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