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Pairing Question



 
 
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  #1  
Old July 31st 03, 05:02 AM
John Fernandez
external usenet poster
 
Posts: n/a
Default Pairing Question

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:

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
  #2  
Old July 31st 03, 05:29 AM
Chris Merli
external usenet poster
 
Posts: n/a
Default Pairing Question


"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



  #3  
Old July 31st 03, 07:15 AM
Greg Wren
external usenet poster
 
Posts: n/a
Default Pairing Question

But the proposed pairing 4-2 (see below) has already been made. Looks like
1-2, 3-5, 4-6 is the best choice.

"Chris Merli" wrote in message
news:[email protected]

"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





  #4  
Old July 31st 03, 07:52 AM
SDchessTDL
external usenet poster
 
Posts: n/a
Default Pairing Question

John,
You should make your first priority pairing players 1 & 2. 3 can
not play them. That leave 4 & 5 who are within 80 points of each
other. 3 has to play 6.
I do not understand your problem. I must admit to pairing without a
computer for more than 25 years. This is simple you are trying to make
it to hard.
Terry Likens


eepmeep (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:

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

  #5  
Old July 31st 03, 12:14 PM
John Fernandez
external usenet poster
 
Posts: n/a
Default Pairing Question

Grant Perks wrote:

What is wrong with:

4 vs 1
5 vs 2
3 vs 6

Keeping 1 and 2 in rating order versus 4 and 5. Especially since 4 and 2 have
already played.

Best,
Grant Perks


Grant,

I agree. That's the pairing that made sense to me when I was looking at
alternatives (it is the only alternative).

However, I can't find a way to justify breaking 29D1, which says to drop Player
4 in this situation. What rule takes higher precedence?

John Fernandez
  #6  
Old July 31st 03, 04:26 PM
John Fernandez
external usenet poster
 
Posts: n/a
Default Pairing Question

Bill Goichberg wrote:

The colors don't work for player 2 either, and 1 vs. 2 in a group of 5 is
hardly a natural pairing. The natural pairings are for 1and 2 to play 4 and
5-if they play each other instead, that is over a 400 point switch for both.


This is much better. Sure, it's a 239 point switch to drop 3 instead of 5,
but
this beats a 400 point switch.


400 point switch? Does this mean I have to figure out pairings BEFORE figuring
out the odd player? Somehow this didn't make sense to me when I was looking at
it.

So what? When all the possibilities are in violation of something, you
choose
the least of evils. A 239 point switch with good colors is far better than a
400+ point switch with bad colors.


I still don't follow exactly where the 400 point thing is coming from, since
determining the odd player is one of the first things it covers.

You can't always do what you "should." No pairing rule is absolute in cases
where the alternative is worse. Also, dropping player 3 is correct even
without considering colors. The fact that it balances colors is just a bonus.

Bill Goichberg


Is this in the rules anywhere? Or is this one of those cases where if I make an
effort to follow the letter of the rules, I'm toast because they are both too
long and they often contradict each other.

I find it amusing, I've been often excoriated for straying from the USCF's
published rules and making the "right ruling", and now I find that when I
attempt to follow USCF's published rules to the letter, that I end up making
the wrong decision.

I admit, I'm very confused here.

John Fernandez
  #7  
Old July 31st 03, 06:22 PM
Fifiela
external usenet poster
 
Posts: n/a
Default Pairing Question

I see your point, but why is pairing 1 vs. 2 so absurd?

Becuase there are other pairings that work.

The GM's get to eat fish for one more round before agreeing to a draw in the
last round after 1. a4 a5 2. h4 h5 1/2-1/2.

You're making this hard because your starting from the bottom and working up.
You should start from the top and work down.
  #8  
Old July 31st 03, 06:32 PM
John Fernandez
external usenet poster
 
Posts: n/a
Default Pairing Question

Alan Fifeld wrote:

The GM's get to eat fish for one more round before agreeing to a draw in the
last round after 1. a4 a5 2. h4 h5 1/2-1/2.


This WAS the last round. It had the benefit of forcing the GMs to fight.

Of course, one of the GMs just stormed out and forfeited. He's no longer
welcome in our tournaments.

John Fernandez
  #9  
Old July 31st 03, 07:21 PM
Fifiela
external usenet poster
 
Posts: n/a
Default Pairing Question

This WAS the last round. It had the benefit of forcing the GMs to fight.

Your tournament was short one round.

If the justification was to force GM's to fight , would you have rejected a
short draw with a double forfeit of both GMs ala Jarecki? Moving players
around in the last round to force "fights" is a bad idea. I did this once or
twice for class players in Opens who wanted to play each other for the class
prize. "Blood" was promised. Instead of the "promised fights" for class
prizes, I usually got lots of quick draws. I get more "fights" if the class
players have to play up or down the last round. Also you denied two other
players a chance to play a GM. Better just to pair top down and let the cards
lie where they fall.

Of course, one of the GMs just stormed out and forfeited. He's no longer
welcome in our tournaments.

No excuse for bad behavior but it was a bad pairing. If I was you, I'd think
about repairing this relationshp (if possible) rather than banning someone
(which is how I read "no longer welcomed").

Not the end of the world; that's why us TD's get "The Power Bucks".
 




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