From Newsgroup: rec.puzzles


R.p has been slow lately. Here are four Oldies-but-Goodies:

(1) How much of the Earth's surface is North of latitude 30-#N, i.e. the thirtieth parallel north. Assume Earth's surface is a perfect sphere.

- - - - - - - - -

For puzzles (2), (3), (4) assume that the team meets during the days before the actual contest, that they discuss the rules, agree to cooperate, and devise an optimal strategy. During the actual contest, no communication among them is allowed except as stated. The colors or numbers appearing on their hats
are chosen at random.
For puzzles (3) and (4) assume each player has a computer with UNLIMITED
memory and computational power.

(2) The four members of the team sit at a table. Each can see the other players' colors (each is red or green), but not his or her own. On a piece
of paper hidden from view, each player writes 'Red', 'Green' or 'Pass.'
The team wins if and only if
. a. at least one player guesses his/her color correctly; and
. b. no player guesses the wrong color.
For example, if three players write 'Pass', and the 4th player has a Green hat, the team wins if and only if that 4th player writes 'Green.'

With optimal strategy, what is the probability that the team wins?
This is tricky. First try it with 3 players. Or two with 'pass' not an option.

- - - - - - - - -

(3) Now instead of a color, a numeral 0 to 9 is written on the back of each player's hat. Instead of 4 players, there are a hundred.
The players are lined up so that the first-to-speak (the first to guess his numeral out loud) can see all numerals except his own, and the k'th-to-speak can hear the (k-1) numerals guessed so far, and see all numerals
except his own and those of the (k-1) who have already guessed.

In turn, each player guesses the numeral on his hat. The team wins if at least 99 of the 100 players guess their numeral correctly.

- - - - - - - - -

(4) This is the same contest as (3) EXCEPT
. a. There are a (countably) infinite number of players. There is a 1st, 2nd and 3rd player as before, and a 4th, 5th player, etc., but this line of
players goes on forever.
. b. Guesses are submitted silently. The k'th player has no information except the (infinitely many) numerals he can see in front of him.
. c. You do assume the Axiom of Choice.

Devise a strategy such that only a finite number of the players will
fail to guess their numeral correctly.

- - - - - - - - -

Cheers,
James

--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sun, 23 Aug 2026 17:44:04 GMT, James Dow Allen wrote:

(1) How much of the Earth's surface is North of latitude 30-#N, i.e. the thirtieth parallel north. Assume Earth's surface is a perfect sphere.

That's an interesting question.

V unir cerivbhfyl gevrq, naq snvyrq, gb trarengr na vzntr eryngrq gb culyybgnkvf, fb V qvqa'g unir zhpu ubcr bs na nafjre. Ohg gura V abgvprq n tbys onyy, ba zl qrfx, ba juvpu V'ir pbybherq va n fznyy ahzore bs
qvzcyrf ...
--
David Entwistle
--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <1787507044-4353@newsgrouper.org>,
James Dow Allen <user4353@newsgrouper.org.invalid> wrote:

(1) How much of the Earth's surface is North of latitude 30-#N, i.e. the >thirtieth parallel north. Assume Earth's surface is a perfect sphere.

This may help (indirectly):

https://upload.wikimedia.org/wikipedia/commons/a/a6/Cicero_Discovering_the_Tomb_of_Archimedes_by_Benjamin_West.jpeg

-- Richard
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From Newsgroup: rec.puzzles


richard@cogsci.ed.ac.uk (Richard Tobin) posted:

/ERROR "unexpected byte sequence starting at index 166: '\xC2'" while decoding/:

Yes. Expecting the degree symbol to be mangled, that's why I wrote
"i.e. the thirtieth parallel north."

In article <1787507044-4353@newsgrouper.org>,
James Dow Allen <user4353@newsgrouper.org.invalid> wrote:

(1) How much of the Earth's surface is North of latitude 30|e-#N, i.e. the >thirtieth parallel north. Assume Earth's surface is a perfect sphere.

This may help (indirectly):

https://upload.wikimedia.org/wikipedia/commons/a/a6/Cicero_Discovering_the_Tomb_of_Archimedes_by_Benjamin_West.jpeg

Yes. Archimedes is often called the greatest mathematical genius to ever live; and the measures of a sphere are among his greatest discoveries.

(A well-known(?) list of the greatest mathematicians in ranked order shows Archimedes as 2nd to Isaac Newton, but that's because the latter man had enormous historical importance while Archimedes -- way ahead of his time --
had rather little.)

-- Richard

-- James
--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On 23/08/2026 18:44, James Dow Allen wrote:

R.p has been slow lately. Here are four Oldies-but-Goodies:

(1) How much of the Earth's surface is North of latitude 30#N, i.e. the thirtieth parallel north. Assume Earth's surface is a perfect sphere.

- - - - - - - - -

For puzzles (2), (3), (4) assume that the team meets during the days before the
actual contest, that they discuss the rules, agree to cooperate, and devise an
optimal strategy. During the actual contest, no communication among them is allowed except as stated. The colors or numbers appearing on their hats
are chosen at random.
For puzzles (3) and (4) assume each player has a computer with UNLIMITED memory and computational power.

(2) The four members of the team sit at a table. Each can see the other players' colors (each is red or green), but not his or her own. On a piece of paper hidden from view, each player writes 'Red', 'Green' or 'Pass.'
The team wins if and only if
. a. at least one player guesses his/her color correctly; and
. b. no player guesses the wrong color.
For example, if three players write 'Pass', and the 4th player has a Green hat,
the team wins if and only if that 4th player writes 'Green.'

With optimal strategy, what is the probability that the team wins?
This is tricky. First try it with 3 players. Or two with 'pass' not an option.

- - - - - - - - -

(3) Now instead of a color, a numeral 0 to 9 is written on the back of each player's hat. Instead of 4 players, there are a hundred.
The players are lined up so that the first-to-speak (the first to guess his numeral out loud) can see all numerals except his own, and the k'th-to-speak can hear the (k-1) numerals guessed so far, and see all numerals
except his own and those of the (k-1) who have already guessed.

In turn, each player guesses the numeral on his hat. The team wins if at least
99 of the 100 players guess their numeral correctly.

- - - - - - - - -

(4) This is the same contest as (3) EXCEPT
. a. There are a (countably) infinite number of players. There is a 1st, 2nd
and 3rd player as before, and a 4th, 5th player, etc., but this line of players goes on forever.
. b. Guesses are submitted silently. The k'th player has no information except the (infinitely many) numerals he can see in front of him.
. c. You do assume the Axiom of Choice.

Devise a strategy such that only a finite number of the players will
fail to guess their numeral correctly.

.s......
..p.....
..o.....
.i......
..l.....
...e....
....r...
.....s..
.....p..
....o...
.....i..
......l.
......r.
.....r..
....s...
...p....
..o.....
..i.....
...l....
..e.....
.r......
.s......
..p.....
..o.....
.i......
..l.....
...e....
....r...
.....s..
.....p..
....o...
.....i..
......l.
......r.
.....r..
....s...
...p....
..o.....
..i.....
...l....
..e.....
.r......

(1) I'll say 1/4, just because I remember some famous result telling us that the area of such a
band around a sphere is the same as the area of the corresponding band of a cylinder of the same
radius. And 30N is "1/2 the way" between the equator and N pole, measured by projecting
perpendicularly onto the earth's axis. (If I hadn't remembered that result I'd have been into
calculating integrals etc.)

(2) Not sure about this one - without any systematic evaluations I reckon a number of approaches
would give a 1/2 chance of winning, but maybe there's something better. The strategies I've thought
of basically get players to bet on a 3-1 split in hat colours, which occurs in 1/2 cases. E.g. if a
player sees 3 red hats, choose green and vice versa. (We could also say that a player seeing RRG
votes R, and vice versa as this still wins for all 3-1 splits. I'd expect there's a similar
approach targetting a 4-0 or 2-2 split(?))

(3) The first player can see 99 hats, and through the 10 possible choices he can make, he can
indicate the mod-10 sum of all those hats. (This will probably be wrong for his own hat, but we're
allowed one failure...). Everybody else can see the mod-10 sum of all 98 of those hats apart from
their own, and so can work out their own hat, given they heard the first playser give the total for
the 99 hats...

(4) I'm pretty sure you are aiming here for the 'mathematical' strategy based on the axiom of
choice [AC]. There is such a 'strategy' going broadly like this:
- form the set of equivalence classes of sequences of hats which are "eventually the same"
(or equivalently which differ at only a finite number of places)
- choose a representative for each class [this is where AC is used]
The above steps are considered part of the players "agreeing their strategy", so all players
somehow know all those (uncountably many) representative choices!
- when in the infinite queue, players look at the hats ahead and "identify" which
equivalence class of hat-sequences they are in.
- Then they guess their hat colour based on their position in the queue and
the corresponding hat colour in the representative hat sequence
that they "agreed" for that class.
This ensures that only finitely many guesses will be wrong.

BUT this isn't really a workable strategy that can be applied by anybody in the queue - it's just a
mathematical argument that such a "strategy" exists (if we accept AC). It looks like you've tried
to mitigate this by saying the players have been given a magical "computer" with "unlimited memory
and computational power". Personally I don't see any reasonable way of interpreting what you've
said as providing something the players could actually /use/ in their strategy. Turing machines
have unlimited memory and computing power, but are of know use whatsoever here. Of course, your
phrase is (deliberately?) vague, and I'm looking forward to seeing whether you can /define/ such a
computer and how exactly the players use it when they're standing in the queue! :)

[Bottom line is I don't consider (4) a proper "puzzle", although (for mathematicians) the maths
involved might be considered interesting. The argument is not exactly "well known", but maybe "well
known by set theorists!"? Much better presentations of the argument can be found online by
searching, um, probably "infinite hat puzzle using AC" or the likes.]

Regards,
Mike.

--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles


Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 23/08/2026 18:44, James Dow Allen wrote:

R.p has been slow lately. Here are four Oldies-but-Goodies:
...
For puzzle[] ... (4) assume each player has a computer with UNLIMITED memory and computational power.

A normal computer has 2^N words of memory, each word M bits, where
N and M are finite and smallish. The preposterously UNLIMITED computer
has 2^F words of F bits each where F is infinite; along with appropriate peripheral hardware. Never mind what hyper-reality must be postulated
to implement such a machine.

(2) ...
With optimal strategy, what is the probability that the team wins?
This is tricky. First try it with 3 players.
Or two with 'pass' not an option.

"Tricky" in an almost perverse way. Do try it first with only 3 players.

- - - - - - - - - - -

I will offer brief comments on Mike's excellent answers,
preserving his spoiler warning.

.s......
..p.....
..o.....
.i......
..l.....
...e....
....r...
.....s..
.....p..
....o...
.....i..
......l.
......r.
.....r..
....s...
...p....
..o.....
..i.....
...l....
..e.....
.r......
.s......
..p.....
..o.....
.i......
..l.....
...e....
....r...
.....s..
.....p..
....o...
.....i..
......l.
......r.
.....r..
....s...
...p....
..o.....
..i.....
...l....
..e.....
.r......

(1) I'll say 1/4, just because I remember some famous result telling us that the area of such a
band around a sphere is the same as the area of the corresponding band of a cylinder of the same
radius. And 30N is "1/2 the way" between the equator and N pole, measured by projecting
perpendicularly onto the earth's axis. (If I hadn't remembered that result I'd have been into
calculating integrals etc.)

Did you need to take a sine?

The famous result is of course by the famous man Richard Tobin mentioned, who may
or may not have famously run through the streets of Syracuse shouting "Eureka!"

(3) ... Everybody else can see the mod-10 sum of all 98 of those hats apart from
their own,

"can see OR has heard"

and so can work out their own hat, given they heard the first playser give the total for
the 99 hats...

(4) I'm pretty sure you are aiming here for the 'mathematical' strategy based on the axiom of
choice [AC]. There is such a 'strategy' going broadly like this:
- form the set of equivalence classes of sequences of hats which are "eventually the same"
(or equivalently which differ at only a finite number of places)
- choose a representative for each class [this is where AC is used]
The above steps are considered part of the players "agreeing their strategy", so all players
somehow know all those (uncountably many) representative choices!
- when in the infinite queue, players look at the hats ahead and "identify" which
equivalence class of hat-sequences they are in.
- Then they guess their hat colour based on their position in the queue and
the corresponding hat colour in the representative hat sequence
that they "agreed" for that class.
This ensures that only finitely many guesses will be wrong.

Yes.

BUT this isn't really a workable strategy ...
... I'm looking forward to seeing whether you can /define/ such a
computer and how exactly the players use it when they're standing in the queue! :)

Does the hypothetical infinite (Aleph-1) computer I mention above, in a hypothetical hyper-reality count? Even as a thought experiment?

[Bottom line is I don't consider (4) a proper "puzzle", although (for mathematicians) the maths
involved might be considered interesting...]

I think it's interesting because it seems to imply that the Axiom of Choice must obviously be false!!

Regards,
Mike.

Cheers,
James.
--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Sun, 23 Aug 2026 17:44:04 GMT, James Dow Allen <user4353@newsgrouper.org.invalid> wrote:

R.p has been slow lately. Here are four Oldies-but-Goodies:

(1) How much of the Earth's surface is North of latitude 30#N, i.e. the >thirtieth parallel north. Assume Earth's surface is a perfect sphere.

- - - - - - - - -

A related puzzle based on a my recent, real life, experince.

A few months back I underwent some eye surgery which involved
the removing of the vitreous humour and, temporarily, replacing
it with a gas bubble. It takes about 6 to 8 weeks for the gas to
be absorbed by the body and replaced with new vitreous humour.
Initially, the gas fills almost 100% of the eyball.

At the start, the bubble's interface with the little liquid inside is
barely visible, but as the gas gets absorbed, a black ring becomes
visible (at the top of the field of vision) and then grows larger
and larger as the bubble's volume approaches the halfway point.
(The ring is really the meniscus formed by the gas-liquied interface.)
The interface is now near or at the iris. Then onwards , as the
bubble continues to shrink, the black ring appears at the bottom
of the field of vision and grows progressively smaller.

Assuming that the rate of absorption of the gas in constant,
what is the rate at which the radius of the ring shrinks?

Assume that the inside of the eyeball is a hollow, perfect
sphere and ignore the volume effects due to the menisucs
(even though it is what causes the main effect in question!).

I found the algebra got more tedious that I had initially
expected.

I can also state that in real life, the last stages of the
bubble's life is very noticebale. After weeks of seeing it
the ring decrease very slowly, suddenly one begins to
notice almost hourly changes!

The corresponding question in James's problem, is
asking what is the radius the circle of the "cap", if
the area reduces at a constant rate. (I have not
worked that out yet!)
--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On 24/08/2026 18:04, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 23/08/2026 18:44, James Dow Allen wrote:

R.p has been slow lately. Here are four Oldies-but-Goodies:
...
For puzzle[] ... (4) assume each player has a computer with UNLIMITED
memory and computational power.

A normal computer has 2^N words of memory, each word M bits, where
N and M are finite and smallish. The preposterously UNLIMITED computer
has 2^F words of F bits each where F is infinite; along with appropriate peripheral hardware. Never mind what hyper-reality must be postulated
to implement such a machine.

(2) ...
With optimal strategy, what is the probability that the team wins?
This is tricky. First try it with 3 players.
Or two with 'pass' not an option.

"Tricky" in an almost perverse way. Do try it first with only 3 players.

ok, thinking about 3 players... (will post later)

- - - - - - - - - - -

I will offer brief comments on Mike's excellent answers,
preserving his spoiler warning.

.s......
..p.....
..o.....
.i......
..l.....
...e....
....r...
.....s..
.....p..
....o...
.....i..
......l.
......r.
.....r..
....s...
...p....
..o.....
..i.....
...l....
..e.....
.r......
.s......
..p.....
..o.....
.i......
..l.....
...e....
....r...
.....s..
.....p..
....o...
.....i..
......l.
......r.
.....r..
....s...
...p....
..o.....
..i.....
...l....
..e.....
.r......

(1) I'll say 1/4, just because I remember some famous result telling us that the area of such a
band around a sphere is the same as the area of the corresponding band of a cylinder of the same
radius. And 30N is "1/2 the way" between the equator and N pole, measured by projecting
perpendicularly onto the earth's axis. (If I hadn't remembered that result I'd have been into
calculating integrals etc.)

Did you need to take a sine?

I just needed to know about equilateral triangles...

The famous result is of course by the famous man Richard Tobin mentioned, who may
or may not have famously run through the streets of Syracuse shouting "Eureka!"

Ah yes, I'm not surprised by that.

(3) ... Everybody else can see the mod-10 sum of all 98 of those hats apart from
their own,

"can see OR has heard"

Right. I had misunderstood the problem, thinking player k could see all players hats apart from his
own. Now I realise he can only see the hats for players in front of him, but like you say, he has
/heard/ the answers given by players 1,2,3...k-1 which were all correct (if they're following the
strategy) so that's good enough for the strategy (modified slightly) to work.

and so can work out their own hat, given they heard the first playser give the total for
the 99 hats...

(4) I'm pretty sure you are aiming here for the 'mathematical' strategy based on the axiom of
choice [AC]. There is such a 'strategy' going broadly like this:
- form the set of equivalence classes of sequences of hats which are "eventually the same"
(or equivalently which differ at only a finite number of places)
- choose a representative for each class [this is where AC is used]
The above steps are considered part of the players "agreeing their strategy", so all players
somehow know all those (uncountably many) representative choices!
- when in the infinite queue, players look at the hats ahead and "identify" which
equivalence class of hat-sequences they are in.
- Then they guess their hat colour based on their position in the queue and >> the corresponding hat colour in the representative hat sequence
that they "agreed" for that class.
This ensures that only finitely many guesses will be wrong.

Yes.

BUT this isn't really a workable strategy ...
... I'm looking forward to seeing whether you can /define/ such a
computer and how exactly the players use it when they're standing in the queue! :)

Does the hypothetical infinite (Aleph-1) computer I mention above, in a hypothetical hyper-reality count? Even as a thought experiment?

Well, how do you intend the players to use it, exactly? :) How does "calculation" work with such a
computer? Etc.. I think you might as well just leave it as a mathematical argument rather than
trying to imagine it being real.

[Bottom line is I don't consider (4) a proper "puzzle", although (for mathematicians) the maths
involved might be considered interesting...]

I think it's interesting because it seems to imply that the Axiom of Choice must obviously be false!!

I don't see it that way - how does the implication go? Perhaps you start from the "fact" that there
can't be any such strategy? Where would that come from?

It's tricky trying to apply real-world intuitions to unphysical infinite scenarios. After all, each
player in the queue is supposed as having access to an infinite amount of information [viz the
infinite sequence of hat colours of those in front of the player] to start with, which is beyond our
physical abilities. How do we even argue about what is possible/reasonable in such made up
scenarios? Well, we can make mathematical arguments, sure, but I don't see anything there saying no
such strategy can exist...

Mike.

--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On 24/08/2026 18:04, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 23/08/2026 18:44, James Dow Allen wrote:

R.p has been slow lately. Here are four Oldies-but-Goodies:
...
For puzzle[] ... (4) assume each player has a computer with UNLIMITED
memory and computational power.

A normal computer has 2^N words of memory, each word M bits, where
N and M are finite and smallish. The preposterously UNLIMITED computer
has 2^F words of F bits each where F is infinite; along with appropriate peripheral hardware. Never mind what hyper-reality must be postulated
to implement such a machine.

(2) ...
With optimal strategy, what is the probability that the team wins?
This is tricky. First try it with 3 players.
Or two with 'pass' not an option.

"Tricky" in an almost perverse way. Do try it first with only 3 players.

For 3 players I see similar sorts of strategies, except they are a little simpler.

Specifically, players can effectively bet on a 2-1 split which occurs with probability 3/4, and in
doing so the team wins with that probability. E.g. the strategy can be:
- if a player sees team mates with hats RR then choose G
- if a player sees team mates with hats GG then choose R
- (otherwise pass)

But I don't feel that's helped me with the 4-player game. (I suppose that I could write a program
to enumerate what I consider plausible strategies, but I'm doubting I'm going to get anything better
than 3/4 for the 3-player game...)

I'm probably overlooking some key insight!


Mike.

--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Mon, 24 Aug 2026 07:40:44 -0000 (UTC), David Entwistle wrote:

That's an interesting question.

AI produces some astonishing gibberish in response to this question...
--
David Entwistle
--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote or quoted:

On Mon, 24 Aug 2026 07:40:44 -0000 (UTC), David Entwistle wrote:

That's an interesting question.

AI produces some astonishing gibberish in response to this question...

Might be a bit spoilerish:

AI was ok here. It explained to me that when one gets higher up to
the north pole, the same vertical increment gives a surface increment
that is larger because the surface inclination has decreased.
But on the other hand, the total surface gets smaller because its
circumference is reduced. And these two effects cancel, so only one
sine effect remains. So, this was the approach of Archimedes.
(That's what the chatbot explained to me.)

Now, I always wanted to better understand just what a "surface
element" is. So here are my thoughts:

The boring general approach of today's calculus uses surface elements.
A surface element is something like the square with the corners
(0, 0), (0, 1), (1, 0), and (1, 1). The surface of this square is 1.
So if we call the height dy and the width dx, it's dx ^ dy = 1,
where the product "^" means "the surface of a rectangle with these
sides".

Now, assume a new y coordinate y' = 2 y. The new coordinates of the
square are now (0, 0)', (0, 2)', (1, 0)', and (1, 2)'. Now, dx=1, but
dy'=2, so the product is 2. But the surface has not increased, just
the coordinates have changed, so this product does not give the real
surface anymore. I call such a product the "formal surface", because
it "formally" gives us the surfaces in the new coordinate system
"formally" applying the rule "width * height" for a rectangle, but
it is not really the surface because it contains a coordinate effect.
So I also call this formal surface the /coordinate surface/.

On the other hand y = y' / 2, and dy = dy' / 2, so the same surface
element dx ^ dy is now written dx ^( dy' / 2 )= (1/2) dx ^ dy'.

Just summing the surface of the area with the coordinates (0, 0)',
(0, 2)', (1, 0)', and (1, 2)' gives twice the surfaces but the new
surface element transforms this back to the surface area in the
original coordinates muliplying it by (1/2), and so we get the
same area as before.

The first coordinate system is distinguished: When it is used, the
surface of a rectangle is the product of its width and height with
no correction needed, because it has the special surface element
dx ^ dy = 1 dx ^ dy with "1" as the scaling factor, and scaling by
one is no scaling at all. I call such a coordinate system the
"reference coordinate system". The surface of a rectangle in the
reference coordinate system is the "real surface" of a rectangle,
not just its formal surface.

So, to measure the surface in any coordinate system, we can take
two steps:
- get the formal surface in that coordinate system
- use the surface element to convert that formal surface into the
real surface.

In an R^3, a Cartesian coordinate system with the orthonormal basis
vectors, each of length 1 (think, "one meter") is our reference
coordinate system. To get the real surface of any rectangle in this
system, we can just multiply its coordinate-width by its coordinate-
height.

But for a sphere embedded in an R^3 we prefer /spherical coordinates/.
However, as the spherical coordinates are not the reference coordi-
nates, to get the real surface of any object, we need to multiply
by an appropriate surface element that converts the spherical
coordinate surface back into a real surface.

When t is the polar angle/colatitude, p is the azimuthal angle and
r is the radius, the surface element for spherical coordinates is

dA = r^2 sin t dt ^ dp.

Qualitatively this makes sense, because the same dp corresponds
to a larger real surface when r is larger. So to convert the
coordinate surface of spherical coordinates to a real surface,
we need to multiply them with this surface element.

The area of any surface of the sphere that is delimited by
ranges [p0, p1] and [t0, t1] of p and t values can then be
calculated as the surface integral

p1 t1
/ /
| | r^2 sin t dt ^ dp,
/ /
p=p0 t=t0

where we use the surface element "r^2 sin t dt ^ dp" to correct
the effect of using the spherical coordinates, so as to get
the real surface of this area.

A "line element" does something similar, but for a length instead
of an area.


--- Synchronet 3.22a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles


Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 24/08/2026 18:04, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 23/08/2026 18:44, James Dow Allen wrote:

(2) ...
With optimal strategy, what is the probability that the team wins?
This is tricky. First try it with 3 players.
Or two with 'pass' not an option.

"Tricky" in an almost perverse way. Do try it first with only 3 players.

For 3 players I see similar sorts of strategies, except they are a little simpler.

Specifically, players can effectively bet on a 2-1 split which occurs with probability 3/4, and in
doing so the team wins with that probability. E.g. the strategy can be:
- if a player sees team mates with hats RR then choose G
- if a player sees team mates with hats GG then choose R
- (otherwise pass)

Beautiful!

But I don't feel that's helped me with the 4-player game. (I suppose that I could write a program
to enumerate what I consider plausible strategies, but I'm doubting I'm going to get anything better
than 3/4 for the 3-player game...)

I assume you DO have a 3/4 solution for the 4-player game?
Gur "gevpx" vf gung gurer vf ab jnl gb vzcebir ba gung.

I'm probably overlooking some key insight!

For N=5 or larger the puzzle becomes horrifically tedious and difficult.
(I'll supply keywords for Google search if you wish.)

You have been warned. If you do pursue N=5 despite my admonition
do NOT call me a sadist.

Sbe svir cynlref gur punapr bs fhpprff vf gjragl-svir bhg bs guvegl-gjb.
Sbe fvk cynlref gur punapr bs fhpprff vf gjragl-fvk bhg bs guvegl-gjb.

Mike.

James.
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From Newsgroup: rec.puzzles

On 26/08/2026 08:08, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 24/08/2026 18:04, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 23/08/2026 18:44, James Dow Allen wrote:

(2) ...
With optimal strategy, what is the probability that the team wins?
This is tricky. First try it with 3 players.
Or two with 'pass' not an option.

"Tricky" in an almost perverse way. Do try it first with only 3 players. >>

For 3 players I see similar sorts of strategies, except they are a little simpler.

Specifically, players can effectively bet on a 2-1 split which occurs with probability 3/4, and in
doing so the team wins with that probability. E.g. the strategy can be:
- if a player sees team mates with hats RR then choose G
- if a player sees team mates with hats GG then choose R
- (otherwise pass)

Beautiful!

But I don't feel that's helped me with the 4-player game. (I suppose that I could write a program
to enumerate what I consider plausible strategies, but I'm doubting I'm going to get anything better
than 3/4 for the 3-player game...)

I assume you DO have a 3/4 solution for the 4-player game?
Gur "gevpx" vf gung gurer vf ab jnl gb vzcebir ba gung.

At the time I previously posted I only had the 1/2 solution I'd previously described.

Since then I've written a (pretty naive) program to search for the best strategies, and for the
3-player game it confirms the best we can do gives a 3/4 chance for the team to win.

For the 4-player game I doubt I will be able to run the program to completion - there are just too
many potential strategies to enumerate. :( But as luck would have it, it's found a handful of 3/4
strategies near the beginning of its search tree - these strategies have one of the players always
passing regardless of what hats that player sees!

But these results don't really improve my understanding.

Is there an obvious reason why we can't improve on 3/4? (I see from your hints below that 3/4 can
be exceeded for 5- or 6-players, so I doubt there's any obvious reason.)

Mike.

I'm probably overlooking some key insight!

For N=5 or larger the puzzle becomes horrifically tedious and difficult. (I'll supply keywords for Google search if you wish.)

You have been warned. If you do pursue N=5 despite my admonition
do NOT call me a sadist.

Sbe svir cynlref gur punapr bs fhpprff vf gjragl-svir bhg bs guvegl-gjb.
Sbe fvk cynlref gur punapr bs fhpprff vf gjragl-fvk bhg bs guvegl-gjb.

Mike.

James.

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From Newsgroup: rec.puzzles

On 26/08/2026 23:18, Mike Terry wrote:

On 26/08/2026 08:08, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 24/08/2026 18:04, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 23/08/2026 18:44, James Dow Allen wrote:

(2)a ...
With optimal strategy, what is the probability that the team wins? >>>>>> This is tricky. First try it with 3 players.
Or two with 'pass' not an option.

"Tricky" in an almost perverse way.a Do try it first with only 3 players. >>>

For 3 players I see similar sorts of strategies, except they are a little simpler.

Specifically, players can effectively bet on a 2-1 split which occurs with probability 3/4, and in
doing so the team wins with that probability.a E.g. the strategy can be: >>> -aa if a player sees team mates with hats RR then choose G
-aa if a player sees team mates with hats GG then choose R
-aa (otherwise pass)

Beautiful!

But I don't feel that's helped me with the 4-player game.a (I suppose that I could write a program
to enumerate what I consider plausible strategies, but I'm doubting I'm going to get anything better
than 3/4 for the 3-player game...)

I assume you DO have a 3/4 solution for the 4-player game?
Gur "gevpx" vf gung gurer vf ab jnl gb vzcebir ba gung.

At the time I previously posted I only had the 1/2 solution I'd previously described.

Since then I've written a (pretty naive) program to search for the best strategies, and for the
3-player game it confirms the best we can do gives a 3/4 chance for the team to win.

For the 4-player game I doubt I will be able to run the program to completion - there are just too
many potential strategies to enumerate. :(a But as luck would have it, it's found a handful of 3/4
strategies near the beginning of its search tree - these strategies have one of the players always
passing regardless of what hats that player sees!

But these results don't really improve my understanding.

I hadn't twigged before, but some of these strategies are just the 3-player strategies in disguise,
with the 4th player ignored (and 4th player always passing). Such a strategy can always be used
with any greater number of players, and yields the same win-rate as the 3-player game strategy.

We can conclude:
a) however many players (>3) there is always a fairly straight-forward strategy
borrowed from the 3-player game giving a 3/4 win-rate, and
b) as the number of players increases, the best strategy win rate cannot go down...

Mike.

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From Newsgroup: rec.puzzles


Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 26/08/2026 08:08, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 24/08/2026 18:04, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 23/08/2026 18:44, James Dow Allen wrote:

(2) ...
With optimal strategy, what is the probability that the team wins? >>>>> This is tricky. First try it with 3 players....

"Tricky" in an almost perverse way. Do try it first with only 3 players. >>

For 3 players I see ...

Specifically, players can effectively bet on a 2-1 split which occurs
with probability 3/4, ...

Beautiful!

But I don't feel that's helped me with the 4-player game.

I assume you DO have a 3/4 solution for the 4-player game?
Gur "gevpx" vf gung gurer vf ab jnl gb vzcebir ba gung.

At the time I previously posted I only had the 1/2 solution I'd previously described.

Since then I've written a (pretty naive) program to search for the best strategies, and for the
3-player game it confirms the best we can do gives a 3/4 chance for the team to win.

For the 4-player game I doubt I will be able to run the program to completion ... :(
But as luck would have it, it's found a handful of 3/4
strategies near the beginning of its search tree
- these strategies have one of the players always
passing regardless of what hats that player sees!

But these results don't really improve my understanding.

The N=3 problem was presented long ago, I think right here in r.p.

Extending it to N=4 was my own "invention". Cutesy because the
solution -- (the ONLY way to get 3/4 ??) -- is to nominate one
player ("George") and instruct him to always pass. A sadistic variation, perhaps
since many will ignore that easy way to get 3/4.

IIRC when the puzzle was presented long ago, some mathematicians pointed out that 3/4 could be exceeded for N>4 using ideas borrowed from Hamming (error-correcting)
codes. Perhaps the puzzle had been cross-posted to sci.math or sci.crypt.

Especially good solutions exist for
- N=3 ... p(Win) = 3/4
- N=7 ... p(Win) = 7/8
- N=15 .. p(Win) = 15/16

These are like Hamming error-correction codes. Except that for N=7
there are 2^4 Correct codes, whereas in this puzzle it is precisely
those 2^4 "correct" codes that result in failure.

I'm probably overlooking some key insight!

You and me both, Mike! What's embarrassing is that 50 years ago
I worked with Hamming codes but now get easily confused.

For N=5 or larger the puzzle becomes horrifically tedious and difficult. (I'll supply keywords for Google search if you wish.)

N=5 or 6 may be sadistic. But BIG KUDOS to anyone who demonstrates
correct strategy for the relatively easy(??) N=7 or N=15.

Mike.

James.

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From Newsgroup: rec.puzzles

richard@cogsci.ed.ac.uk (Richard Tobin) writes:

In article <1787507044-4353@newsgrouper.org>,
James Dow Allen <user4353@newsgrouper.org.invalid> wrote:

(1) How much of the Earth's surface is North of latitude 30|e-#N, i.e. the >>thirtieth parallel north. Assume Earth's surface is a perfect sphere.

This may help (indirectly):

https://upload.wikimedia.org/wikipedia/commons/a/a6/Cicero_Discovering_the_Tomb_of_Archimedes_by_Benjamin_West.jpeg

As expected from 3b1b, this is wonderful:

/But why is a sphere's surface area four times its shadow?/

3Blue1Brown
Shared 7 years ago

https://youtu.be/watch?v=GNcFjFmqEc8

Phil
--
We are no longer hunters and nomads. No longer awed and frightened, as we have gained some understanding of the world in which we live. As such, we can cast aside childish remnants from the dawn of our civilization.
-- NotSanguine on SoylentNews, after Eugen Weber in /The Western Tradition/
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From Newsgroup: rec.puzzles

On 27/08/2026 11:42, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 26/08/2026 08:08, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 24/08/2026 18:04, James Dow Allen wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> posted:

On 23/08/2026 18:44, James Dow Allen wrote:

(2) ...
With optimal strategy, what is the probability that the team wins? >>>>>>> This is tricky. First try it with 3 players....

"Tricky" in an almost perverse way. Do try it first with only 3 players. >>>>

For 3 players I see ...

Specifically, players can effectively bet on a 2-1 split which occurs
with probability 3/4, ...

Beautiful!

But I don't feel that's helped me with the 4-player game.

I assume you DO have a 3/4 solution for the 4-player game?
Gur "gevpx" vf gung gurer vf ab jnl gb vzcebir ba gung.

At the time I previously posted I only had the 1/2 solution I'd previously described.

Since then I've written a (pretty naive) program to search for the best strategies, and for the
3-player game it confirms the best we can do gives a 3/4 chance for the team to win.

For the 4-player game I doubt I will be able to run the program to completion ... :(
But as luck would have it, it's found a handful of 3/4
strategies near the beginning of its search tree
- these strategies have one of the players always
passing regardless of what hats that player sees!

But these results don't really improve my understanding.

The N=3 problem was presented long ago, I think right here in r.p.

Extending it to N=4 was my own "invention". Cutesy because the
solution -- (the ONLY way to get 3/4 ??) -- is to nominate one
player ("George") and instruct him to always pass. A sadistic variation, perhaps
since many will ignore that easy way to get 3/4.

Not so much "ignore" as "not realise", as in my case! (I realised the general principle in the end.)

There are other ways to get 3/4 for the 4-player game, without stealing a 3-player strategy. My
program found a few where no player always passes. It's not found anything better than 3/4, and I
don't expect it to, but OTOH it's not going to complete the search any time soon [like in my
lifetime], so I might as well abort it I suppose... I'm not motivated to spend time improving the
performance, so that's it for now.

Mike.

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From Newsgroup: rec.puzzles


James Dow Allen <user4353@newsgrouper.org.invalid> posted:

(2) The four members of the team sit at a table. Each can see the other players' colors (each is red or green), but not his or her own. On a piece of paper hidden from view, each player writes 'Red', 'Green' or 'Pass.'
The team wins if and only if
. a. at least one player guesses his/her color correctly; and
. b. no player guesses the wrong color.
For example, if three players write 'Pass', and the 4th player has a Green hat,
the team wins if and only if that 4th player writes 'Green.'

With optimal strategy, what is the probability that the team wins?

SPOILERS for N=3 and N=7 follow.
22 Stanza's from Hunting the Snark as spoiler warning space ============================================================

"Just the place for a Snark!" the Bellman cried,
As he landed his crew with care;
Supporting each man on the top of the tide
By a finger entwined in his hair.

"Just the place for a Snark! I have said it twice:
That alone should encourage the crew.
Just the place for a Snark! I have said it thrice:
What I tell you three times is true."

The crew was complete: it included a Boots--
A maker of Bonnets and Hoods--
A Barrister, brought to arrange their disputes--
And a Broker, to value their goods.

A Billiard-marker, whose skill was immense,
Might perhaps have won more than his share--
But a Banker, engaged at enormous expense,
Had the whole of their cash in his care.

There was also a Beaver, that paced on the deck,
Or would sit making lace in the bow:
And had often (the Bellman said) saved them from wreck,
Though none of the sailors knew how.

There was one who was famed for the number of things
He forgot when he entered the ship:
His umbrella, his watch, all his jewels and rings,
And the clothes he had bought for the trip.

He had forty-two boxes, all carefully packed,
With his name painted clearly on each:
But, since he omitted to mention the fact,
They were all left behind on the beach.

The loss of his clothes hardly mattered, because
He had seven coats on when he came,
With three pairs of boots--but the worst of it was,
He had wholly forgotten his name.

He would answer to "Hi!" or to any loud cry,
Such as "Fry me!" or "Fritter my wig!"
To "What-you-may-call-um!" or "What-was-his-name!"
But especially "Thing-um-a-jig!"

While, for those who preferred a more forcible word,
He had different names from these:
His intimate friends called him "Candle-ends,"
And his enemies "Toasted-cheese."

"His form is ungainly--his intellect small--"
(So the Bellman would often remark)
"But his courage is perfect! And that, after all,
Is the thing that one needs with a Snark."

He would joke with hyenas, returning their stare
With an impudent wag of the head:
And he once went a walk, paw-in-paw, with a bear,
"Just to keep up its spirits," he said.

He came as a Baker: but owned, when too late--
And it drove the poor Bellman half-mad--
He could only bake Bridecake--for which, I may state,
No materials were to be had.

The last of the crew needs especial remark,
Though he looked an incredible dunce:
He had just one idea--but, that one being "Snark,"
The good Bellman engaged him at once.

He came as a Butcher: but gravely declared,
When the ship had been sailing a week,
He could only kill Beavers. The Bellman looked scared,
And was almost too frightened to speak:

But at length he explained, in a tremulous tone,
There was only one Beaver on board;
And that was a tame one he had of his own,
Whose death would be deeply deplored.

The Beaver, who happened to hear the remark,
Protested, with tears in its eyes,
That not even the rapture of hunting the Snark
Could atone for that dismal surprise!

It strongly advised that the Butcher should be
Conveyed in a separate ship:
But the Bellman declared that would never agree
With the plans he had made for the trip:

Navigation was always a difficult art,
Though with only one ship and one bell:
And he feared he must really decline, for his part,
Undertaking another as well.

The Beaver's best course was, no doubt, to procure
A second-hand dagger-proof coat--
So the Baker advised it--and next, to insure
Its life in some Office of note:

This the Banker suggested, and offered for hire
(On moderate terms), or for sale,
Two excellent Policies, one Against Fire,
And one Against Damage From Hail.

Yet still, ever after that sorrowful day,
Whenever the Butcher was by,
The Beaver kept looking the opposite way,
And appeared unaccountably shy

============================================================

Just this N=3 may seem amazing. ANY guess has precisely a 50% chance
of being wrong. Assume w.l.o.g. that George guesses. (Someone has to guess or the team loses by default.) George fails 50% of the time. Whatever the other teammates do, they must fail at least 50% of the time!

Or so it would seem.

The trick is to combine failures. If you see two of the same color, guess the other color. Otherwise pass. Here are the results. W means player makes a winning
guess; L, a losing guess.
- RRR - LLL
- RRG - ..W
- RGR - .W.
- RGG - W..
- GRR - W..
- GRG - .W.
- GGR - ..W
- GGG - LLL

See? Half the guesses are indeed wrong, but they're bunched together
and the team wins 6 out of 8 cases. The table of cases can be abbreviated
- 2 x LLL
- 6 x W

To win 7/8 of the time when N=7 we want a table of cases
- 16 x LLLLLLL
- 112 x W

Assign the seven 3-bit labels (001, 010, ..., 111) to the seven players.
(All bit triplets except 000.) Each player sums (via XOR) the labels
on the players he sees wearing Green. 1/8 of the time the total sum will
be 000 -- in those cases we LOSE. But otherwise we win.

Suppose the total seen (all but self) is 000. Self writes "Green." if
green is correct, All others pass (see next paragraph). (If Red,
it is one of the losing configs. In fact EVERY player will announce Green
-- and lose -- in that losing config case.)

Suppose players 001 and 010 are Green and all others are Red. Those players compute XOR sums 010 and 010 respectively and all others see 011.
Everyone Passes, except the
player labeled 011 -- the same as the xor total he sees. That player writes "Red." (Green would give a losing config.)

This is the same mechanic as a single-bit-error correcting Hamming code,
except that the goal is reversed.

Cheers, James



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