Duel tourneys and odd number of players

There's probably already a thread on this, and it might even be mentioned in the FAQ that I just skimmed through; if so, apologies.

I'm just wondering how the match-ups for the second round of, for example, a duel tournament, are decided if an odd number of players qualify for the second round.

This is on the assumption that all players that score 3-3 in the first round make it through to the second.

Oh, reason for asking is that I'm in the May 06 Mega-Duel tournament 0-1400, having registered - perhaps dodgily - as a 1200p player, and now my rating is in the 1500s. So, if the numbers are odd by the time the first round is completed, I'd like to volunteer to be given the chop as I'd feel bad otherwise (I'm at 3-3 after the round). If I'm still in I'll play on though, as I'd hate to harm my rating🙂

Thanks,
Martin

I believe that if an odd number of players get through there will be one group of three.

Originally posted by bambee
I believe that if an odd number of players get through there will be one group of three.

This is correct.

OK, cheers🙂