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