1 (White) vs. 4 (Black) – winner = 1
5 (White) vs. 2 (Black) – winner = 2
3 (White) vs. 6 (Black) – winner = 6
I would have expected (at least, in my "Manual Pairing Brain!) 2nd Round Pairings to be as follows:
2 (White - 1.0) vs. 1 (Black - 1.0)
6 (White - 1.0) vs. 5 (Black - 0.0)
4 (White - 0.0) vs. 3 (Black - 0.0)
But, NOPE!! WinTD paired the 2nd Round thusly:
6 (White - 1.0) vs. 1 (Black - 1.0)
2 (White - 1.0) vs. 3 (Black - 0.0)
4 (White - 0.0) vs. 5 (Black - 0.0)
EXPLANATIONS REQUESTED, PLEASE!!
It is, of course, a “LEGAL” pairing … and I am certainly not going to change legal pairings randomly (that were made by WinTD) just because there is a “better pairing.” But … I can not find a reason why WinTD made such a pairing. Any ideas?
Assuming the tournament in question was the SKP_30-30_Ratings_Booster_2 (which seems likely based on the date and the initials and ratings of the players in the original post), no, there was no draw in the first round.
Check the numeric limits on interchanges and transpositions. If transpositions are zero then it may have kept player 3 as the top of the 0-1 score-group to be paired with the lowest 1-0 that had not already played 3.
Nice detective work … yes, S.K.P. Ratings Booster #2 … … … the draws were IN Rd. 2 (and Rd. 3 – that one was a “King & Rook” vs. “King & Bishop”) … … …
“Pairing Effort” is set to “999” … … … hmmm … perhaps it “OVER-thought” it!
Those would be Harkness pairings if that was chosen in the Pairing Rules preferences. Or “Avoid interchanges” rather than “Limit interchanges”. One thing to note is that the “obvious” pairings aren’t that clear. The natural pairings are 1v2, 6v3, 4v5, but 6 and 3 have played, so either the up or down float has to change. Pairing 6 with 5 is pairing the downfloat with the lowest ranked player in the next score group down so switching the 6 and 2 (which is an adjacent swap) may be seen as superior depending upon how the preferences are set.
I can certainly think of an argument for the pairing that the computer gave:
First, to state the obvious, 1 cannot play 4, 2 cannot play 5, and 3 cannot play 6 because they have already faced each other.
The second rule is that players of equal scores should face each other, but there are three players with a score of 1 and three players with a score of 0, so it follows that one player who has a score of 1 must face one player who has a score of 0. Since player 1 is the only player with a score of 1 who is due black, it follows that he should be paired either with player 2 or with player 6. And since player 4 is the only player with a score of 0 who is due white, it follows that he should be paired either with player 3 or with player 5.
The third rule is that, within each score group, upper half should face lower half. But there are three players in each score group. Player 2 and player 6 are closer to each other in rating than either is to player 1, so there is no particular advantage to either pairing. But player 3 and player 4 have ratings that are very close, so a case can be made that pairing player 4 with player 5 better satisfies the principle of pairing upper half against lower half. This would mean that player 3 would be the score 0 player who should be paired with a score 1 player. Since player 3 has already faced player 6, this would indicate that player 3 should be paired against player 2, which would leave players 1 and 6 to be paired against each other.