What's a boy to do?
 

wrote in message
I don't see how it's (objectively) counter-intuitive,

There are two remaining choices, switch or don't switch. I think 50:50 is
an easy conclusion to draw from that.

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.

No doubt many ROFFians are guilty of that; but that is also the intentional
nature of "brain teasers".

In this case, they are
simply ignoring that there are 3, not 2, boards and therefore, the
chances cannot be 1 in 2.

Not exactly, but similar IMO. There are, in fact, 2 choices remaining. The
failure is that of seeing the two choices as random for the original player.
Most folks, I expect, see the problem from the perspective of your "third
person" for whom they *are* random.

Joe F.

What's a boy to do?
 

rb608 wrote:
wrote in message
I don't see how it's (objectively) counter-intuitive,

There are two remaining choices, switch or don't switch. I think 50:50 is
an easy conclusion to draw from that.

Exactly......and that is precisely what makes the Monty Hall problem
interesting......well, that's a part of it, anyway (more about that in
just a moment). It isn't the math. Hell the math is simple enough
that even I (no math wiz......by ANY stretch of the imagination) have
no trouble at all in understanding and accepting various permutations
of the explanation. Anyone adept at mathematics and who takes a moment
to think it through will invariably come up with the right answer and,
doubtless, find the whole thing rather silly.

Somewhat ironically, it takes a basic knowledge of the fundamental laws
of probability to figure out the wrong answer.....you have to know that
tossing a coin will, in the long run, result in something very close to
half heads, half tails. Anyone who doesn't know this can only
guess.....and is as likely to guess right as wrong.....50:50 chance!
Sweet! There is no doubt in my mind that both Craig Whitaker and
Marylin vos Savant were well aware of this when the former posed the
question and the latter decided to answer it.

At least a couple of people have made references to the rules as I
stated them in posing the problem. In fact, there were NO rules.
There was simply a question about how one should proceed in a precisely
and unambiguously stated situation. Suggestions and speculations about
how to work through more or less similar situations (changing the
"rules") may or may not be interesting in their own right, but they
have nothing whatsoever to do with the original problem. I suspect
that most of them have something or other to do with a certain level of
discomfort engendered by the decidedly counterintuitive correct
solution to the original.

When all is said and done, the whole thing is a trick question. What
makes it exquisitely delicious is that, as stated at the outset, I, the
expositor, was not playing any kind of trick on the player......well
not directly, anyway. No, what tricks the player is his or her own
knowledge of probabilities and a lightning quick recognition of an
absurdly easy problem.

Right, Ken?

O.k., that last bit was just a little unfair. Um......or was it?
After all, Haddon said the pretty much same thing. Did anybody else
see it? A shiny new nickel to the first to point out Haddon's own
sorta nasty little trick. (hint: it's in the quotes......more or less)
:)

Bottom line? The Monty Hall problem really isn't much of a
mathematical puzzle at all. What it IS......in spades......is a
beautifully elegant probe into human psychology!

As for illustrating the logic behind the correct solution, here's my
own humble contribution:

Let us change the scenario a bit. Instead of a single player who gets
to decide whether to change his or her pick after one of the losers is
exposed, let's have TWO players......Toivo and Aino. Toivo gets to
pick one of the three possibilities.....Aino automatically gets the
other two. All three positions are exposed. Any one may be the
winner, but no one should have any difficulty in seeing that the smart
money would bet on Aino. Whether in a single round or in repeated
play, the odds are clearly in his favor to the tune of two to
one.......67% to 33%.....not too roughly. Now, let us suppose that
rather than exposing all the possibilites at once, the expositor turns
over one of the boards at random. How does this change the odds?
Clearly, it has absolutely no effect on the odds. O.k., so, at least
one of Aino's two possibilities HAS TO be a loser.....right? After
all, there are three positions and only one of them is the winner.
Alright, so, if the expositor first turns over one of Aino's
possibilities, which is one of the losers, how does this affect the
odds? Again, it cannot possibly affect the odds......the winner and
both losers are where they are.....NOTHING can affect the fact that
Aino wins two times out of three.....more or less......in the long run.

Now, let's go back to the problem as originally stated. Toivo is the
only player. As long as he sticks with his original choice when given
the option, nothing, in essence, is any different than it was in the
two player game......he loses two times out of three......Aino has
simply become invisible. Obvious......right? Right. O.k., so, what
if Toivo chooses to jump one way one time and another the next? Beats
the **** out of me (and you too, if there's an honest bone in your
body). Ah, but what if Toivo changes his choice EVERY time? Well,
then he quite simply BECOMES Aino!! :)

Wolfgang
dunkenfeld knew.

What's a boy to do?
 

"Wolfgang" wrote in message
oups.com...

rb608 wrote:
wrote in message
I don't see how it's (objectively) counter-intuitive,

There are two remaining choices, switch or don't switch. I think 50:50
is
an easy conclusion to draw from that.

Exactly......and that is precisely what makes the Monty Hall problem
interesting......well, that's a part of it, anyway (more about that in
just a moment). It isn't the math. Hell the math is simple enough
that even I (no math wiz......by ANY stretch of the imagination) have
no trouble at all in understanding and accepting various permutations
of the explanation. Anyone adept at mathematics and who takes a moment
to think it through will invariably come up with the right answer and,
doubtless, find the whole thing rather silly.

Somewhat ironically, it takes a basic knowledge of the fundamental laws
of probability to figure out the wrong answer.....you have to know that
tossing a coin will, in the long run, result in something very close to
half heads, half tails. Anyone who doesn't know this can only
guess.....and is as likely to guess right as wrong.....50:50 chance!
Sweet! There is no doubt in my mind that both Craig Whitaker and
Marylin vos Savant were well aware of this when the former posed the
question and the latter decided to answer it.

At least a couple of people have made references to the rules as I
stated them in posing the problem. In fact, there were NO rules.
There was simply a question about how one should proceed in a precisely
and unambiguously stated situation. Suggestions and speculations about
how to work through more or less similar situations (changing the
"rules") may or may not be interesting in their own right, but they
have nothing whatsoever to do with the original problem. I suspect
that most of them have something or other to do with a certain level of
discomfort engendered by the decidedly counterintuitive correct
solution to the original.

When all is said and done, the whole thing is a trick question. What
makes it exquisitely delicious is that, as stated at the outset, I, the
expositor, was not playing any kind of trick on the player......well
not directly, anyway. No, what tricks the player is his or her own
knowledge of probabilities and a lightning quick recognition of an
absurdly easy problem.

Right, Ken?

O.k., that last bit was just a little unfair. Um......or was it?
After all, Haddon said the pretty much same thing. Did anybody else
see it? A shiny new nickel to the first to point out Haddon's own
sorta nasty little trick. (hint: it's in the quotes......more or less)
:)

Bottom line? The Monty Hall problem really isn't much of a
mathematical puzzle at all. What it IS......in spades......is a
beautifully elegant probe into human psychology!

As for illustrating the logic behind the correct solution, here's my
own humble contribution:

Let us change the scenario a bit. Instead of a single player who gets
to decide whether to change his or her pick after one of the losers is
exposed, let's have TWO players......Toivo and Aino. Toivo gets to
pick one of the three possibilities.....Aino automatically gets the
other two. All three positions are exposed. Any one may be the
winner, but no one should have any difficulty in seeing that the smart
money would bet on Aino. Whether in a single round or in repeated
play, the odds are clearly in his favor to the tune of two to
one.......67% to 33%.....not too roughly. Now, let us suppose that
rather than exposing all the possibilites at once, the expositor turns
over one of the boards at random. How does this change the odds?
Clearly, it has absolutely no effect on the odds. O.k., so, at least
one of Aino's two possibilities HAS TO be a loser.....right? After
all, there are three positions and only one of them is the winner.
Alright, so, if the expositor first turns over one of Aino's
possibilities, which is one of the losers, how does this affect the
odds? Again, it cannot possibly affect the odds......the winner and
both losers are where they are.....NOTHING can affect the fact that
Aino wins two times out of three.....more or less......in the long run.

Now, let's go back to the problem as originally stated. Toivo is the
only player. As long as he sticks with his original choice when given
the option, nothing, in essence, is any different than it was in the
two player game......he loses two times out of three......Aino has
simply become invisible. Obvious......right? Right. O.k., so, what
if Toivo chooses to jump one way one time and another the next? Beats
the **** out of me (and you too, if there's an honest bone in your
body). Ah, but what if Toivo changes his choice EVERY time? Well,
then he quite simply BECOMES Aino!! :)



Interesting illustration of how to visualize the correct strategy, but I
suggest any doubters just make a spinner out of a paperclip and a piece of
paper. Draw a circle, divide it into three 'pizza slices' of approximately
the same area, and hold the clip in the center with a pencil tip. Decide
that one 'pizza slice' is the actual prize, spin the clip to choose your
intitial door, and act out the scenario. The logic behind the '2/3' answer
is instantly and undeniably clear.

Or, to paraphrase Wolfgang's Toivo and Aino situation, the odds of winning
by switching doors after one is revealed is exactly equal to the odds that
your original pick was wrong....2/3.

--riverman

What's a boy to do?
 

"Opus McDopus" wrote in message
...

"asadi" wrote in message
...

If I'm one of the remaining members I'd say your chances were pretty
damned good....

john

Are we goin to remain on non-speaking terms forever?

Op


okay, I'll give up in a couple of hours and try sending and e-mail and we'll
figure out why yours are bouncing...


I'd do it now but I'm busy making sure Vince wins the tontine...

john

What's a boy to do?
 

"asadi" wrote in message
. ..

okay, I'll give up in a couple of hours and try sending and e-mail and
we'll figure out why yours are bouncing...


I'd do it now but I'm busy making sure Vince wins the tontine...

john

You can get me at: or you can call me at:
828-292-9005

I'm always on speakin' terms with you, as far as you know, anyway :~^ )

Don't know why your e-mails would have bounced, but I will check my
kill-file?

Op

What's a boy to do?
 

"Opus McDopus" wrote in message
...

Don't know why your e-mails would have bounced, but I will check my
kill-file?

Op

Nope, you aren't in Kill-file, just the usual suspects?

Op

What's a boy to do?
 

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

You toss three darts at a target. Dart A misses the target, then Dart B
misses by even more. What is the probability that Dart C will miss by
more
than Dart A?

Not enough information. But _don't_ play darts barefoot if you
are prone to dropping things...

Jon.

Wrong on the first count, right on the second. :-)

--riverman

What's a boy to do?
 

Jonathan Cook wrote:
riverman wrote:

You toss three darts at a target. Dart A misses the target, then Dart B
misses by even more. What is the probability that Dart C will miss by more
than Dart A?

Not enough information.

That sounds right to me. However, under the circumstances I can't
shake a nagging suspicion that there's a bad smell somewhere in there
that I haven't noticed yet.

To put it another way, while a given mathematical proposition may be
clear, correct, simple and unambiguous, it does not necessarily follow
that it will be easy (or even possible.....perhaps) to deal with it
clearly, correctly, simply and unambiguously in plain English.....or
any other natural language, for that matter. One of the beauties of
the Monty Hall problem is that it illustrates this divide very nicely.
Each individual who has provided a "real world" example of how to
arrive at the correct solution and why it IS correct undoubtedly feels
that it clear and easy to follow. The fact that several of us have
taken on the task suggests otherwise.

Following, as it does, closely on the heels of the MH problem, this one
is naturally suspect. :)

All of this points to a very fertile field of enquiry that is
characterized (as well as by anything else, I think) by the fact that a
lot of very bright people find anything in mathematics beyond
elementary arithmetic and perhaps a bit of algebra to be a completely
impenetrable mystery. Much of it IS (and has been at least since the
invention of zero) counterintuitive. Anyone who doubts this would do
well to consult Whitehead and Russell.

But _don't_ play darts barefoot if you
are prone to dropping things...

Don't play darts barefoot......period.

Wolfgang

What's a boy to do?
 

Wolfgang wrote:
Jonathan Cook wrote:
riverman wrote:

You toss three darts at a target. Dart A misses the target, then Dart B
misses by even more. What is the probability that Dart C will miss by more
than Dart A?

Not enough information.

That sounds right to me. However, under the circumstances I can't
shake a nagging suspicion that there's a bad smell somewhere in there
that I haven't noticed yet.


A nice paraphrase of what I tell my students: some things are simple if
you know how, and impossible if you don't.

To put it another way, while a given mathematical proposition may be
clear, correct, simple and unambiguous, it does not necessarily follow
that it will be easy (or even possible.....perhaps) to deal with it
clearly, correctly, simply and unambiguously in plain English.....or
any other natural language, for that matter. One of the beauties of
the Monty Hall problem is that it illustrates this divide very nicely.
Each individual who has provided a "real world" example of how to
arrive at the correct solution and why it IS correct undoubtedly feels
that it clear and easy to follow. The fact that several of us have
taken on the task suggests otherwise.

One reason that math terminology exists (other than to exclude, as is
with all jargon), is because many things can be explained unambiguously
in math to others who speak the language, and hopefully solved in that
terminology. If it were possible to put everything into non-math terms
easily, we would need neither math terminology nor prerequisite
knowledge.


Following, as it does, closely on the heels of the MH problem, this one
is naturally suspect. :)

All of this points to a very fertile field of enquiry that is
characterized (as well as by anything else, I think) by the fact that a
lot of very bright people find anything in mathematics beyond
elementary arithmetic and perhaps a bit of algebra to be a completely
impenetrable mystery. Much of it IS (and has been at least since the
invention of zero) counterintuitive. Anyone who doubts this would do
well to consult Whitehead and Russell.

No argument that math can be challenging, although I'm not sure what
you mean by 'bright people'. It takes training to learn math...that old
'Royal Road' thing, but people want it to be intuitive and I suspect
that this is and has always been the core of the problem.

But the thing the MH puzzle does, as well as this one, is it makes
people who are not well-versed in math but who feel like their
intuition is the sacred measure, face their error. When faced with the
contradiction between their instincts and the mathematical reality of
the MH puzzle, only a fool would insist that the math is wrong and
their instincts are correct. Yet many very bright people are fools. :-)

But, as with many things mathematical (especially many things
statistical), the key to the answer is in how you approach the
solution. To answer the darts question, merely rely on the definition
of probablilty: the number of ways to achieve your objective, divided
by the number of possible outcomes. List all the possible arrangements
of how the darts could land, and count how many fit our scenario.
First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome. However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

--riverman

What's a boy to do?
 

riverman wrote:
Wolfgang wrote:
Jonathan Cook wrote:
riverman wrote:

You toss three darts at a target. Dart A misses the target, then Dart B
misses by even more. What is the probability that Dart C will miss by more
than Dart A?

Not enough information.

That sounds right to me. However, under the circumstances I can't
shake a nagging suspicion that there's a bad smell somewhere in there
that I haven't noticed yet.


A nice paraphrase of what I tell my students: some things are simple if
you know how, and impossible if you don't.

Not to put too fine a point on it, but that would not be some
things......that would be pretty much everything.

To put it another way, while a given mathematical proposition may be
clear, correct, simple and unambiguous, it does not necessarily follow
that it will be easy (or even possible.....perhaps) to deal with it
clearly, correctly, simply and unambiguously in plain English.....or
any other natural language, for that matter. One of the beauties of
the Monty Hall problem is that it illustrates this divide very nicely.
Each individual who has provided a "real world" example of how to
arrive at the correct solution and why it IS correct undoubtedly feels
that it clear and easy to follow. The fact that several of us have
taken on the task suggests otherwise.

One reason that math terminology exists (other than to exclude, as is
with all jargon), is because many things can be explained unambiguously
in math to others who speak the language, and hopefully solved in that
terminology.

Sure.

If it were possible to put everything into non-math terms
easily, we would need neither math terminology nor prerequisite
knowledge.

True, we wouldn't need a specialized terminology. Prerequisite
knowledge strikes me as a bit ambiguous......could mean any of several
things. Any way I look at it, though, I can't see a way to do without
it.

Following, as it does, closely on the heels of the MH problem, this one
is naturally suspect. :)

All of this points to a very fertile field of enquiry that is
characterized (as well as by anything else, I think) by the fact that a
lot of very bright people find anything in mathematics beyond
elementary arithmetic and perhaps a bit of algebra to be a completely
impenetrable mystery. Much of it IS (and has been at least since the
invention of zero) counterintuitive. Anyone who doubts this would do
well to consult Whitehead and Russell.

No argument that math can be challenging, although I'm not sure what
you mean by 'bright people'.

Well, take several PhDs vehemently defending the wrong answer to a
particular simple problem, for example. As a class, PhDs are arguably
reasonably bright people. More generally, I guess I mean people who
show an aptitude for dealing with a variety of different kinds of
problems and an ability to articulate their thoughts in a manner that
is comprehensible to other concerned parties.

It takes training to learn math...that old
'Royal Road' thing, but people want it to be intuitive and I suspect
that this is and has always been the core of the problem.

To the extent that people are more interested in solutions than
problems (and I think it's fair to say that most people probably are in
most instances), I suspect that they more often than not would prefer
that the answers come intuitively rather than at the expense of hard
mental labor. This seems to me like a perfectly reasonable position.
Personally, I don't think this is "the" core of the problem that so
many people have with math. I believe it's more complicated than that.
For one thing, people start out as children, naturally and voraciously
curious little beasts.......intellectual sponges eager to suck up
whatever they can. Despite some minor variations among different
tribes and cults, educational institutions and, more particularly, the
cultures they serve tend to be remarkably indistinguishable in their
capacity to crush that curiosity at a tender age......as they have
done, for the most part, for centuries. There are many other problems.

But the thing the MH puzzle does, as well as this one, is it makes
people who are not well-versed in math but who feel like their
intuition is the sacred measure, face their error.

In theory. In practice, most just brush it aside. Ever been to Las
Vegas?

When faced with the
contradiction between their instincts and the mathematical reality of
the MH puzzle, only a fool would insist that the math is wrong and
their instincts are correct.

The Monty Hall problem is simple enough that trusting to intuition is
easily demonstrable as a foolish course of action. That's what makes
it illustrative. That's what makes it interesting. But this is by no
means always the case. Take a look at competing sophisticated economic
theories on the advisability of debt load sometime. Whose math are you
going to believe? Intuition tells me that having more cash than debt
is a good thing.

Yet many very bright people are fools. :-)

All bright people do foolish things, but they are NOT fools.....by
definition.

But, as with many things mathematical (especially many things
statistical), the key to the answer is in how you approach the
solution. To answer the darts question, merely rely on the definition
of probablilty: the number of ways to achieve your objective, divided
by the number of possible outcomes. List all the possible arrangements
of how the darts could land, and count how many fit our scenario.
First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome. However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

I don't see anything in the above about whether the person throwing the
darts is a champion player.....or drunk.....or blind. Remember, we are
looking at a conditional probability; we know very little about the
conditions that apply. At its root, this isn't really a math problem.

Counterintuitive, nicht wahr?

Wolfgang
to the man with a hammer.......

What's a boy to do?
 

On 30 Oct 2006 16:17:56 -0800, "riverman" wrote:


First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Those aren't all the ways...think about it.

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

No, it isn't...think about it.

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome.

Well, perhaps it's because of that, or perhaps because it's
wrong...think about it.

However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

Maybe it would help you get on-target answer-wise if you tied a string
to your finger in exactly the same spot two days in a row...

HTH,
R

--riverman

What's a boy to do?
 

riverman wrote:
Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome. However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

I agree with the basic premise and conclusions as you present them; but
I have to disagree that the six original possibilities are equally
possible as the original problem is stated. If all three darts are
thrown at the same time, and the probabilities are computed based on
the final position of the three darts, then I agree. But if the darts
are thrown sequentially, and with the implied effects of skill and
intent (and OBROFF, sobriety), then I'm unconvinced that the outcomes
are equally random.

I like the parallels of this question with the MH paradox. The
intuitive answer for an individual case is different than the
mathematical answer for a large sample size. I suppose that's what
makes them interesting.

Joe F.

What's a boy to do?
 

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

An interesting problem was recently brought to my attention.

I think I've seen this before but never really thought about it.
Thanks.

I like you so I'm going to make this easier for you," and I remove board
number three to show you that it has a "you lose" tag under it.
Obviously,
the five dollars must be under one of the other two. "So," I say, "would
you like to stick with your original pick, or change your mind?"

It is a given that the game is not rigged in any way and you are not
being
fooled by anything ambiguous or otherwise misleading in the description.

The question.......what should you do?

I agree with everything that's been written here on what the "correct"
solution is, and how Marilyn made a bunch of PhDs and professors look
stupid (not that hard, really), but the problem has been irking me,
and I finally decided why.

You posed the problem _singularly_. One try. Probability is about
expected outcomes over lots of attempts. It breaks down in a singular
event.

Expermental probability is about expected outcomes observed over multiple
attempts, while theoretical probability is about expected outcomes
calculated by counting possible outcomes. But Probabilities in general ARE
about predicted outcomes for singular events.

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
convinced that it is worth switching boards (though I absolutely agree
that over lots of tries it is).


Put the numbers 1-100 in a hat, and draw one. It could be #1, or it might
not. It is correct to say that, the probability of it being #1 is 1/100
based on 100 draws, but as a singular event, its either #1 or its not. But
those two outcomes (that it is or its not) aren't equally likely. Its far
more likely to be something other than #1, so if I drew a number and said
"want to bet a horse that it's #1", you'd refuse the bet, because the
probabilities are well against you.

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? Obviously in this case
it doesn't matter, because I still lose on the times the board with the
money is turned over, so switching or not I'll only win, on average, 50%
of the remaining ones, which is 1/3 of all of them. Randomness doesn't
add information to the problem like purposely turning over the losing
board does. 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.

If we RANDOMLY turned over a board, then it would be the winning board 1/3
of the time. In that case, you have a 0% chance of winning no matter what
you do. If the board we randomly turned over was a losing board, then you'd
have a 50% chance of winning by staying or switching. If the original
problem said 'you pick a board, then Wolfie randomly turns over another
board and it says 'you lose', then you might as well keep your original
choice. But that's not the problem being stated. Monty Hall (or Wolfgang, in
this case) is not RANDOMLY turning over a board: he knows the winning board
and he does NOT turn that over. Its not random, so not only does randomness
not add info to the problem, it doesn't even apply.


Probability is not the right analysis for a singular event. At least
that's what I think now, and now I'm going to bed...maybe I'll be wrong
in the morning ;-)

Yes, you will be. Sorry. :-)

--riverman

What's a boy to do?
 

wrote in message
...
On 30 Oct 2006 16:17:56 -0800, "riverman" wrote:


First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Those aren't all the ways...think about it.

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

No, it isn't...think about it.

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome.

Well, perhaps it's because of that, or perhaps because it's
wrong...think about it.

However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

Maybe it would help you get on-target answer-wise if you tied a string
to your finger in exactly the same spot two days in a row...


Thats not possible.

--riverman

What's a boy to do?
 

On Oct 27, 12:40 pm, "Wolfgang" wrote:
An interesting problem was recently brought to my attention.

Let us say that you and I are standing next to a table on which I have
placed three boards identical in every respect except that each has a
different number painted on it.....1, 2, and 3, respectively.

I say to you that if you turn your back I will place a five dollar bill
under one of the boards and a slip of paper that says "you lose" under each
of the others. You then turn back to face the table and point to or name
the board you think has the five dollar bill under it. If you're right, you
win the five bucks.

We proceed. You pick, say, board number one. I say, "O.k., tell you what,
I like you so I'm going to make this easier for you," and I remove board
number three to show you that it has a "you lose" tag under it. Obviously,
the five dollars must be under one of the other two. "So," I say, "would
you like to stick with your original pick, or change your mind?"

It is a given that the game is not rigged in any way and you are not being
fooled by anything ambiguous or otherwise misleading in the description.

The question.......what should you do?

Wolfgang

It don't matter. Play the game 300 times and you win 100 no matter how
you do it.
Initially your chances are 1/3.
With the give away the chances are expressed (1/2)*(2/3), which is
1/3 same as above. Again if you play 300 times you win 100.
If I remember correctly Poission figured all this out working for a
wealthy French nobelman who loved to gamble.

What's a boy to do?
 

Jonathan Cook wrote:
Wolfgang wrote:

An interesting problem was recently brought to my attention.

I think I've seen this before but never really thought about it.
Thanks.

You're welcome.

I like you so I'm going to make this easier for you," and I remove board
number three to show you that it has a "you lose" tag under it. Obviously,
the five dollars must be under one of the other two. "So," I say, "would
you like to stick with your original pick, or change your mind?"

It is a given that the game is not rigged in any way and you are not being
fooled by anything ambiguous or otherwise misleading in the description.

The question.......what should you do?

I agree with everything that's been written here on what the "correct"
solution is, and how Marilyn made a bunch of PhDs and professors look
stupid (not that hard, really),

I'll risk belaboring a point here because I believe it is an important
one. Ms. Savant did NOT make anyone look stupid. I think she was
certainly aware that many people would immediately jump to the wrong
conclusion.....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. What made people look stupid
wasn't coming up with the wrong answer which was (and is) after all
something akin to falling into the trap of trying to figure out the
answer the question of where to bury the survivors of a plane crash
that occurs smack dab on an international border. What makes people
look stupid is insisting on the wrong answer after the correct (and
simple) one has been revealed and explained. Well, that and, in this
instance, Haddon's nasty little bias trick. I get to keep my shiny new
nickel because no one noticed.....or at least no one pointed
out.....that all of the outraged authors of the quotes he used were
associated with an institution of higher learning, 5 of 6 were
identified as Ph.D.s, and several references were made to mathematics
and mathematicians while none of the authors was identified as such. I
have a hard time believing that this is a representative sample of all
the letters sent to Ms. Savant in response to her exposition of the
Monty Hall problem.

but the problem has been irking me,
and I finally decided why.

You posed the problem _singularly_. One try. Probability is about
expected outcomes over lots of attempts. It breaks down in a singular
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
convinced that it is worth switching boards (though I absolutely agree
that over lots of tries it is).

The trouble here is at least partially one of semantics (I cannot for
the life of me understand why semantics is so widely accepted as a
pejorative term.....but that's another rant altogether). 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. Playing
the game once may APPEAR to be singular if one stresses all the details
about who is involved, what they are wearing today, what they had for
breakfast etc. but, in all its essentials, it is identical to millions
of other events. It is NOT singular. The laws of probabilities
apply......MUST apply......not because this particular avatar is
repeated, but because it is repeatable and myriad others like it in
every essential detail have been repeated often enough for the
mathematically derived probabilities to be confirmed experimentally.

At any rate, you DO believe that probabilities apply to single events
and I can prove it easily. All we have to do is raise the
stakes......we don't even need to calculate the odds with any
precision. You are in a large airplane of a type famous for its
ability to glide like a brick, and the engines fail. You have a
parachute. You have never used a parachute before. I think we may
take it as a given that you are not likely to repeat the
experiment......and your internet connection is too slow for Google be
of much use in finding someone else who has. What do you
do......jump?.....or ride it out?


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.

Obviously in this case
it doesn't matter, because I still lose on the times the board with the
money is turned over, so switching or not I'll only win, on average, 50%
of the remaining ones, which is 1/3 of all of them. Randomness doesn't
add information to the problem like purposely turning over the losing
board does. 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.

Probability is not the right analysis for a singular event. At least
that's what I think now, and now I'm going to bed...maybe I'll be wrong
in the morning ;-)

Jon.

What's a boy to do?
 

Wolfgang wrote:
Jonathan Cook wrote:

....
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.

There was supposed to be more to this......yeah, I know, just what
everybody wanted to hear! :)

I'm still not used to posting on Google. Hit the "post message" button
too soon.

Will try again with the whole thing (I hope) shortly.

Wolfgang

What's a boy to do?
 

Jonathan Cook wrote:
Wolfgang wrote:

An interesting problem was recently brought to my attention.

I think I've seen this before but never really thought about it.
Thanks.

You're welcome.

I like you so I'm going to make this easier for you," and I remove board
number three to show you that it has a "you lose" tag under it. Obviously,
the five dollars must be under one of the other two. "So," I say, "would
you like to stick with your original pick, or change your mind?"

It is a given that the game is not rigged in any way and you are not being
fooled by anything ambiguous or otherwise misleading in the description.

The question.......what should you do?

I agree with everything that's been written here on what the "correct"
solution is, and how Marilyn made a bunch of PhDs and professors look
stupid (not that hard, really),

I'll risk belaboring a point here because I believe it is an important
one. Ms. Savant did NOT make anyone look stupid. I think she was
certainly aware that many people would immediately jump to the wrong
conclusion.....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. What made people look stupid
wasn't coming up with the wrong answer which was (and is) after all
something akin to falling into the trap of trying to figure out the
answer the question of where to bury the survivors of a plane crash
that occurs smack dab on an international border. What makes people
look stupid is insisting on the wrong answer after the correct (and
simple) one has been revealed and explained. Well, that and, in this
instance, Haddon's nasty little bias trick. I get to keep my shiny new
nickel because no one noticed.....or at least no one pointed
out.....that all of the outraged authors of the quotes he used were
associated with an institution of higher learning, 5 of 6 were
identified as Ph.D.s, and several references were made to mathematics
and mathematicians while none of the authors was identified as such. I
have a hard time believing that this is a representative sample of all
the letters sent to Ms. Savant in response to her exposition of the
Monty Hall problem.

but the problem has been irking me,
and I finally decided why.

You posed the problem _singularly_. One try. Probability is about
expected outcomes over lots of attempts. It breaks down in a singular
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
convinced that it is worth switching boards (though I absolutely agree
that over lots of tries it is).

The trouble here is at least partially one of semantics (I cannot for
the life of me understand why semantics is so widely accepted as a
pejorative term.....but that's another rant altogether). 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. Playing
the game once may APPEAR to be singular if one stresses all the details
about who is involved, what they are wearing today, what they had for
breakfast etc. but, in all its essentials, it is identical to millions
of other events. It is NOT singular. The laws of probabilities
apply......MUST apply......not because this particular avatar is
repeated, but because it is repeatable and myriad others like it in
every essential detail have been repeated often enough for the
mathematically derived probabilities to be confirmed experimentally.

At any rate, you DO believe that probabilities apply to single events
and I can prove it easily. All we have to do is raise the
stakes......we don't even need to calculate the odds with any
precision. You are in a large airplane of a type famous for its
ability to glide like a brick, and the engines fail. You have a
parachute. You have never used a parachute before. I think we may
take it as a given that you are not likely to repeat the
experiment......and your internet connection is too slow for Google be
of much use in finding someone else who has. What do you
do......jump?.....or ride it out?


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.

Obviously in this case
it doesn't matter, because I still lose on the times the board with the
money is turned over,

Don't look now, but you've just claimed that it doesn't matter
what you do in one situation because something else might happen in a
different situation. If logic counts for anything, you have just torn
the fabric of the universe asunder.

so switching or not I'll only win, on average, 50%
of the remaining ones, which is 1/3 of all of them.

O.k., I'll stipulate that your analysis of the numbers is correct.
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.

Randomness doesn't
add information to the problem like purposely turning over the losing
board does.

Correct.....and it isn't adding anything to understanding or
elucidating it either.

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.

Probability is not the right analysis for a singular event. At least
that's what I think now, and now I'm going to bed...maybe I'll be wrong
in the morning ;-)

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.

Toivo and Aino are back. This time they're in Vegas and they're
playing a somewhat different game. There are ten thousand boards and
only one of them has a five dollar bill under it. As before, Toivo
gets one pick and Aino gets the rest. They are going to play the game
just one time. You don't get to play at all. You just get to bet on
who is going to win. You get to bet just one time.

Who are you going to bet on?

Wolfgang
and how much would you be willing to risk? :)

o.k......that's all of it.

What's a boy to do?
 

As soon as you lifted #3 and exposed it as "you lose", the problem was
over.

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, the new problem is simply one
of choosing #1 or #2. I do this by saying yes or no.
My probablity of getting the $5 is simply 0.5

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

The dart problem is indeterminate -- not enough information about
unstated variables.

cheers

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

What's a boy to do?
 

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 0.5

O.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.

cheers

Prosit!

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. :(

What's a boy to do?
 

On Tue, 31 Oct 2006 23:43:16 +0800, "riverman" wrote:


wrote in message
.. .
On 30 Oct 2006 16:17:56 -0800, "riverman" wrote:


First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Those aren't all the ways...think about it.

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

No, it isn't...think about it.

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome.

Well, perhaps it's because of that, or perhaps because it's
wrong...think about it.

However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

Maybe it would help you get on-target answer-wise if you tied a string
to your finger in exactly the same spot two days in a row...


Thats not possible.

I won't debate that, but it is possible for 2 darts to be the exact same
distance from a target...

HTH,
R

--riverman

What's a boy to do?
 

wrote:
On Tue, 31 Oct 2006 23:43:16 +0800, "riverman" wrote:


wrote in message
.. .
On 30 Oct 2006 16:17:56 -0800, "riverman" wrote:


First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Those aren't all the ways...think about it.

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

No, it isn't...think about it.

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome.

Well, perhaps it's because of that, or perhaps because it's
wrong...think about it.

However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

Maybe it would help you get on-target answer-wise if you tied a string
to your finger in exactly the same spot two days in a row...


Thats not possible.

I won't debate that, but it is possible for 2 darts to be the exact same
distance from a target...

Profound......or, not a chance in hell. Take your pick.

Wolfgang
who notes that intuition is taking a rather substantial beatting here
lately.

What's a boy to do?
 

"Wolfgang" wrote in message

who notes that intuition is taking a rather substantial beatting here
lately.


point me, in the archives, to the halcyon days when intuitive thinking on
ROFF hit its full stride..........g
Tom
......who notes that spelling always gets a whupping
from all sides, apparently.

What's a boy to do?
 

On 31 Oct 2006 16:31:11 -0800, "Wolfgang"
responded...

Well, there was a 100% probability of _that_...

What's a boy to do?
 

wrote:
On 31 Oct 2006 16:31:11 -0800, "Wolfgang"
responded...

Well, there was a 100% probability of _that_...

Nope.

Would you like me to tell you how to calculate the probability?

Wolfgang

What's a boy to do?
 

"Jonathan Cook" wrote in message
...

You posed the problem _singularly_. One try. Probability is about
expected outcomes over lots of attempts. It breaks down in a singular
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
convinced that it is worth switching boards (though I absolutely agree
that over lots of tries it is).

Probability is not the right analysis for a singular event.

No, that's not true. I think you're confusing that with a different
concept. There's something called "expected value" which averages out the
long run. For example, you win a dollar if you call a coin flip right, and
lose a dollar if you call it wrong. Your expected value is winning (or
losing) $0 (you're going to break even in the long run.) However, if you
only flip one time, that's impossible. You can't break even if you flip one
time (or 3 times, for that matter.) This doesn't change the obvious fact
that the probability is 50% for calling it right even if you flip just once.

What's a boy to do?
 

wrote:
On Tue, 31 Oct 2006 23:43:16 +0800, "riverman" wrote:


wrote in message
.. .
On 30 Oct 2006 16:17:56 -0800, "riverman" wrote:


First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Those aren't all the ways...think about it.

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

No, it isn't...think about it.

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome.

Well, perhaps it's because of that, or perhaps because it's
wrong...think about it.

However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

Maybe it would help you get on-target answer-wise if you tied a string
to your finger in exactly the same spot two days in a row...


Thats not possible.

I won't debate that, but it is possible for 2 darts to be the exact same
distance from a target...

HTH,
R

No its not, its a matter of measurement precision.

--riverman

What's a boy to do?
 

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.

cheers

oz

What's a boy to do?
 

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


wrote:
On Tue, 31 Oct 2006 23:43:16 +0800, "riverman" wrote:


wrote in message
.. .
On 30 Oct 2006 16:17:56 -0800, "riverman" wrote:


First, list all the ways to throw three darts, A B and C.

ABC
ACB
BAC
BCA
CAB
CBA

Those aren't all the ways...think about it.

Remember, we are looking at a conditional probability; dart B has
already landed farther than dart A. So our list of outcomes is limited
to:

ABC
ACB
CAB

No, it isn't...think about it.

Our 'definition of success' is when dart C lands further than dart A,
which is clearly only the first two arrangements. So the probability of
throwing a third dart that lands farther than the first (given the
second dart has already landed farther than the first), is 2/3.

Its an unsettling conclusion, because people want to make the argument
that the distance from the bullseye affects the probability of each
outcome.

Well, perhaps it's because of that, or perhaps because it's
wrong...think about it.

However, every possible distance affects every outcome
equally, so they are all still equally likely, as counterintuitive as
it may be.

Maybe it would help you get on-target answer-wise if you tied a string
to your finger in exactly the same spot two days in a row...


Thats not possible.

I won't debate that, but it is possible for 2 darts to be the exact same
distance from a target...

HTH,
R

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.

HTH,
R

--riverman

What's a boy to do?
 

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.

Here's a real oldie but goodie. You are racing a slow tortoise, and you give
the tortoise a head start. In the first moments, you run quickly to where
the tortoise started from, but in that time it has moved ahead. So you
continue to run to where it has advanced to....but it has moved ahead a bit
more. So you run to where it is AGAIN, but it has yet again moved ahead!
This proves that you cannot win the race, as you cannot catch the tortoise,
right?

:-)
--riverman

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?
 

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

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!