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

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.

"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.

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

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

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

"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.

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

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

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

wrote in message
news
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