bbn.com(ARPA/CSNET/UUCP) harvard!bbn.com!denbeste(UUCP)
Chris Phoenix
>In article <525...@hpcuhb.HP.COM>, do...@hpcuhb.HP.COM (Donald Benson) writes:
>> If asked, "Which would fall faster, a 5lb brick or a 10lb brick, assuming
>> no atmosphere?", I would say, "The 10lb brick".
>Actually, the way to ask the question is "Which would touch the ground first, a
>5lb brick or a 10lb brick, assuming no atmosphere?"
>Given they are dropped at different times, they'll fall at the same speed, but
>the 10lb brick will hit the ground sooner. (!!)
Not quite... Since the earth will be closer to the 10lb brick after it is
dropped, gravity will exert more force on it, causing it to accelerate more.
So at the same time into the fall (before either hits) the 10lb brick will
be moving slightly faster.
I wonder if the QM/gravity wave people have anything to add...
--
Chris Phoenix | "I was afraid of worms! Worms, Roxanne!"
cpho...@csli.Stanford.EDU | "More input! More input!"
Usenet: The real source of the "Tastes Great" vs. "Less Filling" debate.
Disclaimer: Don't mind me, I'm just a student!
The Polymath
Actually, the problem is both more and less subtle than you realize. We
had a heck of a debate over a similar puzzle around here. I didn't
believe the answer until I simulated it in software. It still looks
counter-intuitive. Here's our version:
You're a contestant on Wheel of Fortune (Come on down!). Monty Hall
has just presented you with three doors, behind one of which is a
fabulous prize. You make your choice. Monty opens one of the other
doors. No prize is revealed. He offers you a chance to change your
choice.
Are you better off changing your choice? Does it matter?
The counter-intuitive, but correct, answer is yes, you are more likely to
win if you change your choice. The chances are 1/3 that the prize is
behind your original choice and 2/3 that it's behind the remaining door.
I didn't believe it either, and found the mathematical proof confusing, so
I simulated the situation in software and ran 1 million iterations. I'll
post the program, if anyone's that interested. Believe it or not, like it
or not, it works (and the odds of the key being in the box chosen are 1/6,
assuming it wasn't under one of those turned up).
--
The Polymath (aka: Jerry Hollombe, holl...@ttidca.tti.com) Illegitimati Nil
Citicorp(+)TTI Carborundum
3100 Ocean Park Blvd. (213) 452-9191, x2483
Santa Monica, CA 90405 {csun|philabs|psivax}!ttidca!hollombe
What`s in a name?
>In article <78...@blia.BLI.COM> ru...@blia.BLI.COM (Ruth Bevan) writes:
>>If there are six boxes under one of which a key is hidden, and four of
> [new question suggested]
>There are six boxes on the table and you put your finger on one
>of them. God (or some other omniscient being) overturns four of the
>other boxes to show that they are empty. What is the probability
>that the box you placed your finger on has the key?
>
>Stated like that, the probability is 1/6.
Say what? I assume (and I think that this assumption is reasonable based on
your wording) that after God lifts up the box that what you see (i.e. the
box being empty) is what was in the box before it was overturned (i.e.
nothing). Stated like that, the probability of the key being in the box with
your finger on it is 1/2. There are only two boxes and one of them contains
a key. In the question as stated by the original poster, the probability
could indeed be 1/6 as there is no mention of the observer knowing that four
boxes are empty.
I am making my calculations of the probability from the observer's point of
view (i.e. the "you" in the problem). If I were to calculate the
probability from the point of view of someone who didn't observe the
contents of the overturned boxes, it would be 1/6. In a sense this is
similar to a quantum mechanics problem. Probabilities change with more
information. A probability is a measure which deals with something not
measurable accurately. You don't know if the key is in that box or not, so
mathematicians, in their infinite wisdom developed a system for measuring
the chance that the key is in that box based on how much you know. When all
you know is that it is in one of six boxes, the probability is 1/6. If you
know that it is in one of two boxes the probability is 1/2. If you know
that is in one of one box, then your probability is 1 and you have answered
the question.
Therefore one can say that probability is relative to the observer. An
observer who knows more information will make a different estimate of the
odds than someone who knows less. This is why card sharks win even though
they may not know odds in general any better than their marks. A card shark
knows more about the placement of the cards (because s/he is causing it)
than his mark. Therefore s/he can make a more accurate jusgement of the
probabilities. The mark's judgement, however, (assuming that he knows his
stuff) is equally valid based on the information that s/he knows.
The wording of the question is very important. In the question originally
brought before the network, the probability from the test-takers point of
view is 1/6. The reason is that the test taker is never told the outcome of
the four overturnings. I agree that it is a rather hair-splitting
distinction, and would agree with John Berryhill that nothing of particular
importance should be decided based on the outcome of this test. None the less,
the way the question is originally worded, the answer is 1/6. The way you
choose to reword it, the answer is 1/2.
--mike
--
Mic3hael Sullivan, University of Rochester. "Life is like a sewer, what
Your favorite artificial stupidity software. you get out of it, depends on
Internet: misu...@uhura.cc.rochester.edu what you put into it."
UUCP: ...!rochester!ur-cc!misu_ltd -- Tom Lehrer
David Palmer
>If there are six boxes under one of which a key is hidden, and four of
>those boxes have been overturned, which is the probability that I will
>discover the key when the fifth box is overturned?
>the answer apparently was 1/6
>
Since this question was asked by someone who heard it from someone else,
etc., It might be that the original question was something like:
There are six boxes on the table and you put your finger on one
of them. God (or some other omniscient being) overturns four of the
other boxes to show that they are empty. What is the probability
that the box you placed your finger on has the key?
Stated like that, the probability is 1/6.
David Palmer
pal...@tybalt.caltech.edu
...rutgers!cit-vax!tybalt.caltech.edu!palmer
"Only 10% of the 4000 mile long coastline was affected."
-Exxon's version of the oil spill as reported to stockholders
Donald Benson
with unexpected situations (Aircraft have crashed due to false assumptions;
see COMP.RISKS for more info.)
On a test it is probably inappropriate. At the least, the test should begin
with a warning: "We will try to trick you with some of the questions. Be sure
to answer the questions based on exactly how they are asked. These are trick
questions." Probably, many "straightforward" questions would have tricks
discovered by the students.
If asked, "Which would fall faster, a 5lb brick or a 10lb brick, assuming
no atmosphere?", I would say, "The 10lb brick".
If asked, "A 5lb brick and a 10 lb brick are dropped simultaneously, from
the same height side-by-side. Which reaches the ground first, assuming no
atmosphere?", I would say, "They reach the ground simultaneously."
Do you recognize the critical difference between these two wordings?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
My radical ideas would never be accepted by a staid company like this.
DonB
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Malcolm Sanders
>In article <May.22.20.38...@paul.rutgers.edu> mast...@paul.rutgers.edu (Steve Masticola) writes:
>
>Actually, the problem is both more and less subtle than you realize. We
>had a heck of a debate over a similar puzzle around here. I didn't
>believe the answer until I simulated it in software. It still looks
>counter-intuitive. Here's our version:
>
> You're a contestant on Wheel of Fortune (Come on down!). Monty Hall
> has just presented you with three doors, behind one of which is a
> fabulous prize. You make your choice. Monty opens one of the other
> doors. No prize is revealed. He offers you a chance to change your
> choice.
>
> Are you better off changing your choice? Does it matter?
>
>The counter-intuitive, but correct, answer is yes, you are more likely to
>win if you change your choice. The chances are 1/3 that the prize is
>behind your original choice and 2/3 that it's behind the remaining door.
>
>I didn't believe it either, and found the mathematical proof confusing, so
>I simulated the situation in software and ran 1 million iterations. I'll
>post the program, if anyone's that interested. Believe it or not, like it
>or not, it works (and the odds of the key being in the box chosen are 1/6,
>assuming it wasn't under one of those turned up).
Give us a break. Assuming the key gets placed under one of the
six boxes 'at random' then the probability that it is under the
fifth box is 1/6. period. you don't have to assume anything about
the other outcomes(the probability that it is under any one of
the first four boxes is 4/6). Once you assume that the key isn't
under one of the first four boxes, then you are doing a different
problem--one where the probability of the key being under the
fifth box is 1/2.
I'm sorry,but the 'Let's Make a Deal' result is wrong also.
If you specify, as you did, that Monty's choice reveals no
prize, then the probability that the the prize is behind the
door you chose is the same as the probability that it is
behind the door you *didn't* choose--namely 1/2. I'll bet
your computer simulation was counting the times when Monty
uncovered the prize with his choice as well as the times
he didn't--hence your result of being right on your original
choice 1/3 of the time(true no matter what monty does).
Being right 1/3 of the time means that you are wrong 2/3 of
the time i.e. the prize is behind one of the other two doors, one
of which turns out to be Monty's choice.
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Rahul Dhesi
>If asked, "Which would fall faster, a 5lb brick or a 10lb brick, assuming
>no atmosphere?", I would say, "The 10lb brick".
I would convert the problem to SI units before solving it.
--
Rahul Dhesi <dh...@bsu-cs.bsu.edu>
UUCP: ...!{iuvax,pur-ee}!bsu-cs!dhesi
Mike Jassowski
> In article <10...@cit-vax.Caltech.Edu> pal...@tybalt.caltech.edu.UUCP (David Palmer) writes:
> >>If there are six boxes under one of which a key is hidden, and four of
>
>
> >There are six boxes on the table and you put your finger on one
> >of them. God (or some other omniscient being) overturns four of the
> >other boxes to show that they are empty. What is the probability
> >that the box you placed your finger on has the key?
> >
> >Stated like that, the probability is 1/6.
>
>
> your wording) that after God lifts up the box that what you see (i.e. the
> box being empty) is what was in the box before it was overturned (i.e.
> nothing). Stated like that, the probability of the key being in the box with
> your finger on it is 1/2. There are only two boxes and one of them contains
> a key. In the question as stated by the original poster, the probability
> could indeed be 1/6 as there is no mention of the observer knowing that four
> boxes are empty.
>
1) David Palmer is correct as HIS puzzle is stated (at least I have
myself that he is). Let me try to give a verbal proof (being one
who just got by in probability I don't even like SEEING formal
proofs 8`):
The probability that you picked the wrong box in the first place is
5/6 (I hope nobody disputes *this* 8`).
If you did pick the wrong box, and somebody WHO KNOWS WHERE THE
KEY IS (like 'God (or some other omniscient being)') shows you
four empty boxes, then the odds that the key is in the box you
originally picked is 0% (remember that it is assumed here that
you picked the wrong box).
The probability that you picked the right box in the first place is
1/6.
If you did pick the right box, and somebody WHO KNOWS WHERE THE
KEY IS (like 'God (or some other omniscient being)') shows you
four empty boxes, then the odds that the key is in the box you
picked is 100% (remember that it is assumed here that you picked
the right box).
5/6 th's of the time you have a zero percent chance of having picked
the right box.
1/6 th of the time you have a 100% cahnce of having picked the right
box.
Total odds of picking the right box: 1/6.
This is not so strange when you consider that you picked the box
with no prior knowlege. The wierd part is: if you can change your
mind after the four empty boxes are revealed, and you choose the
box which is left, you have a 5/6 odds of picking the corect box.
2) It is most likely that the wording of the original question was indeed
a Monty Hall type question. If I were pressed for time, and I was
given a Monty Hall type question without having knowlege of the Monty
Hall phenominon, I would have given the most intuitive answer of 1/2.
What these tests are trying to find are either:
a) The people who know the most tricks, or
b) The people who make the fewest assumptions about a problem and
can think on their feet.
With appropriate wording (such as the wording Mr. Palmer used) this would
be a very appropriate question for a mathmatics contest.
--Michael Jassowski (m...@hpesmaj.HP.COM)
stephen victor malikoff
> In article <78...@blia.BLI.COM> ru...@blia.BLI.COM (Ruth Bevan) writes:
> >If there are six boxes under one of which a key is hidden, and four of
> >those boxes have been overturned, which is the probability that I will
> >discover the key when the fifth box is overturned?
> >the answer apparently was 1/6
> >
> The question is stated ambiguously. If you have knowledge of the
> outcome for the first 4 turn-overs, then the anwswer is either
> 0 or 1/2, depending on whether the key has or has not already been
> found. If you do not know the outcome of the first 4 turn-overs,
> then the probability of seeing the key on the 5th turn-over is 1/6.
>
Consider the following experiment.
For j:=1 to 1000000 do
Begin
hide the key randomly;
Turn over four boxes;
Have a good look;
See if its under he fifth box;
If so, count:=count+1
end;
Remember the definition of probability? Its is the long running
relative frequency of an event over all events.
If we were told we knew the outcome of the first four turn-overs,
the probability would be 0 or 1/2 as you say. If we dont know the
outcome, the the probability is 1/6 as I and many others have shown.
-Victor.
Peter Vons
understood. Assume you have a bag with one black ball and 5 white balls in it.
You take one out without looking and keep it in your hand. You have 1/6
chance that you have the black one. There is 5/6 chance that the black ball
is still in the bag. Somebody else takes 4 white balls out of the bag. What
is the chance now that the black ball is in the bag. Of course still 5/6.
So if you keep the ball in your hand you have 1/6 chance to have the black
ball. If you choose for the ball in the bag you have 5/6 chance. OK?
Peter Vons.
c...@violet.berkeley.edu
> I'm sorry,but the 'Let's Make a Deal' result is wrong also.
> If you specify, as you did, that Monty's choice reveals no
> prize, then the probability that the the prize is behind the
> door you chose is the same as the probability that it is
> behind the door you *didn't* choose--namely 1/2.
Well, maybe not ...
Imagine making 99 appearances on the show. You will select the right door
33 times. 66 times the prize is behind one of the other 2 doors.
Whether your selection is right or wrong Monty will always be able to show
you an empty door.
Does the described situation really differ from Monty saying,
"Do you want to stay with door number 1, or do you want both doors 2 and 3?"?
--Brad Sherman( b...@ALFA.Berkeley.EDU )
p.s. Does it matter if Monty knows which is the correct door and does it matter
if you know that he knows?
Stephen P. Smith
% Run this through LaTeX
%
\documentstyle[12pt]{article}
\begin{document}
\newcommand{\logand}{\wedge}
\newcommand{\logor}{\vee}
\newcommand{\given}{\,|\,}
\section*{Random Choice?}
Suppose I choose a number at random from the set of numbers
${1, 2, 3}$. Clearly, $P(1) = 1/3$, $P(2) = 1/3$, and $P(3) = 1/3$.
Now assume that you present me with the following reasoning:
``I am only interested in whether you choose $1$ or not.
I know that you chose only one number, hence I already know
that at least one of the numbers $2$ or $3$ was {\em not} chosen.
Hence, I gain no additional information by your giving me the value of
(one of) the unchosen number among $2$ and $3$. If both were
unchosen, then you may randomly select among them to give
me your answer.''
If I tell you that $3$ was not a chosen number among $2$ and $3$,
why aren't you justified in concluding that your number, $1$, has
$1/2$ a chance of being the chosen number?
\section*{Will I Die?}
There are three prisoners: A, B, and C. A judge decrees that
tomorrow one, and only one, of the prisoners will be executed.
The judge then selects the prisoner to be executed at random,
puts the proper events into motion, and goes home for the night.
Prisoner A, of course, knows that at least one of the other prisoners,
either B or C, will not be executed tomorrow. By accident,
he gets a glimpse of a page of the next day's duty roster, which shows
that prisoner B is scheduled for the chain gang.
He reasons, correctly, that prisoner B will not be the one
executed.
What should be A's estimate of his own probability of execution?
\section*{Will I Die, Part II}
Assume the execution situation is as above.
Now however, prisoner A reasons with a knowledgeable guard as follows:
``I already know that one or both of prisoner B or C will not be executed.
I would like to give one of the ones that lives some of my final
thoughts on prison life. Please give me, at random if necessary,
the name of a prisoner that will live.''
Being a kind-hearted (and truthful) soul,
the prison guard tells prisoner A that he should talk to
prisoner B, since B will not be executed tomorrow.
Is A correct when he comes to the depressing conclusion that
his probability of being executed jumped from $1/3$ to $1/2$
with this new information?
\section*{Solution}
The key to the puzzles is the realization that
the new information you get does not always imply
the complement of an event occurring.
Lets take the last puzzle. Let $P(A)=\,P(B)=\,P(C)=1/3$, the
probability of each prisoner being executed, set by the
random selection process. Let $G(B)$ be the event that
that the guard told prisoner A that B was not to die.
{\em Note that G(B) is not equivalent to the complement
of B, $\bar{B}$}, since $\bar{B}$ is the event $C \logor A$,
while $G(B)$ is the event
$C \logor (A \logand {\sf GuardChoosesB})$.
Now assuming that the event {\sf GuardChoosesB} is a random
choice between itself and the event {\sf GuardChoosesC} (or
at least it is from A's information) then,
\begin{eqnarray*}
P[G(B)] &=& P[C] + P[ A ] \: P[ {\sf GuardChoosesB}] \\
&=& 1/3 + (1/3) \: (1/2) \\
&=& 1/2
\end{eqnarray*}
Using Bayes rule, we have:
\begin{eqnarray*}
P[A \given G(B)] &=& P[G(B) \given A] \: P[A] \over P[G(B)] \\
&=& ({1\over 2}) \: ({1\over 3}) \: ({1\over 2})^{-1} \\
&=& 1/3
\end{eqnarray*}
Thus prisoner's A chance of being executed is still only $1/3$
and does not jump to $1/2$.
Interestingly, A does now know that C's probability of being
executed is $2/3$ since
\begin{eqnarray*}
P[C \given G(B)] &=& P[G(B) \given C] \: P[C] \over P[G(B)] \\
&=& 1 \: ({1\over 3}) \: ({1\over 2})^{-1} \\
&=& 2/3
\end{eqnarray*}
The first of the other two puzzles is analogous to the
one just solved. The event {\em I tell you 3} is
the compound event {\em 2 is chosen} or ({\em 1 is chosen and
I randomly picked 3 to tell you}).
Note that in the middle puzzle, prisoner A has direct
evidence against the event $\bar{B}$ and hence can
conclude that his probability of being executed is
$1/2$.
\end{document}
"Stephen P. Smith"
Chris A. Heitmann
you are not better off changing your choice. You are giving the odds
as 1/3 that the car(or whatever) is behind your choice and 2/3 that it
is behind the one you didn't choose. Actually it has a 1 out of 2 chance
of being behind either one. Your answer implies that because you know
what is behind an unrelated door, you have a better chance of knowing
which of the two relevant doors it is behind.
Thought experiment: Flip two coins. Look at the first coin only. Does
this tell you, or give you some info, about what side of the other coin
is facing up?
Your 2/3 and 1/3 are referring to the original problem, which is no longer
current after you know that one curtain contains nothing. Currently you
have a 50/50 chance of picking the right curtain.
Chris
c...@genrad.com or {decvax,husc6,mit-eddie}!genrad!cah
Mike Gore, Institute Computer Research - ICR
In article <74...@bsu-cs.bsu.edu> dh...@bsu-cs.bsu.edu (Rahul Dhesi) writes:
>In article <525...@hpcuhb.HP.COM> do...@hpcuhb.HP.COM (Donald Benson) writes:
>>If asked, "Which would fall faster, a 5lb brick or a 10lb brick, assuming
>>no atmosphere?", I would say, "The 10lb brick".
>
>I would convert the problem to SI units before solving it.
Of course it would be nice to know the MASS of each brick too ...:-)
# Mike Gore, Technical Support, Institute for Computer Research
# Internet: mag...@watdcsu.waterloo.edu or mag...@watdcsu.uwaterloo.ca
# Bitnet: mag...@watdcsu.bitnet UUCP: uunet!watmath!watdcsu!magore
# These ideas/concepts do not imply views held by the University of Waterloo.
The Polymath
} has just presented you with three doors, behind one of which is a
} fabulous prize. You make your choice. Monty opens one of the other
} doors. No prize is revealed. He offers you a chance to change your
} choice.
}
} Are you better off changing your choice? Does it matter?
}
}The counter-intuitive, but correct, answer is yes, you are more likely to
}
}... I simulated the situation in software and ran 1 million iterations. I'll
}post the program, if anyone's that interested. ...
I'm getting a lot of requests for the above mentioned program and some of
my e-mail replies are getting bounced. It's not long, so I'm posting it
here. My apologies to those in sci.physics who aren't interested, but I'm
getting requests from both groups, so someone is. I've directed
follow-ups to sci.math (which I don't normally read, BTW).
/***************************************************************************/
/* Monty Hall Simulator */
/* */
/* Author: The Polymath (Jerry Hollombe) */
/* */
/* Problem: On a quiz show, you are offered a choice of three doors, */
/* behind one of which is a prize. After making your choice */
/* the MC opens one of the remaining doors, revealing it */
/* concealed no prize. The MC then offers you the opportunity */
/* to stay with your original guess or change your mind. */
/* */
/* Are your chances of winning better, worse or unchanged if */
/* you change your mind? */
/* */
/* Solution: Counter to intuition, your chances of winning are 2/3 if */
/* you change your mind and only 1/3 if you don't. This */
/* program demonstrates this fact by simulating the game */
/* situation a large number of times and recording wins and */
/* losses for each condition. (It can also be worked out */
/* mathematically, but most people won't believe it until */
/* they see it). */
/* */
/* Instructions: */
/* To run enter the program name. It will, by default, run */
/* 2000 games and change on the second guess approximately 50% */
/* of the time, at random. You can also pass it one or two */
/* positional command line parameters. The first is the */
/* number of games to run and the second is the percentage of */
/* times to choose a different door on the second guess. */
/* The parameters default as above. Number of games must */
/* appear before percent of new guesses. */
/* */
/* Example: */
/* m.hall will run with defaults */
/* */
/* m.hall 10000 will run 10,000 games with 50% */
/* new choices. */
/* */
/* m.hall 10000 100 will run 10,000 games with 100% */
/* new choices. */
/* */
/***************************************************************************/
#include <stdio.h>
#define TRUE 1
#define FALSE 0
#define bool short
main(argc, argv)
int argc;
char *argv[];
{
short change_pc; /* Percent of times to change mind (argv[2]) */
short choice; /* Index of chosen door */
char doors[3]; /* Door array, P: prize, O: no prize, R: revealed */
int games; /* Game loop index */
short i; /* General index */
int new_choices = 0; /* Number of new choices */
int no_nc_wins = 0; /* Count of wins on no new choice */
int no_nc_losses = 0; /* Count of losses on no new choice */
int nc_wins = 0; /* Count of wins on new choice */
int nc_losses = 0; /* Count of losses on new choice */
bool rechose; /* New choice flag, TRUE: 2nd choice != 1st */
int tot_games; /* Total number of games to play (argv[1]) */
srandom(getpid()); /* Initialize random number generator */
if (argc > 1) /* Set total games to play */
tot_games = atoi(argv[1]);
else
tot_games = 2000;
if (argc > 2) /* Set percent time to change mind */
change_pc = atoi(argv[2]);
else
change_pc = 50;
/************************/
/* MAIN LOOP */
/* Let the games begin! */
/************************/
for (games = 0; games < tot_games; games++)
{
for (i = 0; i < 3; i++) /* Clear doors array. */
doors[i] = 'O';
doors[random() % 3] = 'P'; /* Place prize at random. */
choice = random() % 3; /* Make first choice. */
do /* Reveal an empty door at random. */
{
i = random() % 3;
if ((doors[i] == 'O') && (i != choice))
doors[i] = 'R';
} while (doors[i] != 'R');
if ((random() % 100) < change_pc) /* Change choice change_pc% of time */
{
rechose = TRUE; /* Flag change of mind */
new_choices++; /* Count change of mind */
for (i = 0; i < 3; i++) /* Make new choice */
{
if ((doors[i] != 'R') && (i != choice))
{
choice = i;
break;
}
} /* end for */
}
else
rechose = FALSE;
if ((doors[choice] == 'P') && (!rechose)) /* Count 1st choice wins */
no_nc_wins++;
if ((doors[choice] == 'P') && (rechose)) /* Count 2nd choice wins */
nc_wins++;
} /* end for */
/********************/
/* End of MAIN LOOP */
/********************/
nc_losses = new_choices - nc_wins; /* Calculate losses */
no_nc_losses = games - new_choices - no_nc_wins; /* Calculate losses */
/*********************/
/* Print out results */
/*********************/
printf ("\nOk, I played %d games. \n", games);
printf ("I made a new choice on the second guess %d times. \n", new_choices);
printf ("With the new choices I won %d times and lost %d. \n", nc_wins, nc_losses);
printf ("Without them I won %d times and lost %d. \n\n", no_nc_wins, no_nc_losses);
printf ("Believe it or not, like it or not, it works! \n\n");
fflush (stdout);
return(0);
} /* end main */
/***************************************************************************/
/* END MONTY HALL SIMULATOR CODE */
/***************************************************************************/
Bill Tanenbaum
< In article <10...@cit-vax.Caltech.Edu> pal...@tybalt.caltech.edu.UUCP (David Palmer) writes:
< >There are six boxes on the table and you put your finger on one
< >of them. God (or some other omniscient being) overturns four of the
< >other boxes to show that they are empty. What is the probability
< >that the box you placed your finger on has the key?
< >
< >Stated like that, the probability is 1/6.
<
< Say what? I assume (and I think that this assumption is reasonable based on
< your wording) that after God lifts up the box that what you see (i.e. the
< box being empty) is what was in the box before it was overturned (i.e.
< nothing). Stated like that, the probability of the key being in the box with
< your finger on it is 1/2. There are only two boxes and one of them contains
< a key. In the question as stated by the original poster, the probability
< could indeed be 1/6 as there is no mention of the observer knowing that four
< boxes are empty.
<
< I am making my calculations of the probability from the observer's point of
< view (i.e. the "you" in the problem).
This is the "Monty Hall" problem which has been discussed to death on the net
at least twice. The probability is indeed 1/6, not 1/2. Since this has
been discussed to death so many times, I would simply suggest to "What's in
a name" that he conduct a simple experiment.
If he then still thinks the probability is 1/2, he should then send me private
mail so we can have a little wager. He will pick one of the six boxes.
I, knowing where the key is, will overturn four empty boxes. He will then
see if the key is under his box. If so, I will give him $20. If not,
he will give me $10. We will play this game one thousand times. If he is
right, he should make around $5000, more than the price of a plane ticket
to Chicago from Rochester. I am awaiting a reply.
--
Bill Tanenbaum - AT&T Bell Labs - Naperville IL att!ihlpb!tan
Jef Poskanzer
#include <stdio.h>
#define BOXES 3
#define GAMES 1000000
#define ZONK 0
#define OPENED 1
#define PRIZE 2
main( argc, argv )
{
int i, j, k, box[BOXES], guess, switchedguess, stickwins, switchwins;
srandom( (int) time( 0 ) );
stickwins = switchwins = 0;
for ( i = 0; i < GAMES; i++ )
{
/* Initialize boxes. */
for ( j = 0; j < BOXES; j++ )
box[j] = ZONK;
box[random() % BOXES] = PRIZE;
/* Make guess. */
guess = random() % BOXES;
/* Monty opens up all but two boxes, all empty ones, and
** not the guessed one. */
for ( j = k = 0; k < BOXES - 2; j++ )
{
if ( j != guess && box[j] == ZONK )
{
box[j] = OPENED;
k++;
}
}
/* Now, stick with original guess or switch to remaining box. */
for ( j = 0; j < BOXES; j++ )
{
if ( j != guess && box[j] != OPENED )
{
switchedguess = j;
break;
}
}
/* Check results. */
if ( box[guess] == PRIZE )
stickwins++;
if ( box[switchedguess] == PRIZE )
switchwins++;
}
printf(
"stick: %d wins out of %d, probability %g\n", stickwins, GAMES,
(float) stickwins / GAMES );
printf(
"switch: %d wins out of %d, probability %g\n", switchwins, GAMES,
(float) switchwins / GAMES );
exit( 0 );
}
- - - - - - - - - -
Results:
stick: 333980 wins out of 1000000, probability 0.33398
switch: 666020 wins out of 1000000, probability 0.66602
With six boxes, equivalent to the black and white balls in a bag:
stick: 166559 wins out of 1000000, probability 0.166559
switch: 833441 wins out of 1000000, probability 0.833441
Verrrrrrry interesting.
---
Jef
Jef Poskanzer j...@helios.ee.lbl.gov ...well!pokey
"Protozoa are small, and bacteria are small, but viruses are smaller than the
both put together."
Bela Lubkin
that may make it seem intuitive:
Monty Hall gives you the choice of 3 doors, behind one of which
is a prize. You chose one at random. Now, obviously, in 1 case
out of 3, you have chosen the prize; in the other two, you
haven't. At this point your chance of having picked the prize has
been "crystalized" at 1/3.
Now Monty opens one of the doors you didn't chose, showing you
that it is empty. This does not change the probability of
finding the prize behind the door you originally chose; it's
still 1/3. Since only one other door remains, the chance that
the prize is behind it MUST be 2/3.
But WHY? Consider: in 1 case out of 3, you've chosen the prize.
Monty can choose either of the other doors and show you that it's
empty; in which case, if you change your mind, you will lose the
prize. The partial probability here is (1/3)*0. In the other 2
cases, you've chosen an empty door. Monty is now CONSTRAINED to
chose the one remaining door that's empty. The third door MUST
have the prize behind it. This partial probability is (2/3)*1.
The total probability that the prize is behind the third door,
the door that you have not chosen and that Monty has not shown
you, is [(1/3)*0] + [(2/3)*1], or [0] + [2/3], or 2/3.
The problem is counterintuitive because there is another related
problem that is more intuitive. That is: Monty Hall gives you
the choice of 3 doors, behind one of which there is a prize. You
choose a door. Monty then RANDOMLY chooses one of the doors you
didn't choose, and the prize HAPPENS not to be behind that door.
What is the probability that it is behind the door you chose? In
this case, the probability is 50%, as intuition suggests. In
reality, the 1/3 probability that Monty chooses a random door and
PICKS THE PRIZE is EDITED OUT by the producers of the program --
or, at least, this is the effect; the reality is that Monty
already knows which door the prize is behind and therefore does
not choose it; that path is effectively PRE-edited out.
Apparently the mind has difficulty that Monty can have
information that >I< don't have; it LOOKS like Monty chose the
"teaser" door randomly and HAPPENED not to hit the prize; so the
probability that >MY< door is the right one is still 50%, and
"I'm going to stick with my choice, damnit!"
None of this, by the way, answers the question: "If I am in this
situation, SHOULD I SWITCH?" Monty or his producers may have a
strategy by which they only show you what's behind a "teaser"
door if you happen to be sitting on the prize. In that case it's
obviously wrong to switch. Only by taking careful notes on
hundreds of hours of game shows could you answer that question...
>Bela<
The Polymath
It's been very amusing to see all the postings making the same mistakes
and wrong assumptions about this problem that I did.
It's been less amusing to be accused of being an idiot by people who've
obviously been too lazy to simulate the game for themselves, let alone
work up a mathematical proof.
I've already posted my software simulation. For those too lazy or
ignorant to understand it (or write their own), here's something you can
try with just four index cards and a pencil:
Use one index card to keep score. The categories are:
Stayed with original choice and won.
Stayed with original choice and lost.
Changed choice and won.
Changed choice and lost.
Use the pencil to tally results.
Mark a 'P' on one side of one card.
Mark an 'N' on one side of each of the remaining cards.
Find a suck ... um ... volunteer to choose while you play Monty Hall.
Play the game:
Shuffle the cards and hold them so only you can see the letters.
Have the volunteer choose one at random.
Reveal a non-chosen card with an 'N' on it.
Offer them the chance to change their minds.
Play the game 50 to 100 times and tally the results of each game
(final choice 'P' is a win, final choice 'N' is a loss).
Let me know how it works out. I and my simulation await your apologies.
(Again, _my_ apologies to sci.physics. I _am_ directing followups to
sci.math. Really).
Davyd Norris
> In article <78...@blia.BLI.COM> ru...@blia.BLI.COM (Ruth Bevan) writes:
>
> Since this question was asked by someone who heard it from someone else,
> etc., It might be that the original question was something like:
>
> There are six boxes on the table and you put your finger on one
> of them. God (or some other omniscient being) overturns four of the
> other boxes to show that they are empty. What is the probability
> that the box you placed your finger on has the key?
>
> Stated like that, the probability is 1/6.
What a load of half-hearted mathematical *crap*. If you look at the start
of the problem as you have posed it, then your answer is valid. But the
moment that even one of those boxes is overturned AND YOU SEE THE CONTENTS
the probability of your outcome has changed, whether or not you put your
pinky on the magic box at the start, middle or end of the gedank.
This is expressible as the following:
Pr(A/B) = Pr(Box 5 has the key/four boxes don't)
^ means given
The formula for this type of probability is given in elementary stats.
texts, which I used when I was 14 ( and that is the reason I have
forgotten how the hell it goes! :-)), but thinking about the problem
as you have posed it will show you that the answer you gave is ridiculous.
I concur with the great majority of the net body in that the answer is
either 1/2 (prior knowledge) or 1/6 ( blind ) , and faced with the
question as worded in NEWS, I would have subconsiously appended the words
"... are overturned and seen to be empty..." to the question, leading me
to the answer of 1/2. ( Which is in fact wrong in this case, if as it
is suggested, the contents of the overturned boxes are still unknown.)
Which leads me to the point that the person who thought up the question
should either be given a free course in logic, or have wires put on their
gonads.
Daffy.
-------------------------------------------------------------------------------
Davyd Norris : "... We use this to do our hard physics sums."
Physics Dept. : "Oh, Albert just used to use the back of an
Monash University, : envelope." - Mrs. Einstein at the opening
Vic, Australia. : of the Smithsonian Computer Institute.
-------------------------------------------------------------------------------
What`s in a name?
understand now. Wording was a big problem. Actually Chris Long was the
first one to post an accurate answer (though I disagreed with him over
e-mail). He said the probability varied between 1/6 and 1/2 depending on
the algorithm Monty (or God) used to uncover the boxes. (Since the
algorithm was not explicitly stated) Now if only there was a way to
represent the probabilities inherent in Monty's brain....
Thanks for clearing it up folks. Personally I think it's an interesting
problem that I hadn't seen before. I wouldn't expect high-school students
to recognize Monty Hall problem right off and get beyond inconsistencies in
the wording though...
Anyway, sorry if I had to jump in and add to a discussion all of you found
uninteresting.
Sorry if this is not a very well said article but I'm cleaning up my usenet
mess and have only an hour.
Next Fall--
--mike
--
Mic3hael Sullivan, University of Rochester. "Life is like a sewer, what
Society for the Incurably Pompous. you get out of it, depends on
David Messer
>had a heck of a debate over a similar puzzle around here. I didn't
>believe the answer until I simulated it in software. It still looks
>counter-intuitive. Here's our version:
>
> You're a contestant on Wheel of Fortune (Come on down!). Monty Hall
> has just presented you with three doors, behind one of which is a
> fabulous prize. You make your choice. Monty opens one of the other
> doors. No prize is revealed. He offers you a chance to change your
> choice.
>
> Are you better off changing your choice? Does it matter?
>
>The counter-intuitive, but correct, answer is yes, you are more likely to
>win if you change your choice. The chances are 1/3 that the prize is
>behind your original choice and 2/3 that it's behind the remaining door.
>
>I didn't believe it either, and found the mathematical proof confusing, so
>I simulated the situation in software and ran 1 million iterations.
I remember this technique. It's called a "Monte Hall Simulation". :-)
--
Paranoia is just good thinking if | David Messer da...@Lynx.MN.Org -or-
everybody IS out to get you. | Lynx Data Systems ...!bungia!viper!dave
Doug Merritt
>>If asked, "Which would fall faster, a 5lb brick or a 10lb brick, assuming
>>no atmosphere?", I would say, "The 10lb brick".
>
>I would convert the problem to SI units before solving it.
For real rigor, convert the bricks to wave functions first.
And the earth, too, of course. Extra points for including the
sun, planets, nearby stars.
Doug
--
Doug Merritt {pyramid,apple}!xdos!doug
Member, Crusaders for a Better Tomorrow Professional Wildeyed Visionary
"Welcome to Mars; now go home!" (Seen on a bumper sticker off Phobos)
Jeffrey Naiman
>In article <10...@cit-vax.Caltech.Edu>, pal...@tybalt.caltech.edu
>(David Palmer) writes:
>> In article <78...@blia.BLI.COM> ru...@blia.BLI.COM (Ruth Bevan) writes:
>>
>> Since this question was asked by someone who heard it from someone else,
>> etc., It might be that the original question was something like:
>>
>> There are six boxes on the table and you put your finger on one
>> of them. God (or some other omniscient being) overturns four of the
>> other boxes to show that they are empty. What is the probability
>> that the box you placed your finger on has the key?
>>
>> Stated like that, the probability is 1/6.
>
>What a load of half-hearted mathematical *crap*. If you look at the start
>of the problem as you have posed it, then your answer is valid. But the
>moment that even one of those boxes is overturned AND YOU SEE THE CONTENTS
>the probability of your outcome has changed, whether or not you put your
>pinky on the magic box at the start, middle or end of the gedank.
Sigh....
The key fact which you are missing is that the boxes that were turned over
were not done so randomly. Then you would be correct. Someone has 'fixed'
the game. He knows where the key is. Therefore the elementary formulas do
not apply rotely to this problem.
Try the following experiment. Have a friend pick a number between 1 and 1000.
You get one guess. Your friend then says that his number is either your
guess or one other number. This is equivalent to 'turning over' the other
998 numbers and showing that they are not true. What is the chance that his
number is the 'other' number? Very high....
>I concur with the great majority of the net body in that the answer is
>either 1/2 (prior knowledge) or 1/6 ( blind ) , and faced with the
>question as worded in NEWS, I would have subconsiously appended the words
>"... are overturned and seen to be empty..." to the question, leading me
>to the answer of 1/2. ( Which is in fact wrong in this case, if as it
>is suggested, the contents of the overturned boxes are still unknown.)
The original post was not worded well. The wording was lost in the
third-hand posting. I'd be interested to discover what the original wording
was. Is the answer 1/6 because you shouldn't assume that you have eyes,
or is it 1/6 because of the Monty phenomenon.
>
>Daffy.
>-------------------------------------------------------------------------------
>Davyd Norris : "... We use this to do our hard physics sums."
>Physics Dept. : "Oh, Albert just used to use the back of an
>Monash University, : envelope." - Mrs. Einstein at the opening
>Vic, Australia. : of the Smithsonian Computer Institute.
>-------------------------------------------------------------------------------
- Jeff Naiman (nai...@yale.edu)
Chris Torek
>... Here's our version:
>
> You're a contestant on Wheel of Fortune (Come on down!). Monty Hall
> has just presented you with three doors, behind one of which is a
> fabulous prize. You make your choice. Monty opens one of the other
> doors. No prize is revealed. He offers you a chance to change your
> choice.
>
> Are you better off changing your choice? Does it matter?
>
>The counter-intuitive, but correct, answer is yes, you are more likely to
>win if you change your choice. The chances are 1/3 that the prize is
>behind your original choice and 2/3 that it's behind the remaining door.
>
If you like, here is a non-mathematical explanation of the difference
between this situation and the one where you make no choice initially,
but are shown what is behind one of the three doors: As posed, *you*
limit which of the doors Hall can open. If---as is more likely---you
have chosen a bad one, Hall must show you the *other* bad one. Only
if you have chosen the one with the prize has he not been constrained.
So:
If you picked a bad one (2/3 chance), Hall shows you the other
bad one, leaving the one unchosen door as the good one.
If you picked the good one (1/3 chance), Hall shows you either
bad one, leaving the one unchosen door as the other bad one.
--
In-Real-Life: Chris Torek, Univ of MD Comp Sci Dept (+1 301 454 7163)
Domain: ch...@mimsy.umd.edu Path: uunet!mimsy!chris