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#1




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




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. 51 42 36 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




Pairing Question
But the proposed pairing 42 (see below) has already been made. Looks like
12, 35, 46 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. 51 42 36 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




Pairing Question

#5




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




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 5if 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




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 1/21/2. You're making this hard because your starting from the bottom and working up. You should start from the top and work down. 
#8




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 1/21/2. 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




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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