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February 17th, 2006

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Stellenbosch Handicap Underway

The first round of the Stellenbosch Handicap has been completed, with three of the 13 players obtaining promotions from their games. Since ranks are “fixed” for handicap tournaments, players promoted during the tournament are usually the ones in the running for the top positions.

Two of the promoted players, Wim (24k) and Andre (28k), will be playing each other in the second round. The other promoted player was Hanno (16k), who will be playing Louis. Jaco unfortunately lost against David by time (they had opted to play with a clock). Handicaps in the second round, which officially takes place on 2 March, range from 2 to 4.

First round results and secound round pairings are available.

Posted by Hugo in Rank Changes, Stellenbosch, Tournaments


This entry was posted on Friday, February 17th, 2006 at 10:44 am and is filed under Rank Changes, Stellenbosch, Tournaments. You can follow any responses to this entry through the comments RSS 2.0 feed. You can skip to the end and leave a response. Pinging is currently not allowed.

5 Responses to “Stellenbosch Handicap Underway”

  1. Hugo says:

    Some people are curious about the pairings for the third round. (E.g. I am.) The winner of Wim vs Andre (which turns out to be Andre, they played early), will play the winner of Hanno vs Louis. The winner of David vs Hugo will play the winner of Heinz vs Ben. These two games will be the “top two boards”, the winner of the tournament will be one of the two winners of these two games, the one with the highest SOS.

    Andre will be playing a 10H or 12H game against Louis or Hanno. For the “second top board”, the worst case is David vs Ben, 9H, the best case is Hugo vs Heinz, 4H – other possibilities are either 6H or 7H.

    There will be many players with 1 victory, should not be a problem to get good pairings. There will be 3 “players” with 0 victories, one of these is the “bye” player. One of these three players will necessarily be playing someone with 1 victory. The pairing-script does not require the bye player to play a 0-victory opponent, it may also play a 1-victory opponent, leaving the two real 0-victory players playing one another.

    This will definately not happen in this tournament though, as, to minimize handicaps, one of the “other two” players will receive the bye, the other one will play against a 1-victory opponent. If two real 0-victory players had to play one another, it would have been worst case Jaco vs Willie (26H), best-case Iain vs Rory (20H) – the bye will definately be used to avoid such pairings. This is why I like odd-number of players in a tournament, the bye just makes the pairings prettier. 😉

    I have not made trial pairings, but I do believe the 0-victory and 1-victory players should have better pairings than 2-victory players, as they have more possible opponents allowing more room for optimising the handicaps. (It’s typically the relatively few 2-victory players and relatively few 0-victory players that suffer bad pairings in the third round.)

  2. Hugo says:

    “Executive summary”: Andre will be playing either Louis or Hanno at 10H or 12H, I believe all other games should be

  3. Hugo says:

    Uh, oops. …other games should be <= 9H

  4. Hugo says:

    Eish, what an idiot. There is a fourth player that may end with 0-victories after the second round, if he loses his game. That’s Ben. So then there’s 4 players with 0-victories, counting the fake “bye” player. In this case, worst case has Jaco playing Ben at 11H, and Iain or Willie obtaining a bye. Best case would have Ben playing Rory at 8H.

    If Ben wins his second round game against Heinz, the above comments are correct (and games other than Andre’s should be

  5. Hugo says:

    So, Rory has beaten Jaco, meaning if Ben loses his game, Jaco will probably be playing Ben at 11H (9H+20komi).

    Let reality strike – the handicaps are likely not going to be particularly pretty. If all games have a result (with one player winning, not both losing, for example), it will be one of the following sets of handicaps, from best to worst:

    [8, 6, 5, 3, 2, 1, 0],
    [10, 5, 4, 3, 2, 1, 0],
    [10, 5, 5, 4, 3, 1, 0],
    [10, 8, 6, 5, 4, 1, 0],
    [10, 8, 8, 8, 7, 5, 0],
    [10, 10, 5, 4, 3, 1, 0],
    [11, 5, 5, 4, 4, 2, 0],
    [11, 10, 5, 5, 4, 1, 0],
    [11, 10, 6, 5, 5, 3, 0],
    [11, 10, 8, 7, 5, 5, 0],
    [12, 11, 7, 5, 4, 1, 0],
    [12, 11, 10, 7, 5, 5, 0],
    [13, 11, 10, 10, 4, 1, 0],
    [13, 12, 11, 8, 4, 1, 0],
    [16, 11, 10, 10, 7, 5, 0],
    [16, 12, 11, 8, 7, 5, 0],

    Those last two aren’t particularly pretty, and happens if Iain beats Willie, Heinz beats Ben and David beats Hugo. Of course, the last 5 aren’t pretty, with half the games at >9 handicap. If the odds were equal, that means 5/16 = 31 percent chance of half the games at >9H. In the grand scheme of things, it’s 3 out of 18 games, i.e. worst case, 17 percent of the tournament games at >9H.

    The solution in the future? Get more people to play! With >8 players, there is a chance that the winner is chosen by SOS. That’s fine. With 16 players, two players should have 3 wins each, and the winner between them is chosen by SOS. Let’s try for 16 players next time! (Or 15 if we want the bye to help the handicaps. Or 20, of course.)

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