The Pac-10 conference’s official word is, in the case of the USC, UA and WSU ending up in three-way tie (all winning Saturday), USC would finish fifth due to its 3-1 record in games between the teams. Then UA and WSU’s tie would be settled under the rules of the two-team tiebreaker. That means, because both would have defeated UW and UCLA once and lost both to Cal, WSU’s two wins over ASU (UA lost twice to the Sun Devils) would break the tie in WSU’s favor and the Cougars would be sixth. THIS IS FROM THE PAC-10 OFFICE as relayed by WSU basketball SID Jessica Schmick.
If the Cougars win and either USC or UA lose, the Cougars are sixth in a head-to-head tie with USC, fifth if its UA. Also, if UA and WSU both lose and USC wins, USC is fifth, Arizona sixth and WSU seventh. If UA, USC and WSU were to all lose, UA would finish fifth, Oregon State sixth, USC seventh and WSU eighth. If USC and WSU lose, OSU is sixth, USC seventh and WSU eighth. And, of course, with UA, OSU and Arizona all playing after the WSU-UW game, nothing will be known for sure until about 7:30 tomorrow night.
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So says Grippi. That sounds about right to me -- when it comes to multi-team tiebreakers, they generally go back to the highest tiebreaker possible for the remaining teams once the initial tiebreaker is broken.
