HDTenEightyP/NetNews huggingface dataset:
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Subject: Re: Revised Monty Hall problem
Organization: Purdue University
Date: Sat, 2 Jan 1993 06:32:57 GMT

In article <1i01n8INN1bb@iskut.ucs.ubc.ca> leon@unixg.ubc.ca (Leon ter Beek) writes:
>There are five doors, behind two of them a car, and behind each of the
>other three doors a goat. You pick a door and the quizmaster (who knows
>what's behind the doors) opens a door with a goat behind it.
>1) Should you change doors?

I assume Monty always opens an unchosen door and reveals a goat, as in the
original puzzle.

Without switching, you have a 2/5 chance of winning a car.
If you switch, and if your original choice was a car, then only one of the
three doors that you could switch to has a car behind it. If your original
choice was a goat, then your probability of winning by switching rises to 2/3.
Since your first choice is a car 2/5 of the time, your probability of winning a
car by using the always-switch strategy is 2/5 * 1/3 + 3/5 * 2/3 = 8/15, which
is greater than 2/5. Therefore, you should switch.

>2) Why is the sum of the probability of changing and not changing not
>equal to one?

Because the probability of winning without switching is 2/5, and
the probability of winning by switching is 8/15. The sum of these is 14/15, and
14/15 is not one. That is why the sum of the probabilities is not one.

Oh, I see. You think this is a contradiction, and you want an explanation of
why it isn't.

It's like the famous bellhop problem that concludes by asking where the other
dollar went. You wave your hands and add up some probabilities, and then you
ask why the sum is not 1.

If you add up the probabilities of disjoint events that cover (allmost all of)
the space, then you will get one. In the three-door problem, you have

	probability of winning by switching + probability of losing by
	switching = 1					[Rule 1]

and also, because the two events are identical, you have

	probability of losing by switching = probability of winning by not
	switching					[Rule 2]

from which it follows that

	probability of winning by switching + probability of winning by not
	switching = 1.					[Rule 3]

In the five-door game, Rule 1 continues to hold. However, Rule 2
is false. It is possible for your original choice to be a car, and
for you to switch and wind up with the other car. It is also possible
for your first choice to be a goat, and for you to switch and wind
up with another goat.  Rule 2 fails because the two events described
are not identical.  Since Rule 3 was derived from Rules 1 and 2,
there is no reason to expect Rule 3 to hold for the five-door
problem. Hence, no contradiction.

--
Dave Seaman
ags@seaman.cc.purdue.edu
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: rec.puzzles
Subject: Re: new year's problem
Date: 2 Jan 93 12:29:00 GMT
Distribution: rec
Organization: CWI, Amsterdam

In article <1993Jan2.1584.7529@channel1> "eliot gelwan" <eliot.gelwan@channel1.com> writes:
 > On 12-31-92 KEITH F. LYNCH  wrote:
 > KF/What property did 1987 have, that no year after it will have
 > KF/until 2013?
 > 
 > Prime numbers
 > 
A very strange idea about prime numbers.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland; e-mail: dik@cwi.nl
From: kvdoel@cs.ubc.ca (Kees van den Doel)
Newsgroups: rec.puzzles
Subject: Re: A nice New Year puzzle.
Date: 31 Dec 1992 17:50:54 -0800
Organization: Computer Science, University of B.C., Vancouver, B.C., Canada
NNTP-Posting-Host: cascade.cs.ubc.ca

In article <1993Jan1.002031.5763@zip.eecs.umich.edu>
kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:

>Given a plane and exactly 3 colors, prove whether or not it is possible
>to assign a color to each point of the plane in such a way that no two points
>exactly 1 inch apart have the same color.

2 points sqrt(3) inch apart have the same color (take 2 points of
colors, say, 2 and 3, 1 inch apart. Form 2 equilateral triangles with
those 2 points, you get 2 new points (sqrt(3) inches apart) which must
have color 1, i.e. the same).  Pick a point, say it has color 1. The
circle with radius 1 around it has color 2 or 3. The circle with radius
sqrt(3) around it has color 1. On this big circle, pick an arbitrary
point and draw a circle of radius sqrt(3) around *it*, which must also
have color 1. The last and the first circles intersect at a point which
therefore must have color 1 *and* color 2 or 3, which is impossible.


Kees
Newsgroups: rec.puzzles
From: "mark longridge" <mark.longridge@canrem.com>
Subject: dotc newsletter
Reply-To: "mark longridge" <mark.longridge@canrem.com>
Organization: Canada Remote Systems
Distribution: rec
Date: 24 Dec 92 02:43:28 EST

Time to update the Domain of the Cube Newsletter again...

If you were a Rubik's Cube freak (or like combinatorial puzzles)
you'll be pleased to here DOTC is now up to issue #6. I've
finally finished the cube chronology after lots of research
into the cube's beginnings. In fact patent information on the
cube goes back to 1960 (William Gustafson's Globe). The
time-line refers to every relavent event in the cube's history,
now updated to include some of Constantin's puzzles and Square-1.
Not to mention a few ideas of Uwe Meffert nobody seems to
have EVER seen before!

Cube Patterns have now been improved upon, making Rubik's
Compendium and Inside Rubik's Cube and Beyond by Bandelow
obsolete (although IRCAB is still an excellent book!)
There are also those who are still making new cube variants...

My cube programs (the best of which run on IBM PCs with at
least EGA) now have nifty new pattern and solving algorithms.
New patterns not in any of the books have been compiled.

Anyone wishing to obtain the new DOTC Catalog can do so by
sending a Self-Addressed envelope to

DOTC Newsletter
259 Thornton Rd N
Oshawa Ontario Canada, L1J 6T2

I'll also field submissions for future issues.
Email me at: mark.longridge@canrem.com
or send it to the above address.

Oh and issue #6 sports the infamous new puzzle idea section!
Puzzles which exist as ideas or simulations only...

-> Mark Longridge, puzzle historian <-
--
Canada Remote Systems  - Toronto, Ontario
World's Largest PCBOARD System - 416-629-7000/629-7044
Newsgroups: rec.puzzles
From: rbarnick@mitre.org (Barnie Barnick)
Subject: Re: Lost solution
Followup-To: rec.puzzles
Organization: MITRE
Date: Thu, 24 Dec 1992 17:42:34 GMT

*****
Tim,
The answer is 4 and 19.  
*****

In article <1992Dec22.193053.24084@bernina.ethz.ch>, timh@igc.ethz.ch (Tim
Harvey) wrote:
> 
> Dear Netters,
> 	A log while ago I collected the following brainteaser;
> 
> ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
> 
> 
> Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce
> any truth from any set of axioms.  Two integers (not necessarily unique) are
> somehow chosen such that each is greater than 1 and less than 100.  Mr. S.
> is given the sum of these two integers; Mr. P. is given the product of these
> two integers.  After receiving these numbers, the two logicians do not
> have any communication at all except the following dialogue:
>    Mr. P.:  I do not know the two numbers.
>    Mr. S.:  I knew that you didn't know the two numbers; I do not know the
>             two numbers.
>    Mr. P.:  Now I know the two numbers.
>    Mr. S.:  Now I know the two numbers.
> "The two numbers" in this dialogue refers to those two integers that were
> mysteriously chosen.
> 
> Given that the above statements are absolutely truthful, what are the two
> numbers?
> ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
> Anyway (and I hope it's not a FAQ) I didn't save the solution!
> 
> Can anyone supply a definitive answer? (Then we can all get back to work)
> Many thanks,
> Tim
From: exujbl@exu.ericsson.se (Jerry Blackerby 214-907-7810)
Newsgroups: rec.puzzles
Subject: Math Problem
Date: 23 Dec 92 13:34:10 GMT
Reply-To: exujbl@exu.ericsson.se
Organization: Ericsson Network Systems, Inc.
X-Disclaimer: This article was posted by a user at Ericsson.
              Any opinions expressed are strictly those of the
              user and not necessarily those of Ericsson.

Keywords: 

Several years ago, Martin Gardner wrote a puzzle/math problem column in
Scientific American.  Several of the columns dealt with the number 4.
He presented variations of two 4's, three 4's, and four 4's with mathematical
symbols to equate numbers from 1 upward.  For example: 4 x 4 x 4 = 64

At one time, he published a long list of these equations.  Does anyone
remember them or know where they may be published in a book.

Thanks,

Jerry


==================================_=====================================
|| Jerry Blackerby          ||   | ~~   ||  Practice random kindness  ||
|| 214-907-7810 (w)         ||  \,  _}  ||  and senseless acts of     ||
|| exujbl@exu.ericsson.se   ||    \(    ||  beauty.                   ||
========================================================================
From: mouse@thunder.mcrcim.mcgill.edu (der Mouse)
Newsgroups: rec.puzzles
Subject: Re: Paper Folding
Date: 23 Dec 92 05:53:35 GMT
Organization: McGill Research Centre for Intelligent Machines

In article <1992Dec17.134838.29102@serval.net.wsu.edu>, wvanweer@wsuaix.csc.wsu.edu (Wayne VanWeerthuizen) writes:
> In article <1992Dec17.120841.14563@thunder.mcrcim.mcgill.edu> mouse@thunder.mcrcim.mcgill.edu (der Mouse) writes:

>> Does the paper have stripes on both sides?  Do the stripes run the
>> same way on both sides?
> Okay, the stripes run on only ONE side of the paper.  The paper can
> start out with any shaped perimeter, but must be flat and one
> continuous piece.

Must it be simply connected?  Is it allowed to have cuts that do not
remove material?

If it is allowed to have cuts that remove no material, I have a
solution in nine unit squares; this can be reduced to six plus epsilon
by trimming three of the squares down to narrow rectangles.

> Oh, and two more provisions.  If the paper is laid back flat after
> having been folded, all the small sections delimited by the creases,
> must be convex, with no holes.  And for two regions to be connected
> they must share at least a line segment at the crease of length 1/2.

Then the three squares can't be trimmed below area 1/2, and my least
area is 7.5 units.  This can be reached as follows:

Given the positive quadrant divided into unit squares, take the squares
with their lower left corners at the following points:

(0,0)	(0,1)	(0,2)
(1,1)	(1,2)	(1,3)
(2,2)	(2,3)	(2,4)

Cut along the line segments (1,1)-(1,2) and (2,2)-(2,3).

Cut out and remove the right-hand half (ie, the rectangle with x values
whose fractional parts are between .5 and 1) of the (0,1), (1,2), and
(2,3) squares.  Crease all the remaining connections the same way (all
mountain or all valley folds).  It should be trivial to fold it into a
cube at this point; if you fold it so that all the partial squares are
hidden, the result should be the answer.  (Assuming, of course, that
the folds are mountain folds for the side with lines on it.)

I hope I got all those coordinates right... :-)

					der Mouse

			    mouse@larry.mcrcim.mcgill.edu
Newsgroups: rec.puzzles
From: Gordon_Wong@mindlink.bc.ca (Gordon Wong)
Subject: Help Needed for Bob Puzzle
Organization: MIND LINK! - British Columbia, Canada
Date: Thu, 31 Dec 1992 03:42:35 GMT

Has anyone seen this puzzle before? What's the answer to it?

Bob opens up the newspaper and reads a story about a woman (Bobbi)
who skiied over a cliff and died while on a vacation with her
husband (Rob). Bob calls up the police and tells them to arrest
Rob for murder. Why?


From: jerryd@hpgrla.gr.hp.com (Jerry Donovan)
Date: Wed, 23 Dec 1992 17:49:25 GMT
Subject: Re: The Rifleman Problem
Organization: Hewlett-Packard, Greeley, CO
Newsgroups: rec.puzzles

>  Does anyone remember the "Rifleman Problem"? It was published (surely not
>  for the first time) in that renowed journal of scientific research
>  Omni, in the early to mid 80's.

>  The problem is this:

>  N riflemen are situated randomly on a plane. At the same instant, each
>  shoots and kills his nearest neighbor. How many are left?

>  My hazy recollection is that Omni said the 1-D case was solved, but the
>  2-D case wasn't. 

With all the great? discussion of how to interpret a plane in the 2-D
case, I'd like to take a step back and ask a dumb question.  What is
the solution for the 1-D case.  I can give examples of 0 or 1 surviver
up to ceiling(N/3).  So I guess the answer is some probable number
which a standard deviation or something.  Am I way off base?  

(Note: it may be obvious that I am not a math major, so please answer with
that thought in mind.)

Thanks,

Jerry Donovan
Newsgroups: rec.puzzles
From: bsn@world.std.com (Brent S Noorda)
Subject: EZ XMAS PUZZLE
Summary: EZ XMAS PUZZLE
Organization: The World Public Access UNIX, Brookline, MA
Date: Thu, 24 Dec 1992 16:17:18 GMT


           A   B   C   D   E

           F   G   H   I   J

           K   M   N   O   P

           Q   R   S   T   U

           V   W   X   Y   Z

Newsgroups: rec.puzzles
From: rbarnick@mitre.org (Barnie Barnick)
Subject: Re: Lost Solution
Followup-To: rec.puzzles
Organization: MITRE
Date: Thu, 24 Dec 1992 20:08:40 GMT

Bhima,
Let me at least get you started.  It has its basis in prime numbers. 
First, Mr. P. knew he didn't know the two numbers because his product was
factorable in more than one way.  That is, had it been factorable by only
two prime numbers he would have had the two numbers.  For example, had Mr.
P's number been 21, he would have known the two numbers were 3 and 7.  Now
in the second statement, Mr. S. responds by saying he knew that Mr. P
didn't know the two numbers because Mr. S's number could be obtained in
many ways and in each there was a number that was not prime.  When Mr. S
goes on to say that he too did not know the two numbers, that told Mr. P of
Mr. S's problem which meant that his number had to contain both a prime and
non-prime number.  In such a scenario and given the rules of the game, 4
and 19 are the only two that work.  Hopefully this makes enought sense to
give you a way to work through it and see for yourself.  I do not take
credit for this solution.      

In article <38v2sf_@rpi.edu>, nittab@aix02.ecs.rpi.edu (Bhima Nitta) wrote:
> 
> Newsgroups: rec.puzzles
> Subject: Re: Lost solution
> Summary: 
> Expires: 
> References: <1992Dec22.193053.24084@bernina.ethz.ch> <rbarnick-241292103910@cosadpm12.mitre.org>
> Sender: 
> Followup-To: 
> Distribution: 
> Organization: Rensselaer Polytechnic Institute, Troy, NY
> Keywords: 
> 
> In article <rbarnick-241292103910@cosadpm12.mitre.org> rbarnick@mitre.org (Barnie Barnick) writes:
> >*****
> >Tim,
> >The answer is 4 and 19.  
> >*****
> 
> 
> Now my interest is piqued. Why is this the solution ?
> 
> Yours ignorantly,
> 
> Bhima
> 
> 
> 
> 
> 
> >
> >In article <1992Dec22.193053.24084@bernina.ethz.ch>, timh@igc.ethz.ch (Tim
> >Harvey) wrote:
> >> 
> >> Dear Netters,
> >> 	A log while ago I collected the following brainteaser;
> >> 
> >> ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
> >> 
> >> 
> >> Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce
Newsgroups: rec.puzzles
From: joel@math.toronto.edu (Joel Chan)
Subject: Re: new year's problem
Organization: Department of Mathematics, University of Toronto
Distribution: rec
Date: Sat, 2 Jan 93 20:53:17 GMT

In article <1993Jan2.160912.2596@zip.eecs.umich.edu> kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:
>In article <1993Jan2.1584.7529@channel1> "eliot gelwan" <eliot.gelwan@channel1.com> writes:
>>On 12-31-92 KEITH F. LYNCH  wrote:
>>
>>KFWhat property did 1987 have, that no year after it will have
>>KFuntil 2013?
>>
>>Prime numbers
>
>Or, no repeated digits.

I think Kanad is right.  Note that there are plenty of prime numbers
between 1987 and 2013...

1987
1993  <-- Hey, that's this year!!
1997
1999
2003
2011
2017

Wait a minute!!!  2013 isn't even prime!!
Note that 3 x 11 x 61 = 2013.

Joel
-- 
Joel Chan, Department of Mathematics	  	 joel@math.toronto.edu
Toronto Blue Jays: 1992 World Series Champions
"History: Those who ignore it are condemned to repeat it.  Math, too."
				    -From the comic strip "Betty"
From: robertw@informix.com (Robert Weinberg)
Newsgroups: rec.puzzles
Subject: Re: Direction of the sun
Date: 29 Dec 92 16:40:22 GMT
Organization: Informix Software, Inc.

In article <1992Dec24.012636.17115@Csli.Stanford.EDU> hiraga@Csli.Stanford.EDU  
(Yuzuru Hiraga) writes:
> 
> # Heard from a Japanese colleague -- Sorry again if this is in the FAQL
> 
> Two boys, A and B, are watching the sunset.
> A points to the sun and says:
> "This is the direction where the sun is right now."
> B replies:
> "But don't you know that the rays take over 8 minutes to reach the Earth?
> The sun is actually already below the horizon."
> Which is correct?  Why? (Make it understandable to a ten year-old.)

This seems to be a question about reality versus observable reality - a question  
which can be asked about any distant object. We see it, we know that something  
happened in "reality" while the light reached us. For instance, as we read this,  
the Sun may "already" have exploded but we won't know about "the event" for another  
8 minutes.

I'm sure (:>) the physicists out there will tell us is no way that we will be aware  
of any consequences due to the explosion until the 8 minutes have passed - even if,  
say, a Sun-orbiting satellite has "already" been vaporized and has stopped sending  
telemetry.

It is not at all different from the question of the 5 blind men observing different  
parts of the elephant and constructing different "realities" of the "actual"  
elephant that is beyond their observation.

Mr. Hiraga's question "simply" highlights the uneasiness of relying on observation  
while realizing that observation is simultaneously referential. This goes on all  
the time. Tree falling in the woods but nobody hears it, Ronnie is really being  
manipulated by Nancy, etc.

> (Make it understandable to a ten year-old.)
Lots of luck! How about, "remember that Christmas present that you didn't get?"

--
+================================+================================+
|         Rob Weinberg           |  Workstation publications tech |
|   Email : robertw@informix.com |        Graphic designer        |
|   Voice : 415-926-6754         |     <imagine your ad here>     |
+================================+================================+
Newsgroups: rec.puzzles
From: hiraga@Csli.Stanford.EDU (Yuzuru Hiraga)
Subject: Re: simple number puzzle
Organization: Stanford University CSLI
Date: Tue, 22 Dec 1992 07:11:52 GMT

In article <1992Dec21.195038.28106@Csli.Stanford.EDU> I wrote:
>What positive integer cannot be expressed as a sum of 2 or more
>consecutive integers?
>
>More specifically, given a positive integer, how many ways can it
>be expressed as a sum of 2 or more consecutive [positive] integers
>(including none)?


Sorry again, as pointed out by some people, all "integers" should be
read as "positive integers" (i.e. natural numbers).
Thanks.

-YH
Newsgroups: rec.puzzles
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Subject: Re: Lost solution
Organization: Purdue University
Date: Fri, 25 Dec 1992 06:10:20 GMT

In article <1992Dec22.193053.24084@bernina.ethz.ch> timh@igc.ethz.ch (Tim Harvey) writes:
>Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce
>any truth from any set of axioms.  Two integers (not necessarily unique) are
>somehow chosen such that each is greater than 1 and less than 100.  Mr. S.
>is given the sum of these two integers; Mr. P. is given the product of these
>two integers.  After receiving these numbers, the two logicians do not
>have any communication at all except the following dialogue:
>   Mr. P.:  I do not know the two numbers.
>   Mr. S.:  I knew that you didn't know the two numbers; I do not know the
>            two numbers.
>   Mr. P.:  Now I know the two numbers.
>   Mr. S.:  Now I know the two numbers.

Given that the original numbers are in the range [2,99], the numbers must be 4
and 13. As a hint to get you started, the following are the possible sums that
Mr. S might have which would allow him to make his first statement:

	11, 17, 23, 27, 29, 35, 37, 41, 47, 53.

For each of these possible sums except one, there is more than one possible
product that Mr. P could have that would be consistent with his second
statement and also with Mr. S's first statement.

This puzzle is in the FAQ, by the way, but the FAQ answer uses incorrect
reasoning. The solution really has very little to do with primes.

--
Dave Seaman
ags@seaman.cc.purdue.edu
Newsgroups: rec.puzzles
From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
Subject: Re: new year's problem
Organization: University of Michigan EECS Dept., Ann Arbor
Distribution: rec
Date: Sat, 2 Jan 1993 17:40:10 GMT

In article <1993Jan2.160912.2596@zip.eecs.umich.edu> kanad@quip.eecs.umich.edu (Kanad Chakraborty) writes:
>In article <1993Jan2.1584.7529@channel1> "eliot gelwan" <eliot.gelwan@channel1.com> writes:
>>On 12-31-92 KEITH F. LYNCH  wrote:
>>
>>KFWhat property did 1987 have, that no year after it will have
>>KFuntil 2013?
>>
>>Prime numbers
>>
>

Oops !  I didn't notice this before -- thhe above answer given by
E. Gelwan is WRONG.  2013 is not a prime number.

>Or, no repeated digits.
>
>>KFWhat property did 1961 have, that no year
>>after it will have until 6009?
>>
>>turn 'em upside down and see.
>>
>>                   |
>>...eliot          -*-              Postlink: ->CHANNEL1 [R/O okay]
>>                   |               Internet: eliot.gelwan@channel1.com
>>
>>---
>>              [in Brookline MA 11:47 pm, Friday 01/01/93]
>>--
>>Channel 1 (R)   Cambridge, MA
>>
>
>Kanad


My answer again : (a) No repeated digits; (b) same as Eliot's answer.

Kanad
Newsgroups: rec.puzzles
From: isaak@aurora.com (Mark Isaak)
Subject: Re: Direction of the sun
Summary: atmospheric refraction
Reply-To: isaak@aurora.com (Mark Isaak)
Organization: The Aurora Group
Date: Fri, 1 Jan 93 22:37:38 GMT

I'm suprised no nitpickers have yet pointed out that the Earth's atmosphere
refracts the sun's light, so that, even if the Earth and sun were stationary
relative to each other, the sun's true position would be below the horizon
when it appeared to be just above it.
-- 
Mark Isaak           "Every generation thinks it has the answers, and every
isaak@aurora.com      generation is humbled by nature." - Philip Lubin
From: elnitsky@math.lsa.umich.edu (Serge Elnitsky)
Newsgroups: rec.puzzles
Subject: Mini-crossword-acrostic-whatever.
Date: 29 Dec 1992 20:05:33 GMT
Organization: University of Michigan, Mathematics Department, Ann Arbor
NNTP-Posting-Host: procyon.math.lsa.umich.edu

  The 7 words given by the clues form a well-known mathematical fact.

  1. Cogito, ergo ___ .      1: _____________
  2. The New York ___ .      2: _____________
  3. Vive la ___ !           3: _____________
  4. First among ___ .       4: _____________
  5. See 3.                  5: _____________
  6. Baseball pos.           6: _____________
  7. Dull, boring people.    7: _____________