How the tiebreaker works for the Bruins

If USC defeats UCLA at the Coliseum on Saturday, the picture is clear: USC would play in the Rose Bowl game as champion of the Pacific 10 Conference.

If UCLA wins, the Pac-10’s Rose Bowl entrant will be decided by the outcome of the Arizona-Arizona State game on Saturday night.

If UCLA and Arizona both win, the Bruins are bound for Pasadena on Jan. 1, and here’s how -- as explained by UCLA sports information director Marc Dellins:

UCLA, Arizona State, USC and the winner of Saturday’s Oregon-Oregon State game would be 6-3 in conference play.

In a four-way tie among USC, UCLA, Arizona State and Oregon, UCLA and Oregon would, at 2-1, have the best record against the other three teams in the tie, and since UCLA beat Oregon the Bruins would win the tiebreaker.

If there’s a four-way tie among USC, UCLA, Arizona State and Oregon State, it’s more complex.

USC, UCLA and Arizona State would all be 2-1 against the other teams in the tie; Oregon State would be 0-3, so the Beavers would be eliminated.

The next tiebreaker is each remaining team’s record against the other two -- still a tie; all would be 1-1.

The next consideration is each team’s record against the next team after the four-way tie in the standings -- which would be Oregon and Arizona, tied for fifth. USC and UCLA would be 1-1; Arizona State would be 0-2, eliminating the Sun Devils.

That takes it to a head-to-head tiebreaker between UCLA and USC . . . and the Bruins would go based on their win over the Trojans.

Really, it’s that simple . . .