1. Joined
    05 Jul '02
    Moves
    4154
    01 Oct '02 09:29
    After the first round, 9 participants will be left in that division. This
    means that one of us will get a bye until the final game, because
    subsequent rounds will still produce an odd number of players.
  2. DonationAcolyte
    Now With Added BA
    Loughborough
    Joined
    04 Jul '02
    Moves
    3790
    01 Oct '02 09:56
    Try this: you all draw lots, and the two players who draw the short straws play each other in
    the '1 1/2th' round. The winner goes on to the second round as normal. Not entirely fair, but
    better than giving someone a bye to the final.
  3. Donationvaknso
    The Ambassador
    Charleston SC. USA
    Joined
    17 Jun '01
    Moves
    141727
    01 Oct '02 23:20
    Thanks for posting. I'll ask Dave (Schliemann) to answer your
    question on how this will be handled.
    John
    Assistant Td Baker Invitational
  4. Joined
    09 Aug '01
    Moves
    54019
    02 Oct '02 20:34
    fyi
    same situation with the 1600 section where three players will qualify
    for the second round.
    tiger
  5. Kirkland, WA
    Joined
    08 May '02
    Moves
    9175
    03 Oct '02 00:03
    It would have seemed to make sense when there were 18 players to
    give 14 byes in the first round, that would have eliminated 2 players
    in the first round bringing the numbers down to 16 - 8 - 4 - 2 - 1...

    now with 9 we would need to have 2 people play off a first-and-a-half
    round (or give 7 byes in the second) to give 8 - 4 - 2 - 1

    of course given that we are all pawn stars (i hope!!) and that there are
    or will be only 9 people left we could all play each other. I think i could
    manage 16 tournament games at once...
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