"Scott Seidman" wrote in message
. 1.4...
"riverman" wrote in :

Thus, all the possible positions of A and C equal all the possible
positions of C and A...


But the question as you pose it has nothing to do with all the possible
positions of dart A-- it has to do with one specific position of dart A!
Let's say that A is 5cm away. Then you are looking for p(C5), which has
a value that depends only on the distribution of dart C.

Whatever high probability exists if A is close is
countered by the low probability if A is far.

True before dart A is thrown, but not after dart A is thrown. Now, you
have a real honest to goodness value for dart A.


Yes, but the possbilities for the value for A are limitless. You cannot
calculate all the different arrangements. Remember, the original question
did not specify where A landed, only that it did.


Thus, all the possible positions of A and C equal all the possible
positions
of C and A....its a combinatoric problem. Specifically because I
DON'T give the position of dart A.

This doesn't mean that you can just ignore the fact that Dart A is stuck
at a precise location in the dartboard, and it's why there isn't enough
info to offer a p-value.


No its not. Its stuck in any of an infinite number of locations. Each one
has a different effect on the probability of B being closer.


If its any comfort, I'm not making this problem up.

Then the person who did got it wrong.


Well, here's the results of one person who ran 1 million trials each, using
uniform, normal, gaussian and random distributions.
http://www.gatago.com/rec/puzzles/25594126.html

He came up with 2/3 every time. I just have an aversion to Monte Carlo
methods, but go ahead and try it yourself and let us know what the result
is.

--riverman