[EMAIL PROTECTED] (Tanguy) wrote in message 
news:<[EMAIL PROTECTED]>...
> I would like to thank you for sharing your thoughts and I would like
> also to answer some issues
> you've raised.
> Firstly, even if the lead is harshly disputed and even if the boats
> are nearly similar, the boats
> are sufficiently away from each other at the finish line that it is
> easy to say who's the winner.
> Thus a tie is impossible.
> Secondly, the order in the distribution of points is not taken into
> account. As you intuited,
> "210", "201", "120", "102", "021", "012" means merely that "of three
> boats, one of them one
> twice, another one won once, and the third lost all three races".
> I am still looking for a way to guess the number of patterns, rather
> than the indivual outcomes
> from a given number of races or trials.
> 
> Tanguy Arzel

OK, three trivial corrections I HAVE to make :)

(1) Louis Vuitton Cup (not Vitton)
(2) Hauraki Gulf (not bay)
(3) Scenarios (not scenarii)

Pedantic, I know...

If I understand your problem correctly, you are not interested in the
boats that actually win the races, but in the different classes of
results. So the "210", "201", "120", "102", "021", "012" are all to be
considered the same thing. Is this right?

If this is the case then we can apply a little graph theory to the
problem. Each boat is a vertex/node/point of the graph and each race
is represented by an edge/line/connection between each of the points.
Each edge is directed, i.e. it points from, say, vertex A to B and not
from B to A. Suppose that an edge directed from A to B means that boat
A beat boat B in a race. Since each boat races every other boat in the
group, there must be an edge connecting each pair of vertices. This is
called a tournament graph (not a surprising name, eh?).

This graph is explained in Eric's world of mathematics

http://mathworld.wolfram.com/Tournament.html

and the sequence of distinct graphs is given by

http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=000568

Since there are only 9 syndicates in the Louis Vuitton Cup, the
sequence given in the previous web page should answer your question.
There also some references given if you want to find an extended
sequence

I hope this helps!

Regards,


Josh
.
.
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