Hi there, when I read rule 623 (Rules on Odd Men), it specifically states that the lowest ranked player in an odd-numbered score group becomes the odd man, and is paired down.
The only explicit exceptions that I can find, are regarding whether the remaining players in that score group can be paired at all, which means they haven't played each other before.
However, please consider this simplified scenario. After 3 rounds, I have 3 players with 2 points. Let's call these players A/B/C. Now, C would normally be the odd man. If A and B have already played each other, then A would play C, and B would become the odd man, etc. In the worst possible case, I could have 3 odd men to pair down, but I digress! ;)
This is the way that I have always done it. Now, let me expand the above example to include previous colours. Assuming that A/B/C have not played each other yet, and their previous colours were BWB/BWB/WBW, if I make C the odd man, then B would end up with colours of BWBB after 4 rounds, and initiating pressure to have WW in his last 2 rounds.
Wouldn't it make more sense to, in effect, interchange the score group before the odd man is removed, in order to have as much colour equalization as feasible? In other words, A would play C, and their colours would end up BWBW/WBWB. I would not do this for colour alternation, and note that this consideration is a separate adjustment from the seemingly widely applied VARIATION 624.1: The odd man may be paired with the next highest-ranked player whom he has not met in the next-lower group and who is due the opposite colour. Again, I read this variation only for colour equalization, not colour alternation, agreed?!
Thanks in advance for anyone who can help out my curiosity. If you do agree with my proposal, please also indicate any official reference, in case I get challenged onsite! :)
The only explicit exceptions that I can find, are regarding whether the remaining players in that score group can be paired at all, which means they haven't played each other before.
However, please consider this simplified scenario. After 3 rounds, I have 3 players with 2 points. Let's call these players A/B/C. Now, C would normally be the odd man. If A and B have already played each other, then A would play C, and B would become the odd man, etc. In the worst possible case, I could have 3 odd men to pair down, but I digress! ;)
This is the way that I have always done it. Now, let me expand the above example to include previous colours. Assuming that A/B/C have not played each other yet, and their previous colours were BWB/BWB/WBW, if I make C the odd man, then B would end up with colours of BWBB after 4 rounds, and initiating pressure to have WW in his last 2 rounds.
Wouldn't it make more sense to, in effect, interchange the score group before the odd man is removed, in order to have as much colour equalization as feasible? In other words, A would play C, and their colours would end up BWBW/WBWB. I would not do this for colour alternation, and note that this consideration is a separate adjustment from the seemingly widely applied VARIATION 624.1: The odd man may be paired with the next highest-ranked player whom he has not met in the next-lower group and who is due the opposite colour. Again, I read this variation only for colour equalization, not colour alternation, agreed?!
Thanks in advance for anyone who can help out my curiosity. If you do agree with my proposal, please also indicate any official reference, in case I get challenged onsite! :)
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