Fishing - What's a boy to do?

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What's a boy to do?

On Wed, 1 Nov 2006 20:47:13 +0800, "riverman" wrote:

wrote in message
.. .
On 31 Oct 2006 21:50:30 -0800, "riverman" wrote:

No its not, its a matter of measurement precision.

No, it isn't. Or in the alternative, if it is, neither you or anyone
else could, as an absolute, measure whether C was farther than A or A
was farther than C. And if the latter is the case, your answer, above,
to your own question would still be incorrect.

Look, Myron, I'm not trying to bust your balls, and I'm not a
mathematician, so I've no idea as to what mathematicians consider
"oldies but goodies" or whatever when it comes to such problems,
puzzles, or whatever they call them. Maybe you forgot to give all the
details. But if you're now making/claiming assumptions you didn't state
originally, that's on you, and your answer as written to your own
question, also as written, is just wrong. Stated as you stated it, yes,
it is entirely possible for 2 (or 3 or 154 or "x") darts to be _exactly_
the same distance, especially in the theoretical "math puzzle" sense,
from a target.

Or, if one is going to operate in the completely practical sense and
take the position that even with the most accurate measuring devices
available, there's still no way to say "absolutely _exactly_ the same
distance," then it is equally impossible to state as an absolute that it
is always possible to determine which dart is further from the target.

Another alternative is that you are now assuming, but didn't then, or
did then and didn't disclose, that the darts are really "points," and
that in one axis, occupy a single, discreet plane. But that brings up a
host of problems for your answer, including the theoretical vs.
practical and/or the accuracy-of-measurement issue.

LOL. Certainly you're busting my balls. At least, I hope so, because
otherwise you sound like you're raving. The probability of two darts landing
a distance that is so close to identical from a target that it is beyond the
ability to be discerned is inversely proportional to the precision of the
measuring device. The more precise our devices, the less likely it is to
happen, and we have some phenominally precise devices, so the likihood of
this happening is relatively zero....that means its so close to zero that it
has no effect on the calculations.

Next, you'll assert that the odds of a coin landing Heads is not 50%,
because we forgot to count the times it lands on its edge. Or gets eaten by
a bird, or something. Those are relatively zero, although a coin landing on
edge is actually possible (I've had it happen twice in my life).

The point of this puzzler was to illustrate that how you approach the answer
is often the key to making something that seems unsolvable, solvable.

Heck, I'm willing to go completely real-world application. I've
witnessed throws for the first throw in which two darts were close
enough to the same distance from the bull that those throwing simply
re-threw. Heck, I, and every other darts-thrower out there, has hit the
rings and dividers - I'm willing to call that "the same distance
target." Take a 12" piece of string, tie loops in both ends, put a dart
through one loop and a marker in the other. Stick the dart into a board
and use the setup as a compass to mark a circular line. I'm willing to
call that "the same distance." There are darts players out there that
could hit such a line fairly consistently. As such, I've no problem
accepting the premise that in any given three-dart string, two could
well hit such a line. And I didn't bring up measurement precision, you
did.

IAC, given the "puzzle" as you stated it, and your answer as you stated
it, your answer was and is wrong.

HTH,
R

What's a boy to do?

On Wed, 1 Nov 2006 21:44:30 +0800, "riverman" wrote:

wrote in message
.. .
On Wed, 1 Nov 2006 20:47:13 +0800, "riverman" wrote:

wrote in message
...
On 31 Oct 2006 21:50:30 -0800, "riverman" wrote:

No its not, its a matter of measurement precision.

No, it isn't. Or in the alternative, if it is, neither you or anyone
else could, as an absolute, measure whether C was farther than A or A
was farther than C. And if the latter is the case, your answer, above,
to your own question would still be incorrect.

Look, Myron, I'm not trying to bust your balls, and I'm not a
mathematician, so I've no idea as to what mathematicians consider
"oldies but goodies" or whatever when it comes to such problems,
puzzles, or whatever they call them. Maybe you forgot to give all the
details. But if you're now making/claiming assumptions you didn't state
originally, that's on you, and your answer as written to your own
question, also as written, is just wrong. Stated as you stated it, yes,
it is entirely possible for 2 (or 3 or 154 or "x") darts to be _exactly_
the same distance, especially in the theoretical "math puzzle" sense,
from a target.

Or, if one is going to operate in the completely practical sense and
take the position that even with the most accurate measuring devices
available, there's still no way to say "absolutely _exactly_ the same
distance," then it is equally impossible to state as an absolute that it
is always possible to determine which dart is further from the target.

Another alternative is that you are now assuming, but didn't then, or
did then and didn't disclose, that the darts are really "points," and
that in one axis, occupy a single, discreet plane. But that brings up a
host of problems for your answer, including the theoretical vs.
practical and/or the accuracy-of-measurement issue.

LOL. Certainly you're busting my balls. At least, I hope so, because
otherwise you sound like you're raving. The probability of two darts
landing
a distance that is so close to identical from a target that it is beyond
the
ability to be discerned is inversely proportional to the precision of the
measuring device. The more precise our devices, the less likely it is to
happen, and we have some phenominally precise devices, so the likihood of
this happening is relatively zero....that means its so close to zero that
it
has no effect on the calculations.

Next, you'll assert that the odds of a coin landing Heads is not 50%,
because we forgot to count the times it lands on its edge. Or gets eaten
by
a bird, or something. Those are relatively zero, although a coin landing
on
edge is actually possible (I've had it happen twice in my life).

The point of this puzzler was to illustrate that how you approach the
answer
is often the key to making something that seems unsolvable, solvable.

Heck, I'm willing to go completely real-world application. I've
witnessed throws for the first throw in which two darts were close
enough to the same distance from the bull that those throwing simply
re-threw. Heck, I, and every other darts-thrower out there, has hit the
rings and dividers - I'm willing to call that "the same distance
target." Take a 12" piece of string, tie loops in both ends, put a dart
through one loop and a marker in the other. Stick the dart into a board
and use the setup as a compass to mark a circular line. I'm willing to
call that "the same distance." There are darts players out there that
could hit such a line fairly consistently. As such, I've no problem
accepting the premise that in any given three-dart string, two could
well hit such a line. And I didn't bring up measurement precision, you
did.

IAC, given the "puzzle" as you stated it, and your answer as you stated
it, your answer was and is wrong.

OK, whats the right answer then?

Give a person a fish, and they'll eat for the day, teach a person to
fish, and you'll morally and financially ruin them...no, wait...that's
not the one I was looking for...aha - teach a person to fish, and
they'll eat for a lifetime.

HTH,
R

--riverman

What's a boy to do?

"Jonathan Cook" wrote in message
...
Wolfgang wrote:

else, why bother with what is in the final analysis a
very simple problem in mathematics? The whole point of the exercise is
that the answer IS counterintuitive.

When something, especially something as simple as this, is
counterintuitive, my curiosity leads me to ask why it is
such. I'm not interested is saying "oh well; intuition is
wrong!" Why is it wrong?

Intuition is wrong in this instance because it leads one to think, falsely,
that after one of the losers has been exposed one is left with two equally
likely choices.

I think I can, mathematically,
explain why, even though it requires the reader to think
outside the box for a moment.

Well, it looks pretty much like you can't explain it mathematically.
Fortuantely, others have already done so......check out the Wikipedia entry
for the Monty Hall problem for a good example. Besides, several people have
already explained it quite adequately in other ways.

event. As a _singular_ event, you either have the right board or not,
there is no "law of averages" to consider. And singularly, I'm not

Playing the
game once constitutes a "single" event, not a "singular" one in any
meaningful sense. The difference is critical. A singular event is
something that happens only once......something like the evolution of
life on Earth, to pick a particularly controversial example.

Exactly my point!

It is?

That _is_ what I meant.

Ah, I see the problem! As is so often the case here, the words you used
provided no clues about what you meant.

So let's use it and
think outside the box for a moment...

Uh oh. "Outside the box" raises all kinds of red flags.....but, do proceed.

At any rate, you DO believe that probabilities apply to single events

Of course I do.

Yes.

But I wrote "singular".

Yes, you did.

Just for a moment
believe that I understand what I wrote.

I do, but you keep insisting that it means something else.

Think of it this way. What if you randomly turned over one of the
remaining boards, and when the board had "you lose" on it, you offered
to let me switch with the other one. Should I?

Yes, absolutely. Despite the fact that introducing a random element
has entirely changed the nature of the problem, as long as the random
pick turns up a loser, the outcome is identical to that of the
original. Changing your pick doubles the odds of winning.

A random pick _can't_ always turn up a loser.

No kidding? But, in this instance you stated quite clearly that it DID turn
up a loser. Thus, everything that follows is exactly as it is in the Monty
Hall problem. You have illustrated exactly nothing.

I did really mean "random".

IT DOESN'T MATTER IF IT'S RANDOM!! IF YOU TURN UP A LOSER THE RESULT IS
EXACTLY THE SAME AS IT IS IN THE MONTY HALL PROBLEM!!!!!

Yes it's a different game.

Yes, it is.

It was just an illustration.

BUT IT DIDN'T ILLUSTRATE ANYTHING!!!!!!

So what? All you've done devise a scenario in which the odds of
winning are 50:50. This has no bearing on the original problem in
which the point of the whole thing is that the odds are NOT 50:50.

Of course. That's my whole point. That _one_ instance of a
_different_ game can look exactly like _one_ instance of the
stated game, and have different odds.

So what?

It was just a motivating example.

I don't know what that means.

But suppose I just play the game once, and the random
turnover displays a "you lose" board. Then it looks just like your
"play once" scenario.

Sure it does. But the important thing is that it plays out like it
too......it doubles the odds of winning.

Not if it is the "random turnover" game.

IT DOESN'T MATTER IF IT'S RANDOM!! IF YOU TURN UP A LOSER THE RESULT IS
EXACTLY THE SAME AS IT IS IN THE MONTY HALL PROBLEM!!!!!

That's the whole point.

We're starting to get arather large stack of whole points here. Would it be
too much to ask you to whittle it down a bit......say, maybe end up with
just one?

Well, you're right about the utility of probabilities in predicting
singular events. But you are wrong in thinking that this is what we
are dealing with.

Ok, let's run with this for a moment. Remember, I'm not trying
to show that switching is wrong; I'm just trying to show _why_
our intuition says 50:50.

Intuition is wrong in this instance because it leads one to think, falsely,
that after one of the losers has been exposed one is left with two equally
likely choices.

I do _accept_ the analysis for the
game as stated, provided it is not _singular_.

Back to singular! Who's on first?

Let's suppose the game is singular, that it is played exactly
once in the entire life of the universe.

Like the airplane scenario that I thought I had proposed but doesn't seem to
be anywhere in evidence? Like the Toivo and Aino go to Vegas sketch that I
also thought I posted but which turns out to have been some sort of
hallucination on my part?

In this case, there is only _one_ configuration of the game.
Since configurations of the game are symmetrical, we can
without loss of generality choose one. So we choose WLL,
meaning the winning prize is behind door #1, and doors
2 and 3 have "you lose" signs behind them. Remember, there
is no appealing to any other "possibilities", because this
is the only game ever played in the history of the universe.

Now, the player has a choice, and Monty Hall has a followup
choice. If the player chooses the winning door, Monty can
reveal either losing door (two choices). But if the player
chooses a losing door, Monty _must_ reveal the other losing
door (1 choice), since he won't of course reveal the winning
door.

The two configurations for initially selecting the winning
door a

C-R
CR-

Where "C" represents the players choice, and R represents the
door that Monty reveals. The two configurations where the
player initially chooses a losing door a

-CR
-RC

Putting these together, there are _two_ configurations where
the player initially chose a winning door, and _two_ configurations
where the player initially chose a losing door. Flipping the
player selection after revealing simply swaps them, so whether
flipping or not flipping, the player has a 50% chance _after_
one door is revealed of winning the prize.

The point behind this is that when we are invited to step up
and play the game, our intuition treats it as a _singular_
event. And it can be mathematically justified as such.

It gets even better. We _can_ look at the non-singular game
and analyze probabilistically expected outcomes. Given that
we know that switching produces 2/3 chance of winning, and
holding produces a 1/3 chance of winning, but our intuition
thinks it is 50-50. So if 50% of the people will switch
and 50% hold, the entire expected outcome of all games is
50% winners and 50% losers (just take .5*.33 + .5*.67). Thus,
our 50-50 intuition will indeed be a self-fulfilling prophecy
on the long-term statistics of the game results).

Good God!......how do you manage to get out of bed in the morning without at
least occasionally slipping off into an alternative universe or something?

Wolfgang

What's a boy to do?

"MajorOz" wrote in message
oups.com...

On Oct 31, 4:30 pm, "Wolfgang" wrote:
MajorOz wrote:
As soon as you lifted #3 and exposed it as "you lose", the problem was
over.Well, not quite.....there was still the matter of making a
choice.....AFTER figuring out what the best choice is.

Now we have a new one:
Two boards -- one with a five and one without.
By asking me if I wish to change my mind,Huh? Who is asking you to
change your mind about what? The scenario,
as stated, gives no hint that you have done, said, or otherwise decided
anything about which to change your mind.

the new problem is simply one of choosing #1 or #2.Huh? What was the
old problem? (um......is anybody else seeing a whole
bunch of words here that aren't showing up on my screen?)

I do this by saying yes or no.What are you saying "yes" or "no" to? Is
it perhaps #1?.......or maybe
#2?.....something invisible to mere mortals?

My probablity of getting the $5 is simply 0.5O.k........if you say so.

SO, in answer to the question: "what do I do", I flip a coin.Toward
what end?

The dart problem is indeterminate -- not enough information about
unstated variables.We await the detailed analysis with bated
breath......or
palpitations......or something.

cheersProsit!

oz -- there's these two trains, heading towards each other with a bee
flying............Huh?

Wolfgang
who is beginning to think that perhaps brother skwalid has a point
after all.......this universe is starting to get a disturbingly skewed
look to it. :(

Poor Wolffie.............check that box in the corner to see if the cat
is alive or dead.

It would be an absolute riot to watch you try to explain something REALLY
complicated........um.....like crossing a street or something.....sometime.
:)

Wolfgang

What's a boy to do?

wrote in message
...
On Wed, 1 Nov 2006 21:44:30 +0800, "riverman" wrote:

OK, whats the right answer then?

Give a person a fish, and they'll eat for the day, teach a person to
fish, and you'll morally and financially ruin them...no, wait...that's
not the one I was looking for...aha - teach a person to fish, and
they'll eat for a lifetime.

HA! HA! HA! HA! HA! HA! HA! HA!

He doesn't have a clue what the question is......much less the right answer!

HA! HA! HA! HA! HA! HA! HA! HA!

Wolfgang

What's a boy to do?

"riverman" wrote in message ...

wrote in message
...
On Wed, 1 Nov 2006 20:47:13 +0800, "riverman" wrote:

wrote in message
...
On 31 Oct 2006 21:50:30 -0800, "riverman" wrote:

No its not, its a matter of measurement precision.

No, it isn't. Or in the alternative, if it is, neither you or anyone
else could, as an absolute, measure whether C was farther than A or A
was farther than C. And if the latter is the case, your answer, above,
to your own question would still be incorrect.

Look, Myron, I'm not trying to bust your balls, and I'm not a
mathematician, so I've no idea as to what mathematicians consider
"oldies but goodies" or whatever when it comes to such problems,
puzzles, or whatever they call them. Maybe you forgot to give all the
details. But if you're now making/claiming assumptions you didn't
state
originally, that's on you, and your answer as written to your own
question, also as written, is just wrong. Stated as you stated it,
yes,
it is entirely possible for 2 (or 3 or 154 or "x") darts to be
_exactly_
the same distance, especially in the theoretical "math puzzle" sense,
from a target.

Or, if one is going to operate in the completely practical sense and
take the position that even with the most accurate measuring devices
available, there's still no way to say "absolutely _exactly_ the same
distance," then it is equally impossible to state as an absolute that
it
is always possible to determine which dart is further from the target.

Another alternative is that you are now assuming, but didn't then, or
did then and didn't disclose, that the darts are really "points," and
that in one axis, occupy a single, discreet plane. But that brings up
a
host of problems for your answer, including the theoretical vs.
practical and/or the accuracy-of-measurement issue.

LOL. Certainly you're busting my balls. At least, I hope so, because
otherwise you sound like you're raving. The probability of two darts
landing
a distance that is so close to identical from a target that it is beyond
the
ability to be discerned is inversely proportional to the precision of the
measuring device. The more precise our devices, the less likely it is to
happen, and we have some phenominally precise devices, so the likihood of
this happening is relatively zero....that means its so close to zero that
it
has no effect on the calculations.

Next, you'll assert that the odds of a coin landing Heads is not 50%,
because we forgot to count the times it lands on its edge. Or gets eaten
by
a bird, or something. Those are relatively zero, although a coin landing
on
edge is actually possible (I've had it happen twice in my life).

The point of this puzzler was to illustrate that how you approach the
answer
is often the key to making something that seems unsolvable, solvable.

Heck, I'm willing to go completely real-world application. I've
witnessed throws for the first throw in which two darts were close
enough to the same distance from the bull that those throwing simply
re-threw. Heck, I, and every other darts-thrower out there, has hit the
rings and dividers - I'm willing to call that "the same distance
target." Take a 12" piece of string, tie loops in both ends, put a dart
through one loop and a marker in the other. Stick the dart into a board
and use the setup as a compass to mark a circular line. I'm willing to
call that "the same distance." There are darts players out there that
could hit such a line fairly consistently. As such, I've no problem
accepting the premise that in any given three-dart string, two could
well hit such a line. And I didn't bring up measurement precision, you
did.

IAC, given the "puzzle" as you stated it, and your answer as you stated
it, your answer was and is wrong.

OK, whats the right answer then?

--riverman

You could have an almost infinite amount of darts the exact same distance
from the center. The only limiting number is how big the circle is from the
center and how big of diameter is the dart. There are an infinite number of
points equidistant from the center point. And it depends on neither the
precision or accuracy of the measurement. And in your measurement of the
distance it would be more accuracy and not precision. Precision only gives
more numbers after the decimal point.

What's a boy to do?

wrote in message
...
On Wed, 1 Nov 2006 20:47:13 +0800, "riverman" wrote:

wrote in message
. ..
On 31 Oct 2006 21:50:30 -0800, "riverman" wrote:

No its not, its a matter of measurement precision.

No, it isn't. Or in the alternative, if it is, neither you or anyone
else could, as an absolute, measure whether C was farther than A or A
was farther than C. And if the latter is the case, your answer, above,
to your own question would still be incorrect.

Look, Myron, I'm not trying to bust your balls, and I'm not a
mathematician, so I've no idea as to what mathematicians consider
"oldies but goodies" or whatever when it comes to such problems,
puzzles, or whatever they call them. Maybe you forgot to give all the
details. But if you're now making/claiming assumptions you didn't state
originally, that's on you, and your answer as written to your own
question, also as written, is just wrong. Stated as you stated it, yes,
it is entirely possible for 2 (or 3 or 154 or "x") darts to be _exactly_
the same distance, especially in the theoretical "math puzzle" sense,
from a target.

Or, if one is going to operate in the completely practical sense and
take the position that even with the most accurate measuring devices
available, there's still no way to say "absolutely _exactly_ the same
distance," then it is equally impossible to state as an absolute that it
is always possible to determine which dart is further from the target.

Another alternative is that you are now assuming, but didn't then, or
did then and didn't disclose, that the darts are really "points," and
that in one axis, occupy a single, discreet plane. But that brings up a
host of problems for your answer, including the theoretical vs.
practical and/or the accuracy-of-measurement issue.

LOL. Certainly you're busting my balls. At least, I hope so, because
otherwise you sound like you're raving. The probability of two darts
landing
a distance that is so close to identical from a target that it is beyond
the
ability to be discerned is inversely proportional to the precision of the
measuring device. The more precise our devices, the less likely it is to
happen, and we have some phenominally precise devices, so the likihood of
this happening is relatively zero....that means its so close to zero that
it
has no effect on the calculations.

Next, you'll assert that the odds of a coin landing Heads is not 50%,
because we forgot to count the times it lands on its edge. Or gets eaten
by
a bird, or something. Those are relatively zero, although a coin landing
on
edge is actually possible (I've had it happen twice in my life).

The point of this puzzler was to illustrate that how you approach the
answer
is often the key to making something that seems unsolvable, solvable.

Heck, I'm willing to go completely real-world application.

Oh, THIS is gonna be fun! :)

I've witnessed throws for the first throw

Who decides who throws first in the throws for first throw? Um.....and how,
for that mater?

in which two darts were close
enough to the same distance from the bull that those throwing simply
re-threw.

How close is close enough? Who decides? How does they decide?

Heck, I, and every other darts-thrower out there, has hit the
rings and dividers - I'm willing to call that "the same distance
target."

You mean that everyone involved in a single game has hit the rings/dividers
in the same round? Otherwise, it's kind of hard to understand what you are
willing to call that "the same distance target." (whatever the hell that may
mean) If so, I'm willing to call that astonishing.

Take a 12" piece of string, tie loops in both ends, put a dart
through one loop and a marker in the other. Stick the dart into a board
and use the setup as a compass to mark a circular line.

I tried this last night. I tied loops in both ends of a twelve inch
string.......well, actually, it may have been about a sixteenth of an inch
long. See, I had to be sure that the string was straight in order to
measure the length accurately. I mean, if you just sort of drop it on the
table next to a ruler it kind of meanders like the Mississippi and cutting
it where it intersects the line segments denoting 0 and the twelve inch mark
results in something closer to 18 inches when you pull the kinks out. O.k.,
so I laid a book on one end at about 0 and then cut it at 12. Um......it
was roughly 13 inches long because there was an inch or so under the book.
So, I turned it around and holding just the teensiest bit under my left
thumbnail at 0 , I pulled gently on the other end and gazed, perplexed, at
the scissors lying on the table in front of me.......I'd run out of hands!
Hm......

Aha! I taped the 0 end and, pulling the other end gently with my right hand
to straighten it out, I picked up the scissors and cut with my left hand
(not an easy trick, I quickly discovered) at about twelve and an eighth,
reasoning that I MUST be stretching it a bit and when I made the cut it
would snap back to just about where I wanted it to be. Like I said, about
12 and a sixteenth.

O.k., so, I tied a loop in each end and ended up with something roughly 7
and three eighths long......hard to say for sure because to have to sort of
pull it strai......well, you know.

Then I put a dart one loop and a marker through the other and stuck the dart
in a board (as per instructions) and then using it like a compass I drew a
circle with the marker, which is what I assumed you meant by "mark a
circular line."

I started this project as soon as I got home from work yesterday.......about
4:15, I think it was......and was finished by about 6:30.

I'm willing to call that "the same distance."

I started this project as soon as I got home from work yesterday.......about
4:15, I think it was......and was finished by about 6:30. I spent the next
three hours looking at it and then went to bed. I got up around 5:00 this
morning and looked at it some more while going through my morning routine.
I try hard to be a responsible safe driver, so I didn't look at it on the
way to work, but I've been glancing at it from ocassionally as time
permitted through the course of the day.

Now, I'm not one to cast uncalled for casual aspersions on anyone else's
observations or judgments......and you can call this thing whatever you want
to......but, to me, it has from the moment I finished it, and it still does
look like a board inscribed with a roughly 14 inch black circle, at the
center of which (more or less) is a small round hole. Mind you, I want to
stress once again that it may actually BE a "the same distance".....hell,
I'm certainly no authority on those.....but it still LOOKS LIKE a board and
a circle and a little tiny hole.

There are darts players out there that
could hit such a line fairly consistently.

Thus demonstrating (if true) that there are darts player out there who could
hit such a line fairly consistently.

As such, I've no problem
accepting the premise that in any given three-dart string, two could
well hit such a line.

As which? Anyway, it comes as a bit of a shock that you would so readily
accept your own premise. I never figured you to be quite that credulous.

And I didn't bring up measurement precision, you
did.

O.k., I know this wasn't addressed to me and thus I need neither confirm nor
deny anything.....but I AM a bit curious. Who accused you of bringing up
measurement precision?

IAC, given the "puzzle" as you stated it, and your answer as you stated
it, your answer was and is wrong.

And that's the double-naught truth!

Hee, hee, hee.

Wolfgang
seriously though.....who do you think you're fooling?
p.s. yeah, it IS a matter of precision in measuring.

What's a boy to do?

"Calif Bill" wrote in message
.net...

You could have an almost infinite amount of darts the exact same distance
from the center.

No, actually, in practice you can't even have two.

The only limiting number is how big the circle is from the center and how
big of diameter is the dart.

English isn't even your second language, is it?

There are an infinite number of points equidistant from the center point.

Ah now, HERE'S a startling revelation! There's going to be quite a buzz in
geometric circles (if you'll pardon the expression) when word of this gets
out!

And it depends on neither the precision or accuracy of the measurement.

What does?

Hm....

Do you have any idea at all what this little charade is about?

And in your measurement of the distance it would be more accuracy and not
precision.

What would be?

Precision only gives more numbers after the decimal point.

And your contention here is that there is no difference we need bother
ourselves about between 6 inches and 6.99.*

Wolfgang
*particularly stupid......even for billie.

What's a boy to do?

wrote in message
...

I don't see how it's (objectively) counter-intuitive, and I think
attempting to get too involved in "math" (beyond the basic) makes it
more, rather than less difficult - for example, if it had been 4 boards,
two were turned over revealing losers, and then the choice to change
were given, to me, common sense indicates the odds say change your pick
because of the same reasons I feel it does with 3. If you must have
"math," I'm fairly sure the formula would be that the odds in favor of
switching are pretty close to if not exactly x-1/x and the odds in favor
of sticking are always exactly 1/x, when x is greater than 2, but I'm
not a mathematician, so ??? Perhaps the odds in favor need to account
for the first pick when x is higher than 3 - such that it isn't quite
x-1/x - but it's always going to be better odds than 1/x. ****, that's
confusing...that's why, IMO, algebra isn't the way to figure this out.

About the only thing I can figure is that it is much like many threads
on ROFF in that most folks, myself included at times, don't always
_read_ what they are "reading," but rather, um, infer from what is
written by what they _think_ is being said. In this case, they are
simply ignoring that there are 3, not 2, boards and therefore, the
chances cannot be 1 in 2.

No comment on any of that. I just wanted to repost it because it may be the
most beautiful thing I've ever seen! :)

Heck, given the "game" as outlined by Wolfgang, there's nothing
presented in the "rules" preventing the person from choosing the
revealed losing board - they were simply offered a chance to change
their pick. It would be the chooser making the obvious choice not to
choose it because they can clearly see they won't win (they don't need
to know that the chance of winning is 0 in 3).

Um.....well, o.k., this may be even beautifuler.

Wolfgang
hoo boy!