"John Fernandez" wrote in message
...
Going into the final round of the 66th New York Masters, we had a bit of a
pairing argument, which I hope you guys can help me with. Here is the
situation, and I'm showing all work.

There were two players at 2.5/3. They play each other, no problem.
There were five players at 2.0/3, and one at 1.5/3. Here's their pairing
and
breakdown:

Player 1 2693 W- B3 W-
Player 2 2647 W4 B- W3
Player 3 2467 B- W1 B2
Player 4 2245 B2 W- B-
Player 5 2228 B6 W- B-
------------------------------
Player 6 (1.5) 2440 W5 B- W-

Just to code for you, when I do - that means the player played someone
else not
in this group.

So, let's take our first stab at making these pairings. We notice that
Player 3
has played 1 AND 2. Therefore, it would seem that the natural pairing is:

2 vs. 1
3 vs. 4
5 vs. 6

But there's a problem. 5 vs. 6 leaves us with an illegal pairing. ALSO, we
have
Top Half vs. Lower Half issues (Rule 27A3). Rule 27A3 tells us to consult
29C1.
29C2 tells us to make transpositions to avoid pairing players.

The very next rule is 29D, which is the odd player, which we need to
determine.
First of all, player 5 is unacceptable as the odd player. We can take the
next
player (Player 4). Therefore, we will propose the following pairings:

2647 WBWW vs. 2693 WBWB
2467 BWBW vs. 2228 BWBB
-----------------------
2245 BWBW vs. 2440 WBWB

The problem with these pairings is that colors aren't great for Player 5.
I
can't possibly improve them, with Player 4 dropping, and Player 5 of
course
can't drop.

Are there any other legal pairings I can come up with that work? The only
way
to make another legal pairing would be to drop Player 3. If I did that,
the
pairings would be as follows:

2228 BWBW vs. 2693 WBWB
2245 BWBW vs. 2647 WBWB
-----------------------
2467 BWBW vs. 2440 WBWB

Yes, this is ideal, but I can't do it, can I? First of all it seems to be
in
violation of 29D1, which tells us how to find the odd player. Second, 29J1
says:

29 D also states that you must consider rather the remaining players can be
paired. In this case it seems that dropping player 3 to the next lowest
group is the best idea. In fact dropping player 3 would be the natural
pairing for a Harkness pairing system where the middle player in a score
group is dropped. Here it just seems the least disruptive choice.

5-1
4-2
3-6

and the colors even work out nicely.



Transpositions and interchanges for the purpose of maximizing the number
of
players who receive their due color should be limited to 80 points.

Clearly I can't make those second pairings because of that rule. Also,
29J2,
which is invoked in the case of Player 5, gives me 200 points of leeway.
However, the difference between Player 3 and Player 4 and 5 is more than
200
points.

In reality, the colors are not all that bad. The only player who has any
"real"
color issues is Player 5, and I don't feel he is enough to "substantially
improve" the colors and exceed the transposition/interchange limits in
29J3.
Yes, I know 29J6 tells me I should avoid this pairing, it also does say
"In the
score group". Our problem is that within the score group this is
impossible.

The upshot, of course, is that the two GMs in the group are playing each
other.
They vehemently argued against this pairing, saying that they should play
the
2200 players, and let the 2467 drop. They argued that the pairings should
be
made the following way:

Player 1- we need to find an opponent. 3 is unacceptable, but 4 is fine.
But
then we have a problem. If we pair 1 vs. 4, we then have to pair 2 vs. 5,
which
is illegal, as they have already played. Therefore, we pair 1 vs. 5.
(Which
seems odd.) Then we pair 2 vs. 4 and drop 3 down to the next group.

I don't think that's a very logical way to make pairings, especially as
there's
repeated color problems throughout the group, and repeated illegal
pairings.

Comments and questions welcome.

Best,
John Fernandez